A force of 14 N acts on a 5 kg object for 3 seconds. a. What is the object’s change in momentum? b. What is the object’s change in velocity?
Answer:The answer can be calculated by doing thefollowing steps;
Explanation:
A 4.0 kg model rocket is launched, shooting 50.0 g of burned fuel from its exhaust at an average velocity of 625 m/s. What is the velocity of the rocket after the fuel has burned?
Answer:
Velocity of rocket will be equal to 7.81 m/sec
Explanation:
We have given mass of the rocket [tex]m_1=4kg[/tex]
Mass of fuel is given [tex]m_2=50gram=0.05kg[/tex] ( As 1 kg is equal to 1000 gram )
Velocity of fuel [tex]v_2[/tex] = 625 m/sec
We have to find the velocity of rocket [tex]v_1[/tex]
From conservation of momentum we know that initial momentum is equal to final momentum '
So [tex]m_1v_1=m_2v_2[/tex], here [tex]m_1[/tex] is mass of rocket [tex]v_1[/tex] is velocity of rocket [tex]m_2[/tex] is mass of fuel and [tex]v_2[/tex] is velocity of fuel
So [tex]4\times v_1=625\times 0.05[/tex]
[tex]v_1=7.81m/sec[/tex]
So velocity of rocket will be equal to 7.81 m/sec
A swimmer is determined to cross a river that flows due south with a strong current. Initially, the swimmer is on the west bank desiring to reach a camp directly across the river on the opposite bank. In which direction shoud the swimmer head?
A. The swimmer should swim southeast.
B. The swimmer should swim south.
C. The swimmer shuold swim northeast.
D. The swimmer should swim due north.
E. The swimmer should swim due east.
Answer:
C. The swimmer should swim northeast.
Explanation:
A swimmer expects to reach to the eastern bank of a river by swimming whose stream flows from north to south. Then in order to reach the directly opposite bank of the river when standing on the western bank, the swimmer must swim towards northeast direction so that the resultant of the velocity of flow and the velocity of swimming results in a net velocity towards the east direction.
You are holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. What is the direction of the force on the charge you are holding?
1. to the east
2. to the southwest
3. to the south
4. to the northeast
Answer:
southwest
Explanation:
Here, all the particles have same charge this means that only repulsive force is acting on the particles according to Coulomb's laws.
One particle is 1 m to my north and another particle is to my east.
The particle to my north will push the positive charge I am holding to the south and the particle to my east will push the positive charge I am holding to the west.
Hence, the resultant direction of the force will be southwest.
A playground toy has four seats, each 6.4kg , attached to very light rods of length r= 1.5m , as seen from below in the figure.
If two children, with masses of 16kg and 23kg , sit in seats opposite one another, what is the moment of inertia about the rotation axis?
Answer:
The moment of inertia about the rotation axis is 145.35 kg.m²
Explanation:
Given:
Mass of each seat = 6.4 kg
length of rods attached to the seat = radius of the seat from center = 1.5m
Moment of inertia = M₁r²
Moment of inertia about the first child:
total mass at that point = mass of the child + mass of the seat
= 16kg + 6.4 kg = 22.4 kg
Moment of inertia at the point where the first child sits = 22.4 *1.5²
= 50.4 kg.m²
Moment of inertia about the second child:
total mass at that point = mass of the child + mass of the seat
= 23kg + 6.4 kg = 29.4 kg
Moment of inertia at the point where the second child sits = 29.4 *1.5²
= 66.15 kg.m²
Moment of inertia about the two empty seats:
Moment of inertia about the two seats = 2(6.4*1.5²)
= 28.8 kg.m²
Moment of inertia about the rotation axis = Moment of inertia about the point of two children + moment of inertia about the two seat
= (50.4 + 66.15 + 28.8) kg.m²
= 145.35 kg.m²
Therefore, the moment of inertia about the rotation axis is 145.35 kg.m²
The moment of inertia about the rotation axis is 145.35 kgm2.
How do you calculate the moment of inertia?Given that the mass m_s of the seat is 6.4 kg, length r of the rod is 1.5 m. The masses of two children is m1 = 16 kg and m2 = 23 kg.
The moment of inertia about the first child is calculated given below.
[tex]MI _1 = (m_s+m_1)r^2[/tex]
[tex]MI_1 = (6.4+16)\times 1.5^2[/tex]
[tex]MI_1 = 50.4 \;\rm kg m^2[/tex]
The moment of inertia about the second child is calculated given below.
[tex]MI _2= (m_s+m_2)r^2[/tex]
[tex]MI_2 = (6.4+23)\times 1.5^2[/tex]
[tex]MI_2 = 66.15 \;\rm kg m^2[/tex]
The moment of inertia about the two empty seats is calculated given below.
[tex]MI_s = 2(m_sr^2)[/tex]
[tex]MI_s = 2 ( 6.4\times 1.5^2)[/tex]
[tex]MI_s = 28.8 \;\rm kgm^2[/tex]
The moment of inertia about the rotation axis is the total sum of the moment of inertia about the point of two children and the moment of inertia about the two seats.
[tex]MI = MI_1 +MI_2+MI_s[/tex]
[tex]MI = 50.4 + 66.15 + 28.8[/tex]
[tex]MI = 145.35 \;\rm kgm^2[/tex]
Therefore, the moment of inertia about the rotation axis is 145.35 kgm2.
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A typical neutron star may have a mass equal to that of the Sun but a radius of only 20 km.(a) What is the gravitational acceleration at the surface of such a star?_______m/s2(b) How fast would an object be moving if it fell from rest through a distance of 16 m on such a star? (Assume the star does not rotate.)_______m/s
Answer:
[tex]331665750000\ m/s^2[/tex]
3257806.62409 m/s
Explanation:
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
M = Mass of Sun = [tex]1.989\times 10^{30}\ kg[/tex]
r = Radius of Star = 20 km
u = Initial velocity = 0
v = Final velocity
s = Displacement = 16 m
a = Acceleration
Gravitational acceleration is given by
[tex]g=\dfrac{GM}{r^2}\\\Rightarrow g=\dfrac{6.67\times 10^{-11}\times 1.989\times 10^{30}}{20000^2}\\\Rightarrow g=331665750000\ m/s^2[/tex]
The gravitational acceleration at the surface of such a star is [tex]331665750000\ m/s^2[/tex]
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 331665750000\times 16+0^2}\\\Rightarrow v=3257806.62409\ m/s[/tex]
The velocity of the object would be 3257806.62409 m/s
(a) the acceleration due to gravity is 3.3 × 10¹¹ m/s²
(b) the final velocity of the object is 3.26 × 10⁶ m/s
Gravitational force:Given that the neutron star has a mass equal to Sun, M = 2 × 10³⁰ kg
Radius of the star is R = 20 km = 20 × 10³
The weight on the surface of a neutron star that has the same mass as our Sun and a diameter of 21.0 km will be 8.29 × 10¹³ N
(a) The gravitational force on the surface of the neutron star is given by:
F = GMm/R²
mg = GMm/R²
where G is the gravitational constant
M is the mass of the body
m is the mass of the person
and, R is the radius of the body
So,
g = GM/R²
g = (6.67 × 10⁻¹¹)( 2 × 10³⁰)/ (210 × 10³)²
g = 3.3 × 10¹¹ m/s²
(b) According to the third equation of motion:
v² = u² + 2gh
v² = 2gh , since u = 0 as the object is at rest initially
v = [tex]\sqrt{2\times3.3\times10^{11}\times16}[/tex]
v = 3.26 × 10⁶ m/s
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Tribes of the Sioux Nation (among other Plains Indians) maintained historical calendars composed of winter counts. Tribe historians would_____________________.
Answer:
Tribe historians would: Depict a significant event for each year on a buffalo or deer skin
Tribes of the Sioux Nation maintained historical calendars composed of winter counts, using pictorial representations to record significant events. Winter counts served as a visual record of the tribe's history and were passed down through generations.
Explanation:Tribes of the Sioux Nation (among other Plains Indians) maintained historical calendars composed of winter counts. Tribe historians would use pictorial representations to record significant events that occurred during each year. Each year, a new pictorial symbol would be added to the count. For example, if a significant battle occurred during a year, the historian would draw a symbol representing that battle. Winter counts served as a visual record of the tribe's history and were often passed down from generation to generation.
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A calorimeter is used to determine the specific heat capacity of a test metal. If the specific heat capacity of water is known, what quantities must be measured?
Answer:
initial and final temperatures of both the water and metal, mass of the metal, and mass of the water
Explanation:
Heat lost by the metal, [tex]Q = mc(t_{2} - t_{1})[/tex]
Heat gained by the water in the calorimeter, [tex]Q_{w} = m_{w}c_{w}(t_{2w} - t_{1w})[/tex]
For energy to be conserved in the system, the heat lost by the metal will equal the heat gain by the water in the calorimeter.
[tex]mc(t_{2} - t_{1}) = m_{w}c_{w}(t_{2w} - t_{1w})[/tex]
Where,
m is the mass of the metal
c is specific heat capacity of the metal
t₂ is the final temperature of the metal
t₁ is the initial temperature of the metal
[tex]m_{w} [/tex] is the mass of the water
[tex]c_{w} [/tex] is specific heat capacity of water
[tex]t_{2w} [/tex] is the final temperature of water
[tex]t_{1w} [/tex] is the initial temperature of water
From the question given, specific heat capacity of the water is known, the quantities to be measured are;
Initial and final temperatures of both the water and metal,
Mass of the metal, and mass of the water
What is the mechanical advantage of the lever if you hold on to the end of the long side and lift something that's placed on the short side?
Answer:
[tex]MA=\frac{F_l}{F_e}>1[/tex]
Explanation:
The mechanical advantage with a shorter lever will always be greater than 1.
It is so because with a longer effort arm we need to apply lesser force to lift a unit mass which is at a shorter distance from the fulcrum. This is in accordance with the conservation of moments.
[tex]F_e\times d_e=F_l\times d_l[/tex]
where:
[tex]F_e=[/tex] force on the effort arm
[tex]F_l=[/tex] force on the load arm
[tex]d_e\ \&\ d_l[/tex] are the lengths of load effort arm and load arm respectively.
So, one factor balances the other keeping the product of the two constant.
And we know that mechanical advantage is :
[tex]MA=\frac{F_l}{F_e}[/tex]
An engineer claims that solar cells for the generation of electrical power can never meet our high demand for electrical energy. The engineer points out that such cells____________________.
Answer:
The sentence is following with "are made from mostly silicon material".
Explanation:
Silicon is extracted and refined by the use of electrical energy. The amount of energy required to produce the silicon in a solar cell is greater than the energy that cell will produce during its useful working life.
A projectile of mass 0.607 kg is shot straight up with an initial speed of 20.1 m/s. (a) How high would it go if there were no air resistance?
Answer:
20.6m
Explanation:
The mass is useless in this case.
Kinetic Energy = Potential Energy
g is acceleration due to gravity.
(1/2)mv² = mgh (m's drop out)
(1/2)v² = gh
(1/2)(20.1²) = 9.81(h)
h = (20.1 x 20.1)/9.81 x 2
20.61 meters
Does air pressure increase or decrease with an increase in altitude?
Answer:
increase
Explanation:
think of climbing a mountain. the higher you go the harder it is to breathe. its because air pressure is increasing
Answer:
decrease.
Explanation:
The air pressure is given by
P = h x d x g
where, h is the height of air column above the surface and g be the acceleration due to gravity and d be the density of air.
As we go up, the height of air column decreases, density of air decreases and acceleration due to gravity also decreases, so the value of pressure decreases at altitudes.
1A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.
The initial horizontal velocity of the soccer ball is 16.5 m/s
Explanation:
When we throw a ball, there is a constant velocity horizontal motion and there is an accelerated vertical motion. These components act independently of each other. Horizontal motion is constant velocity motion.
[tex]v_{x f}=v_{x i}=v_{x}[/tex]
[tex]a_{x}=0[/tex], so [tex]x=v_{i x} t+\left(\frac{1}{2}\right) a_{x} t^{2}[/tex] for horizontal motion
[tex]y=v_{i y} t+\left(\frac{1}{2}\right) a_{y} t^{2}[/tex] for vertical motion
Given:
x = 35 m
[tex]a_{x}=0 \mathrm{m} / \mathrm{s}^{2}[/tex]
Need to find [tex]v_{i x}[/tex]
y = - 22 m
[tex]v_{i y}=0 \mathrm{m} / \mathrm{s}[/tex]
[tex]a_{y}=-9.8 \mathrm{m} / \mathrm{s}^{2}[/tex] (negative sign indicates downward motion)
By substituting all known values, we can solve for 't' value as below
[tex]y=v_{i y} t+\left(\frac{1}{2}\right) a_{y} t^{2}[/tex]
[tex]-22=0(t)+\left(0.5 \times-9.8 \times t^{2}\right)[/tex]
[tex]t^{2}=\frac{-22}{0.5 \times-9.8}=\frac{-22}{-4.9}=4.4897[/tex]
Taking square root, we get t = 2.12 seconds
Now, substitute these to find initial horizontal velocity
[tex]x=v_{i x} t+\left(\frac{1}{2}\right) a_{x} t^{2}[/tex]
[tex]35=v_{i x}(2.12)+\left(0.5 \times 0 \times(2.12)^{2}\right)[/tex]
[tex]35=v_{i x}(2.12)+0[/tex]
[tex]v_{i x}=\frac{35}{2.12}=16.5 \mathrm{m} / \mathrm{s}[/tex]
An empty graduated cylinder weighs 55.26 g. When filled with 48.1 mL of an unknown liquid, it weighs 92.39 g. The density of the unknown liquid is?
The density of the unknown liquid can be calculated by finding the mass of the liquid and dividing it by the volume. The mass of the liquid is 37.13 g and the volume is 48.1 mL. Thus, the density is 0.772 g/mL.
Explanation:The density of a substance is calculated by dividing its mass by its volume. Here, the mass of the unknown liquid can be calculated by subtracting the weight of the empty graduated cylinder from the weight of it filled with the liquid. That is, mass = 92.39 g - 55.26 g = 37.13 g. The volume of the liquid is given as 48.1 mL. Therefore, the density can be calculated as follows:
Density = mass/volume
= 37.13 g / 48.1 mL = 0.772 g/mL
This means the density of the unknown liquid is 0.772 g/mL.
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A rubber ball is dropped from rest from a height h. The ball bounces off the floor and reaches a height of 2h/3. How can we use the principle of the conservation of mechanical energy to interpret this observation?
a) During the collision with the floor, the floor did not push hard enough on the ball for it to reach its original height.
b) Some of the ball’s potential energy was lost in accelerating it toward the floor.
c) The force of the earth’s gravity on the ball prevented it from returning to its original height.
d) Work was done on the ball by the gravitational force that reduced the ball’s kinetic energy.
e) Work was done on the ball by non-conservative forces that resulted in the ball having less total mechanical energy after the bounce.
Please explain how to figure this out
Answer:
Option e.
In this case the work is done on the ball by nonconservative forces that resulted in the ball having less total mechanical energy after the bounce.
Explanation:
This is the type of nonelastic collision when a moving ball hits the ground.Although the conservation of mechanical energy possessed by the ball which is the sum of P.E and K.E., but kinetic energy is not conserved.The non conservative force did the work on the ball that after bouncing lost some of the mechanical energy of that ball. The kinetic energy in the beginning is converted in some other energy like friction and air resistance in this case.A potential energy function for a system in which a two-dimensional force acts is of the form U = 3x5y − 3x. Find the force that acts at the point (x, y).
Answer:
[tex]\overrightarrow{F} = -\left ( 15x^{4}y-3 \right )\widehat{i}-3x^{5}\widehat{j}[/tex]
Explanation:
[tex]U = 3x^{5}y-3x[/tex]
Force is defined as the
[tex]\overrightarrow{F} = -\frac{dU}{dx}\widehat{i}-\frac{dU}{dy}\widehat{j}[/tex]
[tex]\frac{dU}{dx}=15x^{4}y-3[/tex]
[tex]\frac{dU}{dy}=3x^{5}[/tex]
So, force is given by
[tex]\overrightarrow{F} = -\left ( 15x^{4}y-3 \right )\widehat{i}-3x^{5}\widehat{j}[/tex]
A jet fighter pilot wishes to accelerate from rest at 5 ggg to reach Mach-3 (three times the speed of sound) as quickly as possible. Experimental tests reveal that he will black out if this acceleration lasts for more than 5.2 ss . Use 331 m/sm/s for the speed of sound
Answer
given,
acceleration, a = 5 g = 5 x 9.8 = 49 m/s²
speed of sound, v = 331 m/s
speed , v = 3 Mach = 3 x 331 m/s = 993 m/s
time, t = 5.2 s
a) Calculating the time period of black out.
time required by the aircraft to black out
using equation of motion
v = u + a t
initial velocity = 0 m/s
993 = 0 + 49 t
t = 20.26 s
the required time to reach 3 Mach speed is more than the given time hence, pilot will black out.
b) now, calculating the maximum speed it can reach in 5.2 s
using equation of motion
v = u + a t
initial velocity = 0 m/s
v = 0 + 49 x 5.2
v = 254.8 m/s
A 123 kg box is resting on the ground. The coefficient of static friction is 0.34. What force must be applied to the box to start it moving?
Answer:
410.254 N
Explanation:
Force: This can be defined as the product of the mass of a body. The Unit of force is Newton (N)
Deduced from the question,
Force applied to the box to start it moving = Force of friction.
Fₐ = F
Where Fₐ = Force applied to the box to start it moving, F = Force of friction.
But,
F = μR.......................... Equation 1
Where R = normal reaction, μ = coefficient of static friction.
R = W = mg ( on a level surface)
Where m = mass of the box = 123 kg, g = acceleration due to gravity = 9.81 m/s²
R = 123×9.81
R = 1206.63 N.
Also, μ = 0.34
Substituting into equation 1
F = 1206.63×0.34
F = 410.254 N.
Thus the force applied to the box to start it moving = 410.254 N
Assume that all four H-R diagrams below represent a star in different stages of its life, after it starts to fuse hydrogen in its core. Rank the HR diagrams based on when each stage occurs, from first to last. star in the sequencea little outsidemore outsidethen below the line
A star goes through multiple stages of life represented on the H-R Diagram, from the main sequence stage to a red giant, then a brief stable period after a helium flash, back to a giant, and finally, death. The 'main-sequence turnoff' on the H-R diagram marks where stars begin to leave the main sequence. These stages occur faster for more massive stars.
Explanation:Stars progress through various stages on the H-R diagram as they age. Initially, in the main sequence stage, stars start by fusing hydrogen in their cores. The most massive stars evolve into red giants and supergiants less than a million years after they reach the main sequence, moving up and to the right on the diagram. After, lower mass stars also begin their departure from the main sequence.
During the next phase of evolution, a star experiences a helium flash which causes a change in its internal structure, leading to a brief period of stability. Following the exhaust of the central helium, the star turns into a giant again and its position on the H-R diagram shifts to high luminosity and low temperature. During the final stage of the star's life, it depletes its inner resources and begins to die.
This process of evolution is faster for more massive stars, as the rate at which they go through each stage of life is higher. The main-sequence turnoff point on the H-R diagram denotes the location where stars begin to leave the main sequence and transform towards the red giant region. It is lower on the H-R diagram and the mass of the stars is lower in older clusters.
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The ease with which the charge distribution in a molecule can be distorted by an external electrical field is called the __________.
A) electronegativity
B) hydrogen bonding
C) polarizability
D) volatility
E) viscosity
Answer:
C
Explanation:
The answer is polarization. In a normal molecule the positive and negative charges are distributed in a way that it makes a continuous distribution with no net force. but when you apply an external electric force, the opposite charges go to the opposite ends of the molecule making them a polarized molecule.
This is called polarization.
I hope it helps you.
Thank you.
In order to understand the full scope of a disease, we take its occurrence into account. The __________ of a disease is the number of people in a population who develop a disease at a specified time.
a. incidence
b. PREVALENCE
c. endemic infection
d. sporadic infection
Answer: b. PREVALENCE
Explanation:
Prevalence can be defined as the proportion of individuals in a population having a disease or common characteristic. Prevalence of a particular disease is number of people in a population who have develop a disease at a particular time.
It can be expressed in form of proportion Which equals the number of developed cases at a particular time divided by the total number of population.
Life of a Star Cluster. Imagine you could watch a star cluster from the time of its birth to an age of 13 billion years. Describe in one or two paragraphs what you would see happening during that time?
Explanation:
The life span of a star is quite as many more than 10 billion years. During these many years it would pass through different stages of its life, from nebula to ( perhaps, the current stage) supernova to neutron star or Black Hole.
The stages of the life of star over a span of 13 billion year can be summarized as
the transition stages are
Nebula → Blue star → Blue-white super-giant star → Supernova → Neutron star →Black Hole.
It requires 0.30 kJ of work to fully drive a stake into the ground. If the average resistive force on the stake by the ground is 828 N, how long is the stake?
Answer:
Length of the stake will be 0.3623 m
Explanation:
We have given energy required to fully drive a stake into ground = 0.30 KJ = 300 J
Average resistive force acting on the floor is equal to F = 828 N
We have to find the length of the stake
We know that work done is given by
W = Fd, here W is work done , F is average force and d is the length of the stake
So 300 = 828×d
d = 0.3623 m
So length of the stake will be 0.3623 m
The length of the stake is approximately 0.362 meters.
To solve this problem, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. In this case, the work done to drive the stake into the ground is equal to the potential energy gained by the stake when it is fully driven into the ground.
The work done (W) is given as 0.30 kJ, which is equal to 0.30 * 10^3 J.
The resistive force (F) is given as 828 N.
The distance (d) over which the work is done is the length of the stake.
We can use the formula for work:
[tex]\[W = F \cdot d\][/tex]
Rearranging the formula to solve for \(d\):
[tex]\[d = \frac{W}{F}\][/tex]
Substituting the given values:
[tex]\[d = \frac{0.30 \times 10^3 \, J}{828 \, N}\][/tex]
[tex]\[d = 0.362 \, m\][/tex]
You are given the masses, in grams (g), and the volumes, in cubic centimeters (cm3), for a series of substances. Arrange them in order of increasing density. Rank from lowest density to highest density. To rank items as equivalent, overlap them.
The question requires arranging substances by density, a fundamental physics concept, calculated as mass divided by volume. Understanding densities, from light gases to dense neutron stars, is essential in both everyday applications and advanced scientific fields.
Explanation:The question involves arranging substances in order of increasing density based on given masses and volumes. Density is the measure of how much mass is contained in a given volume, and this concept is fundamental in physics. By calculating the density of each substance using the formula density = mass/volume (expressed in g/cm³ for solids and liquids or g/L for gases), we can compare the densities of the substances and arrange them accordingly. It is crucial to recognize that the density of a material is a characteristic property that can help in identifying the material and understanding its behavior in different environments.
Density plays a significant role in various applications, such as determining whether an object will float or sink in a fluid, in material selection for construction, and in understanding the structure of the universe, from the density of air we breathe (1.2 g/L) to the dense core of neutron stars reaching up to 10¹⁵ g/cm³.
To arrange the substances in order of increasing density, we need to calculate the density of each substance using the formula [tex]\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \][/tex]
Then, we compare the calculated densities to determine their relative order.
Here's how the substances could be ranked:
1. Helium gas: Since helium gas is the least dense substance, it will have the lowest density.
2. Cork: Cork typically has a low density compared to other materials.
3. Water: Water has a moderate density, higher than helium gas and cork.
4. Aluminum: Aluminum generally has a higher density than water.
5. Iron: Iron is denser than aluminum, making it the densest substance among the listed ones.
By comparing the calculated densities, we can arrange the substances in increasing order of density from lowest to highest: helium gas, cork, water, aluminum, and iron.
The worked examples of charged-particle motion are relevant to. The worked examples of charged-particle motion are relevant to. a. a transistor. b. a cathode-ray tube. c. magnetic resonance imaging cosmic rays. d. lasers.
Answer:
b. a cathode-ray tube.
John weighs 90 lbs and Jane weighs 60 lbs. They are both sitting on a seesaw. If John is seated 10 feet away from Jane, how far should each be from the fulcrum of the seesaw?
Answer: Jane 6ft from fulcrum
John 4 ft from fulcrum
Explanation: You use the law of moment of force.
That is Clockwise moment equals anti clockwise moment.
Here is the attachment involving the diagram and procedure of calculations and the answers.
The small child should sit 1.30 m away from the pivot point on the seesaw to maintain balance.
The small child should sit 1.30 m away from the pivot point on the seesaw to maintain balance.
Miko is blowing up balloons for her little sisters birthday party one of the balloons slips out of her hand before she ties it off and it shoots through the air until it is empty which statement explains what caused the balloon to accelerate
The balloon accelerated because of Newton's third law of motion. When the air in the balloon is forcefully ejected, the balloon experiences an equal and opposite force, causing it to accelerate in the opposite direction.
Explanation:The acceleration of the balloon when it slips from Miko's hand is due to Newtons third law of motion. This law states that for every action, there is an equal and opposite reaction. In this case, when the air is forcefully ejected or escapes from the balloon, the balloon experiences an equal and opposite force which causes it to shoot off or accelerate in the opposite direction of the escaping air. This is the same principle that explains the propulsion of rockets and the recoil of guns.
You can visualize the movement of the balloon somewhat like the exploding fireworks shown in figure 9.31. The fireworks explode and eject matter in all directions - and its center of mass behaves similarly to the balloon's - accelerated by the ejection of matter. The balloon, like the firework, would keep the parabolic path of flight as long as there was matter (or in the balloon's case, air) pushing it forward. Once all of the air has escaped, the balloon starts to fall due to gravity, like any other free-falling object.
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A circuit has a current of 2.4 A. The voltage is increased to 4 times its original value, while the resistance stays the same. How should the resistance change to return the current to its original value if the voltage remains at its increased amount?
Answer:
Resistance will become 4 times the previous value
Explanation:
We have given current in the circuit i = 2.4 A
According to ohm's law current in the circuit is given by [tex]I=\frac{V}{R}[/tex]
So [tex]2.4=\frac{V}{R}[/tex]............eqn 1
Now voltage is increased to 4 times so new voltage = 4 V
And current in the circuit is same as 2.4 A
We have to fond the resistance so that after increasing voltage current will be same
So [tex]2.4=\frac{4V}{R_{unknown}}[/tex]..........eqn 2
Dividing eqn 1 and 2
[tex]1=\frac{V}{R}\times \frac{R_{unkown}}{4V}[/tex]
[tex]R_{unknown}=4R[/tex]
So resistance will become 4 times the previous value
Answer:
It should increase to four times its value.
Explanation:
A woman is reported to have fallen 161 ft from a building, landing on a metal ventilator box, which she crushed to a depth of 29 in. She suffered only minor injuries. Ignoring air resistance, calculate:(a) the speed of the woman just before she collided with the ventilator and (b) her average acceleration while in contact with the box. (c) Modeling her acceleration as constant, calculate the time interval it took to crush the box.
Answer:
a) 31.02 m/s
b) - 653.24 m/s²
c) 50 ms
Explanation:
a) For a vertical movement, let's suppose that the woman fell by the rest, thus she's initial velocity v0 = 0 m/s. If she makes a distance (S) of 161 ft = 49.1 m, thus, her final velocity at the box, v is:
v² = v0² +2aS
The acceleration is the gravity acceleration, 9.8 m/s².
v² = 2*9.8*49.1
v² = 962.36
v =√962.36
v = 31.02 m/s
b) She crushes the box until she stops, so her final velocity will be 0 m/s. The initial velocity is now 31.02 m/s, and S = 29 in = 0.7366 m
0 = 31.02² + 2a*0.7366
-1.4732a = 962.36
a = - 653.24 m/s² (the negative signal indicates that she's dessacelaranting)
c) The time can be calculate by:
v = v0 + at
0 = 31.02 -653.24*t
653.24t = 31.02
t = 0.05 s = 50 ms
If the magnitude of a charge is twice as much as another charge, but the force experienced is the same, then the electric field strength of this charge is _____ the strength of the other charge.
Answer:
The new electric field strength of this charge is half of the strength of the other charge.
Explanation:
The electric force acting on the charge particle is given by :
F = q E
[tex]E=\dfrac{F}{q}[/tex]
Where
q is the charged particle
E is the electric field
If the magnitude of a charge is twice as much as another charge, q' = 2q, but the force experienced is the same, then the new electric field is given by :
[tex]E'=\dfrac{F}{q'}[/tex]
[tex]E'=\dfrac{F}{(2q)}[/tex]
[tex]E'=\dfrac{1}{2}\times \dfrac{F}{q}[/tex]
[tex]E'=\dfrac{E}{2}[/tex]
So, the new electric field strength of this charge is half of the strength of the other charge. Hence, this is the required solution.
Final answer:
The electric field strength of the charge with twice the magnitude is half the strength of the electric field of the other charge because the force experienced by both is equal.
Explanation:
If the magnitude of a charge is twice as much as another charge, but the force experienced by each is the same, then the electric field strength that the larger charge is in is half the strength of the other charge’s electric field. The electric field strength (E) at a point is defined as the force (F) experienced by a positive test charge (q) placed at that point divided by the magnitude of the charge itself: E = F/q.
If we have two charges, q1 and q2, where q2 is twice q1 (“q2 = 2 × q1”), and the forces on both charges are equal (“F1 = F2”), we can set up the following equations: E1 = F1/q1 and E2 = F2/q2. Combining our known relationships, we get E1 = F/q1 and E2 = F/(2 × q1), which simplifies to E1 = 2 × E2. Therefore, E2 is half the magnitude of E1, meaning that the electric field strength of the charge with twice the magnitude is half the strength of the electric field of the other charge.