In a crash test, a truck with mass 2100 kg traveling at 22 m/s smashes head-on into a concrete wall without rebounding. The front end crumples so much that the truck is 0.62 m shorter than before.

(a) What is the average speed of the truck during the collision (that is, during the interval between first contact with the wall and coming to a stop)?

vavg = ______ m/s



(b) About how long does the collision last? (That is, how long is the interval between first contact with the wall and coming to a stop?)

Δt = ______ s



(c) What is the magnitude of the average force exerted by the wall on the truck during the collision?

Fwall, avg = ______ N



(d) It is interesting to compare this force to the weight of the truck. Calculate the ratio of the force of the wall to the gravitational force mg on the truck. This large ratio shows why a collision is so damaging.

Fwall, avg / mg =


(e) What approximations were necessary in making this analysis? (Select all that apply.)

Neglect the horizontal component of the force of the road on the truck tires.

Assume a nearly constant force exerted by the wall, so the speed changes at a nearly constant rate.

The deceleration of the truck is approximately equal to g.

Answer:

[tex]8\mu C[/tex]

Explanation:

t = Time taken = [tex]2\mu s[/tex]

i = Current = 3 A

q(0) = Initial charge = [tex]2\mu C[/tex]

Charge is given by

[tex]q=\int_0^t idt+q(0)\\\Rightarrow q=\int_0^{2\mu s} 3dt+2\mu C\\\Rightarrow q=3(2\mu s-0)+2\mu C\\\Rightarrow q=6\mu C+2\mu C\\\Rightarrow q=8\mu C[/tex]

The magnitude of the net electric charge of the capacitor is [tex]8\mu C[/tex]

The total power emitted by a radio transmitter can be calculated by first determining the intensity from the electric field and then multiplying by the surface area of the sphere covered by the isotropic radiation.

The total power emitted by a radio transmitter can be calculated by using the relation between the electric field and the intensity of the radiation. The intensity (I) can be obtained by squaring the electric field strength (E) and dividing it by two times the impedance of free space (Z0). The formula is I = E²/2Z0. After finding the intensity, we can multiply it by the surface area of the sphere which is 4πr² (as it's isotropically radiated, it forms a sphere). The formula becomes Power (P) = I * 4πr². Substituting the given values, we calculate the total power.

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Answer:

25.1%

Explanation:

Let's say m is the true weight and M is the weight of the counterweight.

The original distance between the pans and the center is 25.1 cm / 2 = 12.55 cm.  The fulcrum is then moved 1.4 cm towards the counterweight.

Sum of the moments about the fulcrum:

m (12.55 + 1.4) − M (12.55 − 1.4) = 0

13.95 m − 11.15 M = 0

13.95 m = 11.15 M

M / m = 13.95 / 11.15

M / m = 1.251

The true weight is being marked up by 25.1%.

The true weight of the goods being marked up by the shopkeeper is 25.1%.

A lever is a simple machine made up of a beam or rigid rod that pivots at fulcrum. A lever is a rigid body that can rotate on a single point. The lever is classified into three types based on the locations of the fulcrum, load, and effort.

Let the true mass is m.

and  the reading f the mass is M.

The initial distance between the two pan is 25.1 cm.

So, The initial distance between the pan and the fulcrum is  = 25.1 cm / 2 = 12.55 cm.

The fulcrum is moved 1.4 cm towards the counterweight by the dishonest  shopkeeper.

At equilibrium: the moment of force of two hands from fulcrum is equal. Hence,

mg (12.55 + 1.4)  = Mg (12.55 − 1.4)

13.95 m = 11.15 M

M / m = 13.95 / 11.15

M / m = 1.251

Hence, the true weight is being marked up by = {(M-m)/m}×100%

= {(1.251-1)×100%}

= 25.1%.

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Answer:

[tex]W_{total}=4.5*10^{-8}J[/tex]

Explanation:

Remember that electric potential can be written as:

[tex]V=\frac{kQ}{r}[/tex],

Where V is the electric potential, k is Coulomb's constant, Q is a point charge, and r is the distance from the point charge. Also, we can write the electric potential as:

[tex]V=\frac{W}{q}[/tex],

where W is the work made to move a charge from infinitely far apart to a certain distance, and q the point charge were are moving.

From all this we can get an expression for the work:

[tex]W=\frac{kQq}{r}[/tex]

We are going to let

[tex]q_{1}=-1.00nC\\q_{2}=2.00nC\\q_{3}=3.00nC[/tex]

To take the first charge [tex]q_{1}[/tex] from infinitely far apart to one of the vertices of the triangle, since there is no electric field  and charges, we make no work.

Next, we will move [tex]q_{2}[/tex] . Now, [tex]q_{1}[/tex] is a vertice of the triangle, and we want [tex]q_{2}[/tex] to be 20.cm apart from [tex]q_{1 }[/tex] so the work we need to do is

[tex]W_{12}=\frac{kq_{1}q_{2}}{(0.20)}\\\\W_{12}=\frac{(9*10^{9})(-1*10^{-9})(2*10^{-9})}{0.20}\\\\W_{12}=-9*10^{-8}J[/tex]

Now, we move the last point charge. Here, we need to take in account the electric potential due to [tex]q_{1}[/tex] and [tex]q_{2}[/tex]. So

[tex]W=W_{13}+W_{23}\\\\\\W=\frac{kq_{1}q_{3}}{0.20}+\frac{kq_{2}q_{3}}{0.20}\\\\W=q_{3}k(\frac{q_{1}}{0.20}+\frac{q_{2}}{0.20})\\\\W=(3*10^{-9})(9*10^{9})(\frac{-1*10^{-9}}{0.20}+\frac{2*10^{-9}}{0.20})\\\\[/tex]

[tex]W=1.35*10^{-7}J[/tex]

Now, the only thing left to do is to find the total work, this can be easily done by adding [tex]W_{12}[/tex] and [tex]W[/tex]:

[tex]W_{total}=W_{12}+W\\W_{total}=1.35*10^{-7}-9*10^{-8}\\W_{total}=4.5*10^{-8}J[/tex]

Answer:

The number of excess electrons are on the negative surface is [tex]4.80\times10^{10}\ electrons[/tex]

Explanation:

Given that,

Distance =1.5 cm

Side = 22 cm

Electric field = 18000 N/C

We need to calculate the capacitance in the metal plates

Using formula of capacitance

[tex]C=\dfrac{\epsilon_{0}A}{d}[/tex]

Put the value into the formula

[tex]C=\dfrac{8.85\times10^{-12}\times(22\times10^{-2})^2}{1.5\times10^{-2}}[/tex]

[tex]C=0.285\times10^{-10}\ F[/tex]

We need to calculate the potential

Using formula of potential

[tex]V=Ed[/tex]

Put the value into the formula

[tex]V=18000\times1.5\times10^{-2}\ V[/tex]

[tex]V=270\ V[/tex]

We need to calculate the charge

Using formula of charge

[tex]Q=CV[/tex]

Put the value into the formula

[tex]Q=0.285\times10^{-10}\times270[/tex]

[tex]Q=76.95\times10^{-10}\ C[/tex]

Here, the charge on both the positive and negative  plates

[tex]Q=+76.95\times10^{-10}\ C[/tex]

[tex]Q=-76.95\times10^{-10}\ C[/tex]

We need to calculate the number of excess electrons are on the negative surface

Using formula of number of electrons

[tex]n=\dfrac{q}{e}[/tex]

Put the value into the formula

[tex]n=\dfrac{76.95\times10^{-10}}{1.6\times10^{-19}}[/tex]

[tex]n=4.80\times10^{10}\ electrons[/tex]

Hence, The number of excess electrons are on the negative surface is [tex]4.80\times10^{10}\ electrons[/tex]

To solve this problem we will apply the considerations of equilibrium on a body, we will decompose the tensions into their respective components and apply the sum of forces and solving equations for the given system.

According to the graph, the sum of the horizontal components would be

[tex]\sum T_x = 0[/tex]

[tex]TSin35\° -TSin35\° = 0[/tex]

[tex]0 = 0[/tex]

The sum of the vertical components is

[tex]\sum T_y = 0[/tex]

[tex]2TCos35\° - mg = 0[/tex]

[tex]2TCos35\° = 600N[/tex]

[tex]T = \frac{600N}{2Cos35\°}[/tex]

[tex]T = 366.2N[/tex]

Therefore the net force on the new vine is 366.2N

Answer:

Time period at the moon will be 2.828 sec

Explanation:

Let initially the length of the pendulum is l and acceleration due to gravity is g

And time period is given T = 2 sec

So time period of the pendulum is equal to [tex]T=2\pi \sqrt{\frac{l}{g}}[/tex]-------------eqn 1

Now on the planet acceleration due to gravity [tex]g_{new}=\frac{g}{2}[/tex]

So time period of the pendulum at this planet [tex]T_{new}=2\pi \sqrt{\frac{l}{g_{new}}}[/tex]--------eqn 2

Now dividing eqn 1 by eqn 2

[tex]\frac{T}{T_{new}}=\sqrt{\frac{\frac{g}{2}}{g}}[/tex]

[tex]T_{new}=\sqrt{2}T=1.414\times 2=2.828sec[/tex]

So time period at the moon will be 2.828 sec

If the acceleration due to gravity decreases by a half, the period of a simple pendulum increases by √2, or approximately 1.41. Therefore, the period of the pendulum on the moon of Planet X, where gravity is half, would be 2.8 seconds.

The period of a simple pendulum in physics is influenced by its length and the acceleration due to gravity (g). It is calculated by the formula: T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the gravity. To find the effect of a change in g on the period of the pendulum, we can look at the relationship between them in the formula. Because the formula for the period of a pendulum involves the square root of the gravity, if gravity is reduced by a half, the period will increase by a factor of √2, which is approximately 1.41.

Thus, if the pendulum's period on Planet X was 2.0 s, then on its moon, where gravity is halved, the period will be approximately 2.0 s * 1.41 = 2.8 s.

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Answer:

as soon as they touch the sheet, it returns to its initial position due to the decrease in the load on the rubber ball.

Explanation:

When you rub a foam rod that is an insulator with felt it acquires some loads, when separated these two the barrel is loaded

When approaching the loaded rubber rod, suppose positively, to the aluminum ball that is a metal, the positive charge attracts the free electors of the aluminum until a balance of the electric force and the tension of the wire in which it is hung is established The aluminum foil therefore the aluminum foil approaches the rubber ball, due to the electrostatic attraction, the closer the aluminum wave deflects.

When the ball touches the sheet, the free charges of the metal pass to the insulator neutralizing the existing charges, whereby the electric field and the electric force decrease. Consequently the aluminum sheet descending and moving away from the rubber ball.

 In summary, as soon as they touch the sheet, it returns to its initial position due to the decrease in the load on the rubber ball.

Due do the friction charged develop in the ball, the foil and ball attract.

When the ball and rod contact the charge neutralized and the attraction force decreased, which separate the ball and rod.

When the insulating material are rubbed against each other, the electric charge generate between them.

The effect of friction charge between rod and ball leads to,

Hence due to the friction charged develop in the ball, the foil and ball attract.

When the ball and rod contact the charge neutralized and the attraction force decreased, which separate the ball and rod.

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Answer:

[tex]T=366.23\ N[/tex]

Explanation:

Given:

Now follow the schematic to understand the symmetry and solution via Lami's theorem.

The weight of the monkey will be balanced equally by the tension in both the vines:

Using Lami's Theorem:

[tex]\frac{w}{sin\ 70^{\circ}} =\frac{T}{sin\ 145^{\circ}}[/tex]

[tex]\frac{600}{sin\ 70^{\circ}} =\frac{T}{sin\ 145^{\circ} }[/tex]

[tex]T=366.23\ N[/tex]

Answer:

Option-(D):Thermal energy unchanged, temperature decreased.

Explanation:

Thermal energy and Temperature,T:

Thermal energy of the system is remained unchanged while the temperature,T of the objects present inside the system is more associated with the objects fundamental properties which each matter shows or exhibits in nature.

While the thermal energy is the flow or exchange of thermal or heat energy between the different objects in interaction to each other.

Final answer:

Even if two forces are equal in magnitude and opposite in direction, they can still create a net torque if they are applied at different points that are at different distances from the object's pivot point.

Explanation:

In physics, particularly when discussing Newton's third law, it is essential to recognize that while two forces may be equal in magnitude and opposite in direction, they do not necessarily result in equilibrium if they are acting on different systems. When considering the possibility of a net torque, it is crucial to understand that torque is not just about the magnitude of the force, but also its point of application and the distance from the pivot point (or the axis of rotation).

To produce a torque, a force must be applied in such a manner that it causes the object to rotate around a pivot point. If two equal and opposite forces are applied at different points on an object, and those points are at different distances from the pivot point, then a net torque can occur.

For example, imagine a seesaw with equal forces pushing down on either end. If these forces are applied at different distances from the fulcrum (pivot point), the seesaw will rotate due to the net torque, despite the forces being equal and opposite. This demonstrates that even with equal and opposite forces, the effect on the object's rotational motion depends on the forces' points of application and their distances from the pivot point.

Explanation:

For the given question, distance is given on x-axis and force is given on y-axis. Both distance and force are showing an interval of 5 with starting point 0.

Hence, work in 5 m to 12.5 m is equal and opposite to 12.5 m to 20 m. Since, work is equal then it means that kinetic energy at 5 m will also be equal to the kinetic energy at 20 m.

Therefore, we can conclude that the kinetic energy of the 2 kg object when d equals 20 m is the same as when d is most nearly  5 m.

Final answer:

The velocity of the 2 kg object when d equals 20 m is most nearly 20.4 m/s.

Explanation:

The equation for kinetic energy is K = 1/2mv^2, where K is the kinetic energy, m is the mass, and v is the velocity of the object. To find the velocity when d equals 20m, we can use the equation K = 1/2mv^2 and rearrange it to solve for v. We know the mass of the object is 2kg and the kinetic energy is the same, so we can set up the equation as follows:

1/2(2kg)(v)^2 = 1/2(2kg)(20.4m/s)^2

Simplifying the equation, we find that v is approximately 20.4m/s. Therefore, when d equals 20m, the velocity of the object is most nearly 20.4m/s.

Answer:

none

Explanation:

a. the unit of height does not carry a square( m^2 is unit of area)

b. time cannot be in m/s. it is unit of speed

c. m/s^2 are the units of acceleration

None of the options are correct on the basis of dimensional analysis.

(a)The height of the Transamerica Pyramid is 332 m².

the dimension of height is not correct, it must be m instead of m².

(b)The time duration of a fortnight is 66 m/s.

the dimension of time is s ( seconds) not m/s (meter/second).

(c) The speed of the train is 9.8 m/s².

The dimentsion of speed is m/s (meter/second) not m/s².

Hence, none of the options are correct.

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Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

Answer:

d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

V = [tex]\frac{k*q}{r}[/tex]

We have two charges, q₁ and q₂, and we need to find a point between them, where the electric potential due to them, be zero.

If we call x to the distance from q₁, the distance from q₂, will be the distance between both charges, minus x.

So, we can find the value of x, adding the potentials due to q₁ and q₂, in such a way that both add to zero:

V = [tex]\frac{k*q1}{x}[/tex] +[tex]\frac{k*q2}{(1.63m-x)}[/tex] = 0

⇒[tex]k*q1* (1.63m - x) = -k*q2*x[/tex]:

Replacing by the values of q1, q2, and k, and solving for x, we get:

⇒ x = (2.22 μC* 1.63 m) / 7.89 μC = 0.46 m from q1.

Explanation:

Given:

source temperature T1= 825° C = 1099 K

Sink temperature T2= 15°C = 288 K

Heat Supplied to the engine Qs= 3200 KW

a) Efficiency of a Carnot engine is

[tex]\eta =1-\frac{T_1}{T_2}[/tex]

[tex]=1-\frac{1099}{288}[/tex]

=0.7379

=73.79%

b)

[tex]\eta= \frac{Power delivered}{power recieved}[/tex]

let W be the powe delivered

[tex]0.7379= \frac{W}{3200}[/tex]

W= 2361.408 KW

c) Heat rejected to the cold temperature Qr= Qs-W

= 3200-2361.408=838.60 KW

d) Entropy change in the sink

[tex]\Delta S= \frac{Qr}{T_{sink}}[/tex]

=838.60/288= 2.911 W/K

Answer:

a. [tex]\eta_{th}=0.7377=73.77\%[/tex]

b. [tex]P=2360655.7377\ W[/tex]

c. [tex]Q_o=839344.2622\ W[/tex]

d. [tex]\Delta s=2914.3897\ J.K^{-1}[/tex]

Explanation:

Given:

a.

Since the engine is a Carnot engine:

[tex]\eta_{th}=1-\frac{T_L}{T_H}[/tex]

[tex]\eta_{th}=1-\frac{288}{1098}[/tex]

[tex]\eta_{th}=0.7377=73.77\%[/tex]

b.

Now the power delivered:

since here we are given with the rate of heat transfer

[tex]P=\eta_{th}\times Q_i[/tex]

[tex]P=0.7377\times 3200000[/tex]

[tex]P=2360655.7377\ W[/tex]

c.

rate of heat rejection to the sink:

[tex]Q_o=Q_i-P[/tex]

[tex]Q_o=3200000-2360655.73770491808[/tex]

[tex]Q_o=839344.2622\ W[/tex]

d.

[tex]\Delta s=\frac{Q_o}{T_L}[/tex]

[tex]\Delta s=\frac{839344.2622}{288}[/tex]

[tex]\Delta s=2914.3897\ J.K^{-1}[/tex]

Answer:

A. [tex]\omega=11.1121\ rad.s^{-1}[/tex]

B. [tex]f=1.7685\ Hz[/tex]

C. [tex]T=0.5654\ s[/tex]

Explanation:

Given:

A)

for a spring-mass system the frequency is given as:

[tex]\omega=\sqrt{\frac{k}{m} }[/tex]

[tex]\omega=\sqrt{\frac{56.8}{0.46}}[/tex]

[tex]\omega=11.1121\ rad.s^{-1}[/tex]

B)

frequency is given as:

[tex]f=\frac{\omega}{2\pi}[/tex]

[tex]f=\frac{11.1121}{2\pi}[/tex]

[tex]f=1.7685\ Hz[/tex]

C)

Time period of a simple harmonic motion is given as:

[tex]T=\frac{1}{f}[/tex]

[tex]T=0.5654\ s[/tex]

Answer:

B. QC > 0; QH < 0

Explanation:

Given that there are two reservoir of energy.

Sign convention for heat and work :

1.If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.

2. If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.

From hot reservoir heat is going out that is why it is taken as negative

[tex]Q_H<0[/tex]

From cold reservoir heat is coming inside the reservoir that is why it is taken as positive

[tex]Q_C>0[/tex]

That is why the answer will be

[tex]Q_H<0[/tex] ,[tex]Q_C>0[/tex]

In the context of energy transfers with hot and cold reservoirs, the sign convention is

Given that there are two reservoir of energy. Sign convention for heat and work :

1) If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.

2) If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.

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Answer:

The element X is chlorine.

The element M is Ytterium.

Ytterium (III) chloride is the correct name for [tex]YCl_3[/tex]

Explanation:

Molecules of [tex]X_2, N = 8.92\times 10^{20} [/tex]

Moles of [tex]X_2, n= ?[/tex]

[tex]N=n\times N_A[/tex]

[tex]n=\frac{N}{N_A}=\frac{8.92\times 10^{20}}{6.022\times 10^{23} mol^{-1}}[/tex]

n = 0.001481 mol

Mass of [tex]X_2, m=0.105 g[/tex]

Molar mass of [tex]X_2=M[/tex]

[tex]m=M\times n[/tex]

[tex]M=\frac{m}{n}=\frac{0.105 g}{0.001481 mol}[/tex]

M = 70.87 g/mol

Atomic mass of X = [tex]\frac{70.87 g/mol}{2}=35.435 g/mol[/tex]

X is a chlorine atom.

The compound [tex]MX_3[/tex] consists of 54.47% X by mass.

M' = Molar mass of compound [tex]MX_3[/tex]

Percentage of X in compound = 54.47%

[tex]54.47\%=\frac{3\times 35.435 g/mol}{M'}\times 100[/tex]

M' = 195.16 g/mol

Atomic mass of M = a

[tex]195.16 g/mol=a+3\times 35.435 g/mol[/tex]

[tex]a = 195.16 g/mol-3\times 35.435 g/mol[/tex]

a = 88.86 g./mol

The element M is Ytterium.

Ytterium (III) chloride is the correct name for [tex]YCl_3[/tex]

The element X is Chlorine and the element M is Ytterium.

Ytterium Chloride is the correct name for [tex]MX_3[/tex] that is [tex]YCl_3[/tex]

Number of molecules of [tex]X_2={8.92*10^{20}[/tex]

Number of moles of X:

[tex]n=\frac{8.92*10^{20}}{6.023*10^{23}}[/tex]  

n = 0.001481 mol = 0.105g

Molar mass of [tex]X_2[/tex] molecule

[tex]M= 0.105/0.001481\\\\M=70.87g/mol[/tex]

Therefore molar mass of X atom = molar mass of [tex]X_2[/tex]/2

m = 35.435g/mol

X is a chlorine atom.

Given that the compound  consists of 54.47% X by mass.

Let M' = Molar mass of compound [tex]MX_3[/tex]

Percentage of X in compound = 54.47%

54.47 = {amount of X}/{mola mass of compound}*100

[tex]54.47 = \frac{3*35.435}{M_{'} }*100\\\\M^{'}= 195.16g/mol[/tex]

let Atomic mass of M = A

A = 195.16 - 3 × 35.435

A = 88.86 g/mol

The element M is Ytterium.

Therefore the compound [tex]MX_3[/tex] is Ytterium Chloride [tex]YCl_3[/tex]

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Answer:

The speed of electron is [tex]v=4.52\times 10^6\ m/s[/tex] and the speed of proton is 2468.02 m/s.

Explanation:

Given that,

Electric field, E = 560 N/C

To find,

The speed of each particle  (electrons and proton) 46.0 ns after being released.

Solution,

For electron,

The electric force is given by :

[tex]F=qE[/tex]

[tex]F=1.6\times 10^{-19}\times 560=8.96\times 10^{-17}\ N[/tex]

Let v is the speed of electron. It can be calculated using first equation of motion as :

[tex]v=u+at[/tex]

u = 0 (at rest)

[tex]v=\dfrac{F}{m}t[/tex]

[tex]v=\dfrac{8.96\times 10^{-17}}{9.1\times 10^{-31}}\times 46\times 10^{-9}[/tex]

[tex]v=4.52\times 10^6\ m/s[/tex]

For proton,

The electric force is given by :

[tex]F=qE[/tex]

[tex]F=1.6\times 10^{-19}\times 560=8.96\times 10^{-17}\ N[/tex]

Let v is the speed of electron. It can be calculated using first equation of motion as :

[tex]v=u+at[/tex]

u = 0 (at rest)

[tex]v=\dfrac{F}{m}t[/tex]

[tex]v=\dfrac{8.96\times 10^{-17}}{1.67\times 10^{-27}}\times 46\times 10^{-9}[/tex]

[tex]v=2468.02\ m/s[/tex]

So, the speed of electron is [tex]v=4.52\times 10^6\ m/s[/tex] and the speed of proton is 2468.02 m/s. Therefore, this is the required solution.

Answer:

1.897825 seconds

Explanation:

L = Length of pendulum = 3.58 m

g = Acceleration due to gravity = 9.81 m/s²

The amplitude of a pendulum is given by

[tex]T=2\pi \sqrt{\dfrac{L}{g}}\\\Rightarrow T=2\pi\sqrt{\dfrac{3.58}{9.81}}\\\Rightarrow T=3.79565\ s[/tex]

It can be seen that the time period does not depend on amplitude

Time period includes the time to go from extreme left to right and back to extreme left.

So, time taken to go from extreme right to left is [tex]\dfrac{T}{2}=\dfrac{3.79565}{2}=1.897825\ s[/tex]

The period of a simple pendulum is determined by its length and the acceleration due to gravity. It is independent of other factors such as mass and amplitude.

The period of a simple pendulum is determined by its length and the acceleration due to gravity. The period is independent of other factors such as mass and amplitude. For a small amplitude of less than 15 degrees, the period of a simple pendulum can be calculated using the formula:

T = 2π√(L/g)

where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.

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The force constant of the spring is 1.932 N/m. The maximum mass of the glider can be calculated using Newton's second law. The magnitude of the maximum acceleration of the glider can be calculated using the equation for acceleration in SHM.

In simple harmonic motion (SHM), the force exerted by an ideal spring is directly proportional to the displacement from its equilibrium position. Let's calculate the force constant of the spring by using Hooke's Law. We know that the force exerted by the spring is given by F = kx, where F is the force, k is the force constant, and x is the displacement. At the maximum displacement, the force exerted is F = k * 2.05 m. At the minimum displacement, the force exerted is F = k * 1.06 m. Equating these two forces, we get k * 2.05 = k * 1.06. Solving this equation, we find the force constant of the spring to be k = 2.05 / 1.06 = 1.932 N/m.

The maximum mass of the glider can be found by calculating the maximum force that can be exerted by the spring. The maximum force is given by F = k * x, where F is the force, k is the force constant, and x is the maximum displacement. Plugging in the values, we get F = 1.932 N/m * 2.05 m = 3.961 N. Using Newton's second law, F = m * a, where F is the force, m is the mass, and a is the acceleration, we can calculate the maximum mass. Rearranging the equation gives m = F / a. Since the acceleration in SHM is given by a = 4π^2 * x / T^2, where a is the acceleration, x is the amplitude, and T is the period, we can plug in the given values to find the maximum mass.

The magnitude of the maximum acceleration of the glider can be found by using the equation a = 4π^2 * x / T^2, where a is the acceleration, x is the amplitude, and T is the period. Plugging in the given values, we can calculate the magnitude of the maximum acceleration.

Answer:

New force between them will become [tex]\frac{1}{36}[/tex] times

Explanation:

Let the charge on both the object are [tex]q_1\ and\ q_2[/tex] and distance between them is is given 1 cm

So r = 1 cm = 0.01 m

Electric force between them is given F

According to Coulomb's between two charges is given by

[tex]F=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^2}=\frac{Kq_1q_2}{r^2}[/tex]

According to question [tex]F=\frac{Kq_1q_2}{0.01^2}=\frac{Kq_1q_2}{10^{-4}}[/tex]-----------eqn 1

Now distance is increased by 5 cm so new distance = 5+1 = 6 cm = 0.06 m

So new force [tex]F_{new}=\frac{Kq_1q_2}{0.06^2}=\frac{Kq_1q_2}{36\times 10^{-4}}[/tex]------------------eqn 2

Comparing eqn 1 and eqn 2

[tex]F_{new}=\frac{F}{36}[/tex]

The main reasons for having few impact craters on Earth compared to moon is due to the presence of atmosphere, water bodies and tectonic activities.

Explanation:

Impact craters are formed on the surface of any object when another very high velocity object hits the surface of the particular object. As force exhibited by any object is directly proportional to the product of mass and acceleration of the object.

In case of moon or other natural satellites, there are no means of reducing the acceleration or mass of the incident high velocity object. But in earth, the presence of different layers of atmosphere causes sufficient friction to reduce the size as well as acceleration of the impeding high velocity object.

Then the presence of larger volume of water bodies on the Earth's surface reduces the chance of hitting on land areas. The third important factor reducing the impact craters are the variation in tectonic plates which helps in keeping only recent impacts on the Earth's surface and erase the old impacts.

So the presence of different layers of atmosphere, water bodies and tectonic plats are the main reasons for having fewer impact craters on Earth compared to the Moon.

The voltage across the capacitor is 20.79 V.

Capacitance is a measure of how much charge can be stored in a capacitor per unit potential difference. The voltage across the capacitor can be calculated using the formula V = Q/C, where V is the voltage, Q is the charge, and C is the capacitance. In this case, the charge is 1.85×10-7 C and the capacitance is 8900 pF. Converting the capacitance to farads, we get C = 8900 × 10-12 F. Plugging in these values, we have V = 1.85×10-7 C / (8900 × 10-12 F), which simplifies to V = 20.79 V.

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To solve this problem we will start from the constants given on the two celestial bodies, that is, the earth and the moon. From there we will seek the balance of forces, where the gravitational force is zero. That distance will remain incognito and will be the one we will seek to find, this is

Given that,

[tex]M_E =5.97*10^24 kg[/tex]

[tex]m_m = 7.36*10^22 kg[/tex]

[tex]r_E = 6370 km[/tex]

[tex]r_m = 1737 km[/tex]

The distance between Earth and Moon

[tex]x = 384403*10^3 m[/tex]

As the net gravitational force is zero we have that

[tex]\frac{GMm'}{d^2} = \frac{Gmm'}{(x -d)^2}[/tex]

[tex]\frac{M}{d^2} = \frac{m}{(x -d)^2}[/tex]

Replacing with our values

[tex]\frac{(5.97*10^{24} )}{d^2} = \frac{(7.36*10^{22})}{(x -d)^2}[/tex]

[tex]\frac{(x- d)^2}{d^2}= 0.0123[/tex]

[tex]x^2 + d^2 - 2xd = 0.0123 d^2[/tex]

Replacing x as the distance betwen moon and earth,

[tex](384403 x 10^3)^2+d^2-2(384403 x 10^3)d=0.0123 d^2[/tex]

Solving the polynomial we have that

[tex]d =4.3387*10^8 \text{ or } 3.4506*10^{8}[/tex]

So the minimum value is [tex]3.4506*10^{8}m[/tex]

Final answer:

The acceleration due to gravity at the altitude of the Columbia shuttle is approximately 8.95 m/s^2, which is about 91% of Earth's surface gravity.

Explanation:

To determine the acceleration due to gravity at the altitude at which Columbia is orbiting, we apply the universal law of gravitation, which is represented by the formula:

F = G Mm/r²

From this formula, we can derive the expression for acceleration due to gravity (g):

g = GM/r²

Given:

Radius of Earth (rE) = 6.4 × 106 m

Altitude of Columbia (h) = 0.3 × 106 m

Mass of Earth (M) = 6.0 × 1024 kg

Gravitational constant (G) = 6.67408 × 10-11 m3kg-¹s-²

Total distance from the center of Earth (r) = rE + h

The resulting equation for the acceleration due to gravity at the altitude of Columbia is:

g = (6.67408 × 10-11 m3kg-1s-2) × (6.0 × 1024 kg) / (6.7 × 106 m)²

Upon calculating, this yields an acceleration due to gravity of approximately 8.95 m/s². This acceleration is about 91% of the gravitational acceleration at Earth's surface, which is 9.8 m/s².

Assuming the centripetal force is provided by the Coulomb attraction between the electron and the proton we have that,

[tex]F_c = k \frac{e^2}{R^2}[/tex]

Here,

k = Coulomb's Constant

R = Distance

e = Electron charge

And by the centripetal force,

[tex]F_c = m_e R\omega^2[/tex]

[tex]m_e[/tex] = Mass of electron

R = Radius

[tex]\omega[/tex] = Angular velocity

Equation both expression,

[tex]k \frac{e^2}{R^2} = m_e R\omega^2[/tex]

Replacing,

[tex]R^3 = \frac{(9*10^9)(1.6*10^{-19})^2}{(9.11*10^{-31})(10^{6})^2}[/tex]

[tex]R^3 = 2.52908*10^{-10} m[/tex]

[tex]R = 6.32394*10^{-4}m[/tex]

Therefore the radius of the electron orbit is [tex]6.32394*10^{-4}m[/tex]

Answer:

The time it would take to sail around the world in a sailboat is 1000 hour.        

Explanation:

It is given that,    

The circumference of the Earth, [tex]d=10^4\ mile[/tex]

The speed of a sailboat, [tex]v=10\ mile/hour[/tex]

We need to find the time it would take to sail around the world in a sailboat. It can be calculated using the definition of velocity. It is given by :

[tex]v=\dfrac{d}{t}[/tex]

[tex]t=\dfrac{d}{v}[/tex]

[tex]t=\dfrac{10^4\ mile}{10\ mile/hour}[/tex]

t = 1000 hour

So, the time it would take to sail around the world in a sailboat is 1000 hour. Hence, this is the required solution.

Final answer:

The percent uncertainty of a blood pressure measurement of 120 mm Hg with an uncertainty of ±2 mm Hg is 1.67%. Using this percentage, the uncertainty in a blood pressure measurement of 71 mm Hg would be ±1.2 mm Hg, after rounding to one decimal place.

Explanation:

To calculate the uncertainty in a blood pressure measurement of 71 mm Hg, given the percent uncertainty is the same as that for a measurement of 120 mm Hg with an uncertainty of ± 2 mm Hg, we first find the percent uncertainty of the original measurement.

The percent uncertainty is calculated by dividing the uncertainty by the actual measurement and then multiplying by 100 to get a percentage:

Percent Uncertainty = (Uncertainty / Measurement) × 100

For the measurement of 120 mm Hg, the percent uncertainty is:

(2 mm Hg / 120 mm Hg) × 100 = 1.67%

Assuming the same percent uncertainty for a blood pressure measurement of 71 mm Hg, you would use the same percentage to find the uncertainty:

Uncertainty = Percent Uncertainty × Measurement / 100

Uncertainty in 71 mm Hg = 1.67% × 71 mm Hg / 100 = 1.187 mm Hg

Since we typically round to the nearest whole unit or significant digit the uncertainty would be ± 1.2 mm Hg (rounded to one decimal place).

Answer:

π*R²*E

Explanation:

According to the definition of electric flux, it can be calculated integrating the product E*dA, across the surface.

As the electric field E is uniform and parallel to the hemisphere axis,  and no charge is enclosed within it, the net flux will be zero, so, in magnitude, the flux across the opening defining the hemisphere, must be equal to the one across the surface.

The flux across the open surface can be expressed as follows:

[tex]\int\ {E} \, dA = E*A = E*\pi *R^{2}[/tex]

As E is constant, and parallel to the surface vector dA at any point, can be taken out of the integral, which is just the area of the surface, π*R².

⇒Flux = E*π*R²

Final answer:

The electric flux through a hemisphere surface is calculated using Gauss's Law, where the electric field lines are parallel to the area element vector on the surface. The flux through a closed spherical surface is independent of its radius. Choosing a Gaussian surface with appropriate symmetry aids in evaluating the flux integral.

Explanation:

The magnitude of the electric flux through the hemisphere surface can be calculated using Gauss's Law. For a hemispherical surface of radius R in a uniform electric field,