A 2.0-g piece of Mg reacts with HNO3. Which conditions would produce the GREATEST reaction rate
0.5 M HNO3 at 20 C
0.5 M HNO3 at 40
1.0 M HNO3 at 20
1.0 M HNO3 at 40
Answer:
1.0 M HNO3 at 40°C
Explanation:
Rate of chemical reaction: This can be defined as the number of moles of reactant, converted or product formed per unit time.
Factors that affect rate of chemical reaction:
(a) Temperature: Generally, an increase in temperature increase the rate of chemical reaction by (1) increasing the number of particles with energy equal to or greater than the activation energy, (2) Increasing the average speed of all the reactant particles, due to greater kinetic energy, leading to higher frequency of collision.
(b) Concentration: An increase or decrease in the concentration of the reactant will result to a corresponding increase or decrease in the effective collision of the reactant and hence in the reaction rate.
other factors that affect the rate of chemical reaction are
(i) Nature of the reactant
(ii) Surface area of reactant
(iii) presence of light
(iv) presence of catalyst.
From the question above,
The condition with the highest temperature and concentration will produce the GREATEST reaction rate.
And that is 1.0 M HNO3 at 40 °C
A curve plotted as a function of frequency defining the sound pressure level required to give equal loudness is?
Answer:
equal loudness curve
Explanation:
The curves of equal loudness, first established by Munson and Fletcher in 1930 and subsequently recalculated by Robinson and Dadson, show the relationship that must exist between the frequencies and intensities (or sound pressure) of two sine sounds to be perceived equally loudly. , that is, with the same loudness.
The sinusoidal sounds contained along each curve have the same loudness. This frequency dependence would be mainly due to the transfer characteristics of the external and middle ear. It should also be noted that as the sound pressure level increases, the curves become flatter, that is, the frequency dependence is smaller as the sound pressure level increases.
The loudness level of any sound (complex) is determined by comparing its loudness with that of a sinusoidal sound.
A curve plotted as a function of frequency defining the sound pressure level required to give equal loudness is known as an equal-loudness curve. These curves use the phon unit to illustrate how different sound pressure levels are needed at various frequencies for a sound to be perceived as equally loud.
Explanation:The curve plotted as a function of frequency defining the sound pressure level required to give equal loudness is known as an equal-loudness curve. These curves are critical for understanding how the human ear perceives sound at different frequencies and intensity levels, using a unit called a phon to express loudness numerically. Phons and decibels are defined to be the same at 1000 Hz, which serves as a standardized point of reference. The equal-loudness curves show that at different frequencies, different sound pressure levels are required for a sound to be perceived as equally loud. This phenomenon underlines the non-linear nature of human hearing across the frequency spectrum.
Large numbers of people have compared the loudness of sounds at different frequencies and sound intensity levels to determine these curves. Each curve is labeled with its loudness in phons, and all sounds on a given curve are perceived as equally loud. This concept is vital for various applications, including the design of audio equipment, hearing aids, and soundproofing materials, to ensure sound is produced or mitigated in a manner consistent with human loudness perception.
With each bounce off the floor, a tennis ball loses 23% of its mechanical energy due to friction. When the ball is released from a height of 2.5 m above the floor, what height will it reach after the third bounce?
110 mm
110 cm
11 cm
150 cm
Answer:
110 cm
Explanation:
Gradpoint
You have an empty 20 oz. soda bottle and you blow air over the opening to excite a fundamental standing wave. Now, you slice off the bottom of the bottle (it’s plastic) without changing its length very much. You blow over the opening and excite a fundamental standing wave in the bottle with its bottom end open. The frequency of the standing wave in the second case:______________________________.
Answer:
The frequency of the standing wave in the second case is higher than that in the first case
Explanation:
The frequency and wavelength of a wave are related.
The moment you sliced the bottle, you've reduced the wavelength of the bottle.
When wavelength decreases, frequency increases and vice versa.
So, When frequency increases in the second case, more wave crests pass a fixed point each second. That means the wavelength shortens. So, as frequency increases, wavelength decreases. The opposite is also true—as frequency decreases, wavelength increases.
boy pulls a 5.0-kg sled with a rope that makes a 60.0° angle with respect to the horizontal surface of a frozen pond. The boy pulls on the rope with a force of 10.0 N; and the sled moves with constant velocity. What is the coefficient of friction between the and the ice?
(a) 0.09
(b) 0.12
(c) 0.18
(d) 0.06
(e) 0.24
Answer:
0.1
Explanation:
mass, m = 5 kg
θ = 60°
Force, F = 10 N
velocity is constant , it means the net force is zero.
So, the component of force along the surface is equal to the friction force
FCosθ = friction force
10 x cos 60 = μ x m x g
where, μ is the coefficient of friction
5 = μ x 5 x 9.8
μ = 0.1
Thus, the coefficient of friction is 0.1
An ideal gas is enclosed in a cylinder with a movable piston on top of it. The piston has a mass of 8,000 g and an area of 5.00 cm2 and is free to slide up and down, keeping the pressure of the gas constant.
a. How much work is done on the gas as the temperature of 0.280 mol of the gas is raised from 15.0 degrees C to 27.0 degrees C?
b. What does the sign of your answer to part (a) indicate?
A. The surroundings do positive work on the gas.
B. The gas does positive work on its surroundings.
C. There is no work done by the gas of the surroundings.
Answer:
Explanation:
Given
mass of gas [tex]m=8000 gm=8 kg[/tex]
Area of Piston [tex]A=5 cm^2[/tex]
[tex]T_1=15^{\circ}C[/tex]
[tex]T_2=27^{\circ}C[/tex]
no of moles [tex]n=0.28[/tex]
Work done on the gas is given by
[tex]W=-P\Delta V[/tex]
[tex]P\Delta V[/tex] can also be written as [tex]nR\Delta T[/tex]
as PV=nRT
[tex]W=-nR\Delta T[/tex]
[tex]W=-0.28\times 8.314\times (27-15)[/tex]
[tex]W=-27.93 J[/tex]
negative sign indicates that work is done on the system i.e. surrounding done a positive work on the gas
A stone is thrown with a speed v0 and returns to earth, as the drawing shows. Ignore friction and air resistance, and consider the initial and final locations of the stone. Which one of the following correctly describes the change ΔPE in the gravitational potential energy and the change ΔKE in the kinetic energy of the stone as it moves from its initial to its final location?
A. ΔPE = 0 J and ΔKE = 0 J
B. ΔPE is positive and ΔKE is negative
C. ΔPE = 0 J and ΔKE is positive
D. ΔPE is negative and ΔKE is positive
E. ΔPE = 0 J and ΔKE is negative
Answer:
If the stone is thrown from the ground, the correct answer is A. If it is thrown from a height h, the correct answer is D.
Explanation:
Hi there!
I can´t see the drawing but let´s assume that initially, the stone is on the ground level. If that is the case, initially, the potential energy will be zero and when it returns to Earth it will also be zero. The potential energy depends on the height of the stone. If the final and initial height of the stone is zero, then the change in potential energy will also be zero:
ΔPE = final PE - initial PE
ΔPE = m · g · hf - m · g · hi (where hf and hi are the final and initial height respectively)
ΔPE = m · g (hf - hi)
ΔPE = m · g (0)
ΔPE = 0
Initially, the kinetic energy (KE) of the stone is the following:
KE = 1/2 · m · v0²
As the stone goes up, the kinetic energy is transformed into potential energy; but as the stone starts to fall, the acquired potential energy is transformed again into kinetic energy, so that the final and initial kinetic energy of the stone is the same.
Then:
ΔKE = final KE - initial KE = 0 (because final KE = initial KE).
Then, the correct answer is A.
Always ΔKE = -ΔPE due to the conservation of energy. Potential energy can´t be acquired by the stone if there is no loss of kinetic energy and vice-versa.
Let´s assume now that the stone is thrown from a height hi to the ground.
The final potential energy will be zero (becuase h = 0) but the initial PE will be:
PE = m · g · h1
Then:
ΔPE = final PE - initial PE = 0 - m · g · h1
Then ΔPE will be negative.
The initial kinetic energy will be:
KE = 1/2 · m · v0²
But the final kinetic energy will be equal to the initial kinetic energy plus the loss of potential energy (remember: if potential energy decreases, another type of energy has to increase, in this case, kinetic energy and vice-versa):
ΔKE = final KE - initial KE
ΔKE = 1/2 · m · v0² + m · g · h1 - 1/2 · m · v0²
ΔKE = m · g · h1
Then ΔKE will be positive and the correct answer would be D.
A solid cone is 10 cm high. where is its center of mass?
Answer:
The center of mass of a cone is located along a line. This line is perpendicular to the base and reaches the apex. The center of mass is a distance 3/4 of the height of the cone with respect to the apex.
Explanation:
The center of mass of a solid cone is located one-third of the way up from the base.
Explanation:The center of mass of a solid cone is located at one-third of its height from the base. In this case, the cone is 10 cm high, so the center of mass is located 10 cm * (1/3) = 3.33 cm from the base.
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Object A has a position as a function of time given by rA(t) = (3.00 m/s)t i ^ + (1.00 m/s2)t2j^. Object B has a position as a function of time given byrB(t) = (4.00 m/s)ti^ + (-1.00 m/s2)t2j^. All quantities are SI units. What is the distance between object A and object B at time t = 3.00 s?A) 3.46 m B) 15.0 m C) 18.3 m D) 34.6 m E) 29.8 m
Explanation:
Given that
[tex]rA(t)=(3.00 m/s)t\hat{i}+ (1.00 m/s^2)t^2\hat{j}\texttt{ and }rB(t)=(4.00 m/s)t\hat{i}+ (-1.00 m/s^2)t^2\hat{j}[/tex]
We need to find distance when t = 3 s
Substituting t = 3 s
[tex]rA(t)=(3.00 m/s)\times 3\hat{i}+ (1.00 m/s^2)\times 3^2\hat{j}=9\hat{i}+9\hat{j}\\\\rB(t)=(4.00 m/s)\times 3\hat{i}+ (-1.00 m/s^2)\times 3^2\hat{j}=12\hat{i}-9\hat{j}[/tex]
[tex]\texttt{Displacement = }12\hat{i}-9\hat{j}-(9\hat{i}+9\hat{j})=3\hat{i}-18\hat{j}[/tex]
[tex]\texttt{Magnitude = }\sqrt{3^2+(-18)^2}=18.3m[/tex]
Option C is the correct answer.
The distance between object A and object B is approximately 18.248 meters. (Choice C)
How to calculate the distance between two objects
In this question we must apply the concepts of vector difference, dot product and norm to determine the distance between objects A and B, in meters:
[tex]r_{B/A} = \sqrt{(\vec r_{B}-\vec r_{A})\,\bullet\,(\vec r_{B}-\vec r_{A})}[/tex] (1)
Where:
[tex]\vec r_{A}[/tex] - Vector distance of object A, in meters.[tex]\vec r_{B}[/tex] - Vector distance of object B, in meters. [tex]r_{B/A}[/tex] - Distance of B relative to A, in meters.If we know that [tex]\vec r_{A} = (3\cdot t, t^{2})\,\left[m\right][/tex], [tex]\vec r_{B} = (4\cdot t,-t^{2})\,\left[m\right][/tex] and [tex]t = 3\,s[/tex], then the distance of B relative to A is:
[tex]r_{B/A}=\sqrt{t^{2}+4\cdot t^{4}}[/tex]
[tex]r_{B/A} = t\cdot \sqrt{1+4\cdot t^{2}}[/tex]
[tex]r_{B/A} = 3\cdot \sqrt{1+4\cdot 3^{2}}[/tex]
[tex]r_{B/A} \approx 18.248\,m[/tex]
The distance between object A and object B is approximately 18.248 meters. (Choice C) [tex]\blacksquare[/tex]
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the standard free energy of formation for liquid ethanol is -174/9 kj/mol and that for gaseous athanol is -168.6 kj/mol. calculate the vapour pressure of ehtanol at
Answer:
P=133.71mmHg
Explanation:
the standard free energy of formation for liquid ethanol is -174/9 kj/mol and that for gaseous ethanol is -168.6 kj/mol. calculate the vapour pressure of ethanol at
assumption:
is that temperature is at 25C, at standard pressure of 1bar(750mmHg)
ethanol is an ideal gas
The free energy of ethanol liquid does not vary with pressure,
C2H5OH(l)⟶C2H5OH(g)
free energy of formation on the reactant side is -174.9 kj/mol
fro the product side is -168.6 kj
∅Gvap-∅G(l)=-168.6kj/mol-(-174.9kj/mol)
+6.3kj/mol
∅G=∅Gvap+RTlnK-∅Gliq
∅G=0
0=+6.3kj/mol+8.314Jk/mol/k(298)InK
-6.3/(RT)=Lnk
taking the exponential of both sides
[tex]e^{-6300/(8.314*298)} =K[/tex]
0.178=k
k=p/[tex]p^{0}[/tex]
P^0=refers to the pressure of ethanol vapour at its standard state
partial pressure , which is 750 mmHg
P=0.178*750
P=133.71mmHg
Final answer:
The vapor pressure of ethanol at a given temperature can be calculated using the Clausius-Clapeyron equation.
Explanation:
The vapor pressure of a substance is related to its standard free energy of formation and temperature. To calculate the vapor pressure of ethanol at a given temperature, we can use the Clausius-Clapeyron equation:
ln(P₂/P₁) = △Hvap/R × (1/T₁ - 1/T₂)
Where P₁ and P₂ are the vapor pressures at temperatures T₁ and T₂ respectively, △Hvap is the enthalpy of vaporization, R is the gas constant, and ln denotes the natural logarithm.
By substituting the given values for △Hvap = -174/9 kJ/mol, T₁ = 20.0 °C (293 K), and T₂ = desired temperature, we can solve for P₂.
High energy environments are most likely to contain which one of the following?
a.large particles such as gravel
b.silt-sized particles
c.manganese nodules
d.cosmogenous sediments
e.clay-sized particles
Answer:
a.large particles such as gravel
Explanation:
A high energy environment is composed of the Aquas sedimentary environment and features the high energy of turbulent waves and currents in motion and contains the large parts of gravel that is an aggregation of rocks fragments and are formed as result of the weathering and erosion of rocks.High energy environments are typically associated with larger particles such as gravel due to the ability of these environments to move larger matter. Smaller particles are generally found in lower energy environments.
Explanation:High energy environments, such as fast-flowing rivers or coastal areas with strong waves, are most capable of moving larger particles. So, among these options, the environments with high energy are most likely to contain large particles such as gravel (a). The other particles listed (silt, manganese nodules, cosmogenous sediments, clay) require less energy to be moved, so they are likely to be found in lower energy environments.
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What is the value of g on the surface of Saturn? Assume M-Saturn = 5.68×10^26 kg and R-Saturn = 5.82×10^7 m.Choose the appropriate explanation how such a low value is possible given Saturn's large mass - 100 times that of Earth.A) This low value is possible because the density of Saturn is so low.B) This low value is possible because the magnetic field of Saturn is so weak.C) This low value is possible because the magnetic field of Saturn is so strong.D) This low value is possible because the density of Saturn is so high.
Answer:
Approximately [tex]\rm 11.2 \; N \cdot kg^{-1}[/tex] at that distance from the center of the planet.
Option A) The low value of [tex]g[/tex] near the cloud top of Saturn is possible because of the low density of the planet.
Explanation:
The value of [tex]g[/tex] on a planet measures the size of gravity on an object for each unit of its mass. The equation for gravity is:
[tex]\displaystyle \frac{G \cdot M \cdot m}{R^2}[/tex],
where
[tex]G \approx 6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2[/tex].[tex]M[/tex] is the mass of the planet, and[tex]m[/tex] is the mass of the object.To find an equation for [tex]g[/tex], divide the equation for gravity by the mass of the object:
[tex]\displaystyle g = \left.\frac{G \cdot M \cdot m}{R^2} \right/\frac{1}{m} = \frac{G \cdot M}{R^2}[/tex].
In this case,
[tex]M = 5.68\times 10^{26}\; \rm kg[/tex], and[tex]R = 5.82 \times 10^7\; \rm m[/tex].Calculate [tex]g[/tex] based on these values:
[tex]\begin{aligned} g &= \frac{G \cdot M}{R^2}\cr &= \frac{6.67\times 10^{-11}\; \rm N \cdot kg^{-2} \cdot m^2\times 5.68\times 10^{26}\; \rm kg}{\left(5.82\times 10^7\; \rm m\right)^2} \cr &\approx 11.2\; \rm N\cdot kg^{-1} \end{aligned}[/tex].
Saturn is a gas giant. Most of its volume was filled with gas. In comparison, the earth is a rocky planet. Most of its volume was filled with solid and molten rocks. As a result, the average density of the earth would be greater than the average density of Saturn.
Refer to the equation for [tex]g[/tex]:
[tex]\displaystyle g = \frac{G \cdot M}{R^2}[/tex].
The mass of the planet is in the numerator. If two planets are of the same size, [tex]g[/tex] would be greater at the surface of the more massive planet.
On the other hand, if the mass of the planet is large while its density is small, its radius also needs to be very large. Since [tex]R[/tex] is in the denominator of [tex]g[/tex], increasing the value of [tex]R[/tex] while keeping [tex]M[/tex] constant would reduce the value of [tex]g[/tex]. That explains why the value of [tex]g[/tex] near the "surface" (cloud tops) of Saturn is about the same as that on the surface of the earth (approximately [tex]9.81\; \rm N \cdot kg^{-1}[/tex].
As a side note, [tex]5.82\times 10^7\rm \; m[/tex] likely refers to the distance from the center of Saturn to its cloud tops. Hence, it would be more appropriate to say that the value of [tex]g[/tex] near the cloud tops of Saturn is approximately [tex]\rm 11.2 \; N \cdot kg^{-1}[/tex].
concept is gravitational force. Saturn's low value of g on its surface is possible because of its low average density, which is less than the density of water.
The value of g on the surface of Saturn is determined by its mass and radius. The low value of g on Saturn is possible because of its low average density. Saturn's mass is much larger than Earth's, but its density is much lower, resulting in a lower value of g on its surface.
The low density of Saturn, which is only 0.7 g/cm³, is less than the density of water. This means that Saturn would be light enough to float if placed in water. Therefore, despite its large mass, the low density of Saturn allows for a low value of g on its surface.
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The following is a limitation of a bar chart schedule. a. Scheduling information is presented at a summary level b. activity interrelations are not considered c. ease of preparation d. ease of interpretation e. None of the above
Answer:
b. activity interrelations are not considered
Explanation:
A bar chart schedule or Gantt chart is used to visualize task that scheduled over time.
The element lead (Pb) has a density 11.3 times that of water. Copper (Cu) has a density 7.9 times the density of water. A 5 kg mass of lead and a 5 kg mass of copper are both completely submerged in a bucket of water. Which mass has the LARGER buoyant force acting on it?
A) The buoyant force on the lead mass is larger.
B) The buoyant force on the copper mass is larger.
C) The buoyant force is the same on both masses.
Answer:
The answer is B
Explanation:
Density of the element lead (Pb) is:
[tex]d_{Pb} =11,3kg/dm^3[/tex]
Density of the element Copper (Cu) is:
[tex]d_{Cu} =7,9kg/dm^3[/tex]
First we need o find the volume of both materials:
[tex]V_{Pb}=5/11,3=440cm^3[/tex]
[tex]V_{Cu}=5/7,9=630cm^3[/tex]
And the buoyant forces on elements are:
[tex]P_{Pb}=440*1*9,81/1000=4,32N[/tex]
[tex]P_{Pb}=630*1*9,81/1000=6,18N[/tex]
A certain radioactive isotope placed near a Geiger counter registers 120 counts per minute. If the half-life of the isotope is one day, what will the count rate be at the end of four days?
Answer:
The count rate at the end of four days will be 7.5 counts per minute.
Explanation:
First it is important to know that half-life is the time to a piece of radioactive material to decay 50%. So if we know we start with 120 counts per minute and we already know the half-life of the isotope is 1 hour we expect that past 1 hour the material decays 50% (it's halved) so we will count 60 counts per minute, now if we wait another hour 60 counts will decay in to 30 counts per minute and so on. That should be translate to a math equation as:
final counts = initial counts * [tex](\frac{1}{2})^{\#half\,life\,periods}[/tex]
After 4 days we have 4 half-life periods passed so:
final counts= 120 counts per minute * [tex](\frac{1}{2})^{4}[/tex]
final material = 7.5 counts per minute
Which of the following is defined as an area of the body surface that is innervated by a single spinal nerve?
A. Transverse process
B. Malar
C. Dermatome
D. Spinous process
Answer:
Dermatome. (Ans. C).
Explanation:
Dermatome is defined as the area of the human anatomy skin which is supplied by single spinal sensory nerve root. At the spinal cord these spinal sensory nerve enter the nerve root, and the branches of spinal sensory reach to the periphery of the body.
The sensory nerve which is present in the periphery of the body are the type of nerve which helps to transmit signals from sensation such as pain, temperature, etc. to the spinal cord from some specific area of the anatomy.
A soccer ball is kicked from point Pi at an angle above a horizontal field. The ball follows an ideal path before landing on the field at point Pf . Which of the following statements is true when the ball is at point X?
A. The horizontal velocity vector points to the right & equals v cos θ.
B. The vertical velocity points up & equals v sin θ.
C. The acceleration vector points up.
D. The total velocity equals the initial velocity v.
Answer:
A. The horizontal velocity vector points to the right & equals v cos θ.
Explanation:
The motion describes a parabolic path, where the horizontal speed is constant and the horizontal velocity vector always points to the right and equals v*cos θ.
A ball is fixed to the end of a string, which is attached to the ceiling at point P. As the drawing shows, the ball is projected downward at A with the launch speed v0. Traveling on a circular path, the ball comes to a halt at point B. What enables the ball to reach point B, which is above point A? Ignore friction and air resistance.
Answer:
The ball's initial kinetic energy
The ball comes to a stop at B. At this point its initial kinetic energy is converted into potential energy
Explanation:
A ball is fixed to the end of a string, which is attached to the ceiling at point P. As the drawing shows, the ball is projected downward at A with the launch speed v0. Traveling on a circular path, the ball comes to a halt at point B. What enables the ball to reach point B, which is above point A? Ignore friction and air resistance.
From conservation of energy which states that energy can neither be created nor be destroyed, but can be transformed from one form to another.
Ki+Ui=Kf+Uf
Ki=initial kinetic energy
Ui=initial potential energy
Kf=final kinetic energy
Uf=final potential energy
we know that [tex]\frac{1}{2} mu^{2} +mgha=\frac{1}{2} mv^{2} +mghb[/tex]
m=mass of the ball
ha=downward height a
hb=upward height b
u=initial velocity u
v=final velocity v, which is 0
g=acceleration due to gravity
v=0 at final velocity
1/2mu^2+mgha=0+1/2mv^2
ha=hb+Ki/mh
From the above equation, we can conclude that the ball's initial kinetic energy is responsible for making the ball reach point B.
Point B is higher than point A from the motion gained by the ball
Examine the five words and/or phrases and determine the relationship among the majority of words/phrases. Choose the one option that does not fit the pattern. A. abyssal clay B. calcareous ooze C. coarse lithogenic sediment D. manganese nodule E. siliceous ooze
Answer:
The Correct Option is C (coarse lithogenic sediment)
Explanation:
This is because all other options are found on sea floors, except for coarse lithogenic sediments that are as a result of erosion on land. Both calcareous and siliceous ooze are common sea sediments. Abyssal clay and manganese nodule are red clay and rock concretion respectively found in the bottom of the sea.
Evidence that the cosmic background radiation really is the remnant of a Big Bang comes from predicting characteristics of remnant radiation from the Big Bang and comparing these predictions with observations. Four of the five statements below are real. Which one is fictitious?
Answer:
B) The cosmic background radiation is expected to contain spectral lines of hydrogen and helium, and it does.
Explanation:
If a solution surrounding a cell is hypotonic relative to the inside of the cell, in which direction will water move?
Answer:
If the cell is placed in a surrounding solution which is hypotonic in nature.
Then the water from outside of the cell to the inside of the cell. The water will keep on moving from the outside of the cell to the inside of the cell.
The flow of water will take place until the outside environment of the cell and the inside of the cell becomes equal.
The flow of water will take place from the outside of the cell to the inside of the cell.
Answer:
The direction of motion of water molecules will be into the cell.
Explanation:
A hypotonic solution is one which has the less concentration of solute in the solvent as compared to the solution on the other side of the semi-permeable membrane. This creates an osmotic pressure gradient across the semi-permeable membrane which is responsible for the flow of water molecules across the membrane until the concentration becomes equal for both the solutions.(A semi-permeable membrane is a sheet or a plane barrier which does not allows the molecules over certain size to pass through it. Here the membrane does not allows the molecules larger in size than that of water molecules to pass through it.)At a fabrication plant, a hot metal forging has a mass of 67.2 kg and a specific heat capacity of 438 J/(kg C°). To harden it, the forging is quenched by immersion in 786 kg of oil that has a temperature of 37.1 °C and a specific heat capacity of 2950 J/(kg C°). The final temperature of the oil and forging at thermal equilibrium is 58.3 °C.
Assuming that heat flows only between the forging and the oil, determine the initial temperature in degrees Celsius of the forging.
Answer:
Tm = 1,728.38 °C
Explanation:
mass of metal forging (Mm) = 67.2 kg
specific heat capacity of metal forging (Cm) = 438 J/kg°C
initial temperature of metal forging (Tm) = ?
final equilibrium temperature (Te) = 58.3 °C
mass of oil (Mo) = 786 kg
specific heat capacity of oil (Co) = 2950 J/kg°C
temperature of oil (To) = 37.1 °C
Mm × Cf × (Tm - Te) = Mo × Co × (Te - To)
Tm = [tex]\frac{Mo x Co x (Te - To) }{Mm x Cf}[/tex] + Te
Tm = [tex]\frac{786 x 2950 x (58.3 - 37.1) }{67.2 x 438}[/tex] + 58.3
Tm = 1,728.38 °C
You are in an airplane going down a runway currently going 24 m/s and accelerating at 8m/s for 800 meters until take off. How long does it take the plane to go those 800 meters?
Answer:
Time taken to cover the distance 11.45 s.
Explanation:
The given parameters:
Initial Velocity(u)=24 m/s
Acceleration (a)=8 [tex]m/s^{2}[/tex]
Distance or Displacement(s)=800 m
Displacement or Distance is equal in the above because it travels in a straight line.
We have to apply Second Equation of Motion,
s=ut+[tex]\frac{1}{2} at^{2}[/tex]
800=24t + 4[tex]t^{2}[/tex]
200=6t + [tex]t^{2}[/tex]
[tex]t^{2}[/tex] + 6t - 200=0
Solving, the quadratic equation to find out the roots, we get that the possible values of t will be 11.45 s and a negative value.
The negative value will be neglected as time cannot be negative.
Hence, the time taken is 11.45 s.
A large building has an inclined roof. The length of the roof is 54.0 m and the angle of the roof is 17.0° below horizontal. A worker on the roof goes of a hammer from the peak of the roof. Starting from rest, it slides down the entire length of the roof with a constant acceleration of 2.87 m/s2. After leaving the edge of the roof, it falls a vertical distance of 46.5 m before hitting the ground.
a. How much time does it take the hammer to fall from the edge of the roof to the ground?
b. How far horizontally does the hammer travel from the edge of the roof until it hits the ground?
Answer:
a) [tex]t=2.6\ s[/tex]
b) [tex]s=43.7747\ m[/tex]
Explanation:
Given:
length of inclined roof, [tex]l=54\ m[/tex]
inclination of roof below horizontal, [tex]\theta=17^{\circ}C[/tex]
acceleration of hammer on the roof, [tex]a_r=2.87\ m.s^{-2}[/tex]
height from the lower edge of the roof, [tex]h=46.5\ m[/tex]
Now, we find the final velocity when leaving the edge of the roof:
Using the equation of motion:
[tex]v^2=u^2+2.a_r.l[/tex]
[tex]v^2=0^2+2\times 2.87\times 54[/tex]
[tex]v=17.6057\ m.s^{-1}[/tex]
The direction of this velocity is 17° below the horizontal.
∴Vertical component of velocity:
[tex]v_y=v.sin\ \theta[/tex]
[tex]v_y=17.6057\times sin\ 17^{\circ}[/tex]
[tex]v_y=5.1474\ m.s^{-1}[/tex]
a.
So, the time taken to fall on the ground:
[tex]h=ut+\frac{1}{2} g.t^2[/tex]
here:
initial velocity, [tex]u=v_y=5.1474\ m.s^{-1}[/tex]
putting respective values
[tex]46.5=5.1474\times t+0.5\times 9.8\times t^2[/tex]
[tex]t=2.6\ s[/tex]
b.
Horizontal component of velocity, [tex]v_x=v.cos\ \theta=17.6057\ cos\ 17^{\circ}=16.8364\ m.s^{-1}[/tex]
Since there is no air resistance so the horizontal velocity component remains constant.
∴Horizontal distance from the edge of the roof where the hammer falls is given by:
[tex]s=v_x.t[/tex]
[tex]s=16.8364\times 2.6[/tex]
[tex]s=43.7747\ m[/tex]
Using equations of motion, it's calculated that it takes about 3.07 seconds for the hammer to fall to the ground from the roof. The hammer also travels approximately 89.2 meters horizontally from the roof to the ground.
Explanation:The question is related to physics concepts of motion under gravity and kinematics. For simplicity, let's ignore air resistance.
Part a:The time it takes for the hammer to fall to the ground from the edge of the roof can be calculated using the equation of motion: h = 0.5gt^2, where h is the height, g is the acceleration due to gravity (9.8 m/s²), and t is the time. Solving the equation for time (t): t = sqrt(2h/g). Substituting the given values, we get t = sqrt((2*46.5)/9.8) ≈ 3.07 s.
Part b:The horizontal distance travelled by the hammer can be calculated using the formula: distance = speed × time. The horizontal speed of the hammer when it falls off the roof will be the same speed it had just as it left the roof due to the roof slope acting on it with constant acceleration. This can be gotten from the equation v = u + at, where u is the initial velocity (0 m/s), a is the acceleration (2.87 m/s²) and t is the time. The time here is the time it takes for the hammer to slide down the roof, gotten by time = distance/speed = 54.0/v. Solving all these gives a distance of approximately 89.2 m.
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At an uncontrolled intersection, when must the car on the right yield to the car on the left?
Answer: the car on the right should yield to the car that arrived first. That is When the car on the left arrives first
Explanation:
It must be noted that the law did not grant the 'right-of-way'. The law only says when the right of way must be yielded. The law does state who must yield the right of way neither does the law give right of way to anyone.
Yielding the right of way to another vehicle simply means that you are letting them go before you in a traffic situation.
Therefore, When two vehicles approaches an intersection without no traffic signs or signals, (that is, an uncontrolled intersection) the two vehicles must slow down. Always Yield to vehicles already in the uncontrolled intersection and drivers who arrive at the uncontrolled intersection before you.
The vehicle on the left should always yield to the right of the way to the vehicle on the right. The driver with ''right-of-way'' must pay attention to avoid a collision.
Beginning at the NW corner of the intersection of Pine & 675, thence north 950 feet, thence west 380 feet, thence south 950 feet, thence east 380 feet. Is this an acceptable metes and bounds description?
Answer:
this description is valid for mediadle displacement, bone is an acceptable description
Explanation:
The description of a person's position must be done with a position vector. These vectors must have magnitude, a given direction and a starting point.
In the description this has a starting point corner NO of pine and 675.
Each displacement occurs with respect to the previous one, indicating the magnitude of the displacement and its direction.
After analyzing this description is valid for mediadle displacement, bone is an acceptable description
To increase the speed at which Google Analytics compiles reports, what action could be taken?
Answer:
Answer Choose “Faster response” in the sampling pull-down menu
Explanation:
At the top of the report, below the date range selector, select faster response,
This option uses a smaller sampling size to give you faster results. In order to get a better understanding lets define sampling, according to Google analytics website
In data analysis, sampling is the practice of analyzing a subset of all data in order to uncover the meaningful information in the larger data set.
For example, if you wanted to estimate the number of trees in a 100-acre area where the distribution of trees was fairly uniform, you could count the number of trees in 1 acre and multiply by 100, or count the trees in a half acre and multiply by 200 to get an accurate representation of the entire 100 acres.
To speed up Go_ogle Analytics reports, you can reduce the date range and complexity of the report, or increase server resources.
Explanation:To increase the speed at which Go_ogle Analytics compiles reports, several actions could be taken. First, one can reduce the amount of data that is being processed by adjusting the date range of the reports. By analyzing a smaller chunk of data, the report can be generated much faster.
Secondly, one can adjust the complexity of the report. The more complex the report, the more data needs to be processed, hence it will take longer. Reducing the complexity and only focusing on the key metrics can help speed up the report compilation.
Lastly, one can increase the capacity of their server resources. If a company has the means, it can invest in dedicated server resources for Go_ogle Analytics, allowing it to process reports faster.
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According to quantum physics, measuring velocity of a tiny particle with an electromagnet
A. Has no effect on the velocity of the particle.B. Affects the velocity of the particle.
Answer:
Option A.
Explanation:
In quantum physics there is a law to relate the position and the momentum of the particle, it says that if we know with precision where is a quantum particle, we can not know the momentum of this particle, in other words, the velocity of the particle. So, when we measure the velocity of the particle we find the correct value of the particle, but we can not determine with accuracy where is the particle. This law is known as the Heisenberg's uncertainty principle and, its expressed as follows:
[tex] \Delta x \Delta p \geq \frac{h}{4 \pi} [/tex]
where Δx: is the position's uncertainty, Δp: is the momentum's uncertainty and h: is the Planck constant.
Therefore, the correct answer is A: measuring the velocity of a tiny particle with an electromagnet has no effect on the velocity of the particle. It only affects the determination of the particle's position.
I hope it helps you!
Calculate the magnitude of the flux of a constant electric field of 5.00 N/C in the z-direction through a rectangle with area 4.00 m2 in the xy-plane. (a) 0 (b) 10.0 N m2/C (c) 20.0 N m2/C (d) more information is needed
Answer:
The magnitude of the flux is [tex]2.00 N m^2/C[/tex]
Explanation:
The electric flux through a planar area is defined as the product of electric field and the component of the area perpendicular to the field.
Electric flux = Electric field * Area * (angle between the planar area and the electric flux)
The equation is
[tex]\phi = E A cos(\theta)[/tex]
Where:
[tex]\phi[/tex]is the Electric Flux
A is the Area
E is the Electric field
[tex]\theta[/tex] is angle between a perpendicular vector to the area and the electric field
Now substituting the values,
[tex]\phi = 5.00 \times 4.00 \times cos(0)[/tex]
[tex]\phi = 5.00 \times 4.00 \times 1[/tex]
[tex]\phi = 2.00 N m^2/C[/tex]
The flux of a constant electric field in the z-direction through a rectangle in the xy-plane is zero, because the angle between the direction of the electric field and the direction of the normal to the area is 90 degrees, which makes the dot product zero.
Explanation:To calculate the magnitude of the flux of an electric field, we use the equation: Φ = E . A where Φ is the electric flux, E is the electric field, and A is the area of the surface. The dot (.) represents a dot product, which means we consider the angle between the field and the area. In this problem, the electric field (E) is given as 5.00 N/C and the area of the rectangle (A) is 4.00 m². Also, because the electric field is in the z-direction (up and down), and the rectangle is in the xy-plane (flat), the angle between the field and the area is 90 degrees.
However, the dot product for angles of 90 degrees is zero because cos(90°) = 0. So, regardless of the magnitudes of the electric field and the area, the flux is zero because Φ = E . A = EAcos(90°) = 0. Therefore, the correct answer is (a) 0.
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A) A motorist traveling at 12 m/s encounters a deer in the road 41 m ahead. If the maximum acceleration the vehicle’s brakes are capable of is −7 m/s 2 , what is the maximum reaction time of the motorist that will allow her or him to avoid hitting the deer? B) If his or her reaction time is 1.1333s, how fast will (s) he be traveling when(s) he reaches the deer?
Answer:
A) Reaction Time = 2.73 seconds.B) He will not reach the deer.Explanation:
A) Initial speed, u = 12 m/s
Acceleration, a = -7 m/s²
Final velocity, v = 0 m/s
We have equation of motion v² = u² + 2as
0² = 12² + 2 x -7 x s
s = 10.29 m
He need 10.29 m to stop.
Remaining distance = 43 - 10.29 = 32.71 m
We have
Remaining distance = Reaction Time x Initial velocity
32.71 = Reaction Time x 12
Reaction Time = 2.73 seconds.
B) Distance traveled in 1.1333 s = 1.1333 x 12 = 13.60 m
Remaining distance = 41 - 13.6 = 27.4 m
Initial speed, u = 12 m/s
Acceleration, a = -7 m/s²
Displacement, s = 27.4 m
v² = 12² + 2 x -7 x 27.4
v² = -239.6
Not possible.
Motorist will not hit deer.
He will not reach the deer.