A large truck breaks down out on the road and receives a push back to town by a small compact car.

Pick one of the choices A through F below which correctly describes the forces that the car exerts on the truck and the truck exerts on the car for each of the questions. You may use an answer more than once or not at all.

A. The force exerted by the car pushing against the truck is equal to that exerted by the truck pushing back against the car.

B. The force exerted by the car pushing against the truck is less than that exerted by the truck pushing back against the car.

C. The force exerted by the car pushing against the truck is greater than that exerted by the truck pushing back against the car.

D. The car's engine is running so it exerts a force as it pushes against the truck, but the truck's engine isn't running so it can't exert a force back against the car.

E. Neither the car nor the truck exert any force on each other. The truck is pushed forward simply because it is in the way of the car.

F. None of these descriptions is correct.

If the car and truck are moving at cruising speed, which choice(s) below are true?

1. The car is pushing on the truck, but not hard enough to make the truck move.

2. The car, still pushing the truck, is speeding up to get to cruising speed.

3. The car, still pushing the truck, is at cruising speed and continues to travel at the same speed.

4. The car, still pushing the truck, is at cruising speed when the truck puts on its brakes and causes the car to slow.

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:

1

Explanation:

Answer:

a)   I = √ (2G m³ (1/2r³ - 1/R)), b) I = √ (8 G m³ (1/2r -1/R))

Explanation:

.a) The relation of the Impulse and the moment is

                    I = Δp = m [tex]v_{f}[/tex] - m v₀

We can use Newton's second law with force the force of universal attraction

                  F = ma

                  G m m / r² = m a

                  dv / dt = G m / r²

Suppose re the direction where the spheres move is x

                 dv/dx  dx/dt = G m / x²

                  dv/dx  v = G m / x²

                  v dv = G m dx / x²

We integrate

                   v² / 2 = Gm (-1 / x)

We evaluate this integra from the lower limit v = 0 for x = R to the upper limit, where the spheres v = v and x = 2r are touched

                  v² / 2-0  = G M (-1 / R + 1 / 2r)

                  v = √ [2Gm (1 /2r - 1/ R) ]

The impulse on the sphere is

                 I = m vf - m v₀

                 I = m vf - 0

                 I = m √ (2Gm (1 / 2r-1 / R)

                 I = √ (2G m³ (1/2r³ - 1/R))

b) during the crash each sphere arrives with a velocity v and leaves with a velocity –v, the same magnitude but opposite direction

                      I = m [tex]v_{f}[/tex]- m v₀

                      I = m v - m (-v)

                      I = 2mv

                      I = 2m √ (2Gm (1 / 2r-1 / R)

                      I = √ (8 G m³ (1/2r -1/R))

We calculate the impulse received by each sphere before they make contact and during their elastic collision.

In this scenario, we have two identical hard spheres separated by a distance R. The spheres are released from rest and allowed to collide under the influence of their gravitational attraction. We need to find the magnitude of the impulse received by each sphere before they make contact and during their elastic collision.

(a) Before the spheres make contact, the magnitude of the impulse received by each sphere can be found using the equation:

Impulse = Change in momentum = Mass x Change in velocity.

Since both spheres are released from rest, their initial velocities are zero. Therefore, the change in velocity is the final velocity. Using the equation:

Final velocity = sqrt(2 x G x m / R).

Substituting the values, we can calculate the magnitude of the impulse received by each sphere before they make contact.

(b) During their elastic collision, the magnitude of the impulse received by each sphere can be found using the equation:

Impulse = Change in momentum = Mass x Change in velocity.

Since the spheres collide elastically, there is no change in velocity. Therefore, the magnitude of the impulse received by each sphere during their contact is zero.

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To solve this problem we will apply the linear motion kinematic equations. With the equation describing the position with respect to acceleration (gravity), initial velocity and time, we will find time.

Our values are,

[tex]h = 2.5m[/tex]

[tex]x = 0.9m[/tex]

1) We know that, the time, the ball in the air depends on the maximum height.

Therefore the height is calculated as

[tex]h = v_{yi}t+\frac{1}{2} gt^2[/tex]

There is not initial velocity then replacing,

[tex]2.5 = 0t + \frac{1}{2}9.8t^2[/tex]

[tex]t = \sqrt{\frac{2(2.5)}{9.8}}[/tex]

[tex]t = 0.714s[/tex]

By symmetry, the time taken to reach ground is equal to the time taken to reach the maximum height, then

[tex]T = t+t[/tex]

[tex]T = 2*0.714s[/tex]

[tex]T = 1.426s[/tex]

2 ) In the time t=1.428s, the ball moves horizontally a distance x = 0.9

Then the horizontal velocity,

[tex]v_x = \frac{x}{t} = \frac{0.9m}{1.428s}[/tex]

[tex]v_x = 0.6302m/s[/tex]

3 ) The ball's speed just before the second bounce is found as

[tex]v_y = v_{yi} +gt_{down}[/tex]

Replacing we have

[tex]v_y = 0+(9.8)(0.714)[/tex]

[tex]v_y = 6.9972m/s[/tex]

4 ) The angle of the velocity vector with respect to the ground right after the first bounce is

[tex]\theta = tan^{-1} (\frac{v_y}{v_x})[/tex]

[tex]\theta = tan^{-1} (\frac{6.9972}{0.6302})[/tex]

[tex]\theta = 84.8\°[/tex]

Answer:

[tex]F_1=2.53\ 10^{-4} \ N[/tex]

The net force goes to the right

Explanation:

Electrostatic Force

Let's consider the situation where 2 point charges q1 and q2 are separated by a distance d. An electrostatic force appears between them whose magnitude can be computed by the Coulomb's formula

[tex]\displaystyle F=\frac{k\ q_1\ q_2}{d^2}[/tex]

Where k is the constant of proportionality

[tex]k=9.10^9\ Nw.m^2/c^2[/tex]

Two equally-signed charges repel each other, two opposite-signed charges attract each other.

We need to find the total net force exerted on q1 by q2 and q3. We're assuming the charges are placed to the right of the origin, so the distribution is shown in the figure below.

Since q3 repels q1, its force goes to the right, since q2 attracts q1, its force goes to the right also, thus the total force on q1 is :

[tex]F_1=F_{31}+F_{21}[/tex]

It's directed to the right

Let's compute the individual forces. q3 is separated 2 cm from q1, so d=0.02 m

[tex]\displaystyle F_{31}=\frac{9.10^9\ 4.5\ 10^{-9}\ 1.5\ 10^{-9}}{0.02^2}[/tex]

[tex]F_{31}=0.000151875\ N[/tex]

[tex]\displaystyle F_{21}=\frac{9.10^9\ 3\ 10^{-9}\ 1.5\ 10^{-9}}{0.02^2}[/tex]

[tex]F_{21}=0.00010125\ N[/tex]

[tex]F_1=F_{31}+F_{21}=0.000151875\ N+0.00010125\ N=0.000253125 \ N[/tex]

Expressing the force in scientific notation

[tex]\boxed{F_1=2.53\ 10^{-4}\ N}[/tex]

The net force goes to the right

Solution: The magnitude of the net force acting on q₁ is : Fn = 25.312 [N]

and the direction is in the direction of the positive x axis

The electric force between two charges q₁ and q₂, is according to Coulombs´law as:

F₂₁ = K ×  q₁ × q₂ / d₁₂²

In that equation

F₂₁  is the force exerted by charge q₂ on the charge q₁

It is pretty obvious that F₂₁  = F₁₂, and the force is of rejection if the charges are of the same sign or attraction if they are of opposite signs

Now in this case we calculate F₂₁

F₂₁ = [9×10⁹ × ( 1.5)×10⁻⁹ × (3)×10⁻⁹] /(0.02)²

F₂₁ = 40.5 × 10⁻⁹/ 4 × 10⁻⁴

F₂₁ = 10.125 × 10⁻⁵ [N]

F₂₁  is in the direction of the positive x   ( attraction force)

And

F₃₁ =  [9×10⁹ × ( 1.5)×10⁻⁹ × (4.5)×10⁻⁹] /(0.02)²

F₃₁ = 15.188× 10⁻⁵  [N]

F₃₁ is in the direction of the positive x ( rejection force)

Then the net force is  Fn  = 10.125 + 15.188  

Fn = 25.312 [N]    in direction of x positive

Answer: FR=2.330kN

Explanation:

Write down x and y components.

Fx= FSin30°

Fy= FCos30°

Choose the forces acting up and right as positive.

∑(FR) =∑(Fx )

(FR) x= 5-Fsin30°= 5-0.5F

(FR) y= Fcos30°-4= 0.8660-F

Use Pythagoras theorem

F2R= √F2-11.93F+41

Differentiate both sides

2FRdFR/dF= 2F- 11.93

Set dFR/dF to 0

2F= 11.93

F= 5.964kN

Substitute value back into FR

FR= √F2(F square) - 11.93F + 41

FR=√(5.964)(5.964)-11.93(5.964)+41

FR= 2.330kN

The minimum force is 2.330kN

To determine the magnitude of force F to minimize the resultant FR of three forces, we can use the rule for finding the magnitude of a vector. By taking the square root of the sum of the squares of the components, we can find the magnitude of F.

The question is asking to determine the magnitude of force F so that the resultant FR of the three forces is as small as possible. To find the magnitude of the resultant force, we need to use the rule for finding the magnitude of a vector. We take the square root of the sum of the squares of the components. In this case, we have F = F₁ + F₂, and we can plug in the values given to find the magnitude of F.

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Answer:

A: 5.67*10^-5 N

B: 3.49*10^-5 N

C: -9.16*10^-5 N

Explanation:

[tex]\\A\\F_{A} = F_{BA} + F_{CA} =G*\frac{m_{b}*m_{a}}{r^2_{AB}} + G*\frac{m_{c}*m_{a}}{r^2_{AC}} \\\\= (6.67*10^(-11))*(\frac{363*517}{0.5^2} + \frac{154*363}{0.75^2})\\\\= 5.67*10^-5 N \\\\B\\\\F_{B} = -F_{BA} + F_{CB} =-G*\frac{m_{b}*m_{a}}{r^2_{AB}} + G*\frac{m_{c}*m_{b}}{r^2_{BC}} \\\\= (6.67*10^(-11))*(-\frac{517*363}{0.5^2} + \frac{517*154}{0.25^2})\\\\= 3.49*10^-5\\\\[/tex]

[tex]F_{C} = -F_{CA} - F_{CB} =-G*\frac{m_{b}*m_{c}}{r^2_{BC}} - G*\frac{m_{c}*m_{a}}{r^2_{AC}} \\\\= (6.67*10^(-11))*(-\frac{517*154}{0.25^2} - \frac{154*363}{0.75^2})\\\\= -9.16*10^-5[/tex]

Answer

Pressure, P = 1 atm

air density, ρ = 1.3 kg/m³

a) height of the atmosphere when the density is constant

   Pressure at sea level = 1 atm = 101300 Pa

   we know

   P = ρ g h

   [tex]h = \dfrac{P}{\rho\ g}[/tex]

   [tex]h = \dfrac{101300}{1.3\times 9.8}[/tex]

          h = 7951.33 m

height of the atmosphere will be equal to 7951.33 m

b) when air density decreased linearly to zero.

  at x = 0  air density = 0

  at x= h   ρ_l = ρ_sl

 assuming density is zero at x - distance

 [tex]\rho_x = \dfrac{\rho_{sl}}{h}\times x[/tex]

now, Pressure at depth x

[tex]dP = \rho_x g dx[/tex]

[tex]dP = \dfrac{\rho_{sl}}{h}\times x g dx[/tex]

integrating both side

[tex]P = g\dfrac{\rho_{sl}}{h}\times \int_0^h x dx[/tex]

[tex] P =\dfrac{\rho_{sl}\times g h}{2}[/tex]

 now,

[tex]h=\dfrac{2P}{\rho_{sl}\times g}[/tex]

[tex]h=\dfrac{2\times 101300}{1.3\times 9.8}[/tex]

  h = 15902.67 m

height of the atmosphere is equal to 15902.67 m.

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:

-7 C

Explanation:

Assuming that object other than the styrofoam balls was part of the charge transfer. In order to maintain charge balance, the initial charge of the system must equal the ending charge. If all balls were uncharged initially, the ending charge on the third ball must be:

[tex]C_1+C_2+C_3 = 0\\3+4+C_3=0\\C_3=-7[/tex]

There must be -7 C of charge on the third ball.

Answer:

- Contact 1 with 3 ,  initial charge of 1.8 C.

- contact 1 with 2 and then 1 with 3 , first body should have 3.6 C

Explanation:

The excess charge on a body is distributed evenly throughout the body.

We can have two different configurations:

- Contact 1 with 3

When the third body was touched with the first, the initial charge was distributed between the two, so that when each one separated, it had half the charge, in this configuration the first body should have an initial charge of 1.8 C.

- contact 1 with 2 and then 1 with 3

Another possible configuration of the exercise is that the first body touches the second and the charge decrease to the half and then touches the third where it again decreases by half, so that the first body only gives it every ¼ of its initial load.

Therefore in this configuration if the third body has a load of 0.9C the first body should have 3.6 C

Answer: The answer is 258

Explanation: The distance between the nuclei of the atoms in a diatomic molecule X2 is measured to be 516pm

Therefore the atomic radius of X2 is

516/2=258

Answer: 21.48m

Explanation:

Using the inverse square law, the intensity of sound is inversely proportional to the square of the distance.

Intensity ∝ 1/distance²

let l represent intensity and d represent distanc

l₁/l₂ = d₂²/d₁² ......1

given;

l₁ = 75dB

l₂ = 65dB

d₁ = 20 m

substituting into eqn 1, we have;

75/65 = d₂²/20²

d₂² = 75 × 20²/65

d₂² = 461.54

d₂ = √461.54

d₂ = 21.48m

Answer:

62.64 RPM.

Explanation:

Given that

m= 4.6 g

r= 19 cm

μs = 0.820

μk = 0.440.

The angular speed of the turntable = ω rad/s

Condition just before the slipping starts

The maximum value of the static friction force =Centripetal force

[tex]\mu_s\ m g=m\ \omega^2\ r\\ \omega^2=\dfrac{\mu_s\ m g}{m r}\\ \omega=\sqrt{\dfrac{\mu_s\ g}{ r}}\\ \omega=\sqrt{\dfrac{0.82\times 10}{ 0.19}}\ rad/s\\\omega= 6.56\ rad/s[/tex]

[tex]\omega=\dfrac{2\pi N}{60}\\N=\dfrac{60\times \omega}{2\pi }\\N=\dfrac{60\times 6.56}{2\pi }\ RPM\\N=62.64\ RPM[/tex]

Therefore the speed in RPM will be 62.64 RPM.

Answer:

Horizontal component of velocity will be 61.28 ft/sec

Vertical component of the velocity will be 51.423 ft/sec

Explanation:

We have given velocity of football v = 80 ft/sec'

Angle at which football is thrown [tex]\Theta =40^{\circ}[/tex]

Now we have to fond the horizontal and vertical component of the velocity

Horizontal component of velocity [tex]v_x=vcos\Theta =80\times cos40^{\circ}=80\times 0.766=61.28ft/sec[/tex]

Vertical component of the velocity [tex]v_y=vsin\Theta =80\times sin40^{\circ}=51.423ft/sec[/tex]

Answer:

[tex]49.6\times 10^5\ kg[/tex]

[tex]48.6\times 10^6\ N[/tex]

[tex]80.1\times 10^5\ N[/tex]

[tex]49.6\times 10^5\ kg[/tex]

Explanation:

[tex]1\ slug=14.5939\ kg[/tex]

[tex]340\times 10^3\ slug=340\times 10^3\times 14.5939=4961926\ kg[/tex]

The mass in SI unit is [tex]49.6\times 10^5\ kg[/tex]

Weight would be

[tex]W=mg\\\Rightarrow W=4961926\times 9.81\\\Rightarrow W=48676494.06\ N[/tex]

The weight in SI unit is [tex]48.6\times 10^6\ N[/tex]

[tex]1\ ft/s^2=0.3048\ m/s^2[/tex]

[tex]5.30\ ft/s^2=5.30\times 0.3048=1.61544\ m/s^2[/tex]

[tex]W=mg\\\Rightarrow W=4961926\times 1.61544\\\Rightarrow W=8015693.73744\ N[/tex]

The weight on the moon is [tex]80.1\times 10^5\ N[/tex]

The mass of an object is same anywhere in the universe.

So, the mass of the rocket on Moon is [tex]49.6\times 10^5\ kg[/tex]

The velocity of the object at t=3 seconds is calculated by differentiating the given position function and substituting t=3 into the derived velocity function, resulting in a value of 0.25 m/s.

This problem is about finding the velocity of an object given a position function x(t)=4t²/(t+1), which describes the object's position at any time 't'. To find the velocity at a specific time, you need to take the derivative of the position function, which gives you a velocity function v(t).

The derivative of x(t) with respect to 't' is v(t) = 4t(t+1)-4t²/ (t+1)² = 4/(t+1)². So, when t0=3, the velocity v(3) = 4/(3+1)² = 4/16 = 0.25 m/s.

In conclusion, at t = 3 seconds, the velocity of the object is 0.25 m/s.

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The question seeks to determine the velocity of an object at a specific time given its position function. However, the position function provided in the question appears to contain a typo or error, therefore it is not possible to calculate the velocity without the correct function.

The question involves calculating the velocity of an object at a given time t based on its position function. To find the velocity at time t = 3 seconds, we'll need to take the derivative of the position function. This is a principle from physics dealing with kinematics. However, the function provided, which is x(t) = 4t2t+1, seems to be incorrect or has a typo as it is not usable in its current form. Assuming that the correct position function might have been x(t) = 4t^2/(2t+1), the derivative or velocity function would be v(t) = d/dt [x(t)]. Nevertheless, since we do not have a correct function to work with, we cannot find a specific numerical value for the velocity at t = 3 seconds.

Answer:

4009.21658986 Hz

Explanation:

v = Speed of sound in air = 348 m/s

L = Length of tube = 2.17 cm

Fundamental frequency for a tube where one end is closed and the other end open is given by

[tex]f=\dfrac{v}{4L}\\\Rightarrow f=\dfrac{348}{4\times 2.17\times 10^{-2}}\\\Rightarrow f=4009.21658986\ Hz[/tex]

The fundamental frequency of the canal is 4009.21658986 Hz

Answer:

(a) The x component of the net electrostatic force on charge q3 is 0.195N

(b) The x component of the net electrostatic force on charge q3 is 0.021N

Explanation:

The force of q1 on q3 is a repulsive force and those of q2 and q4 on q3 are attractive forces. From second diagram at the bottom of the page in the attachment below it can be seen that the force of q1 on q3 is directed vertically downward and has on y component while the force of q4 on q3 is directed to the right and has on x component. The force of q2 on q3 has both x and y components. The full solution can be found in the attachment below.

Thank you for reading.

The electric force between two charged bodies at rest is called as electrostatic force.

The values are the following:

The x component of the net electrostatic force = 0.17 N

The y component of the net electrostatic force = 0.045 N

For component X

[tex]= (8.99 \times 10^{9} N.m2/ C^{2} \dfrac{(2.0\times10^{-7} )^2} {(0.050 m)^2}(-2 +\frac{2}{2\sqrt{2} } )\\ = 0.17 N[/tex]

X= 0.17 N

For component Y

Y = 0.045 N.

Hence, X= 0.17 N and Y = 0.045 N.

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Explanation:

Given that,

Mass of the raindrops, [tex]m=0.5\ g=0.0005\ kg[/tex]

Charges, [tex]q=1\ mC=10^{-3}\ C[/tex]

Distance between raindrops, d = 1 cm = 0.01 m

(a) The force due to motion of raindrop is balanced by the electric force between charges. It is given by :

[tex]ma=\dfrac{kq^2}{r^2}[/tex]

a is the acceleration of the raindrop

[tex]a=\dfrac{kq^2}{r^2m}[/tex]

[tex]a=\dfrac{9\times 10^9\times (10^{-3})^2}{(0.01)^2\times 0.0005}[/tex]

[tex]a=1.8\times 10^{11}\ m/s^2[/tex]

(b) It is clear that the acceleration is too large. It can break apart the rain drop.

(c) If the charge is of the factor of [tex]10^{-8}\ C[/tex], the charge would be more reasonable.            

Hence, this is the required solution.  

To find the acceleration of the charged raindrops, we can use Coulomb's Law and Newton's second law. The acceleration is found to be 1.8 x 10^13 m/s^2, which seems unreasonable for raindrops. The assumption of the raindrops acquiring charges of 1.00 mC is likely responsible for this result.

To calculate the acceleration of the two charged raindrops, we can use Coulomb's Law, which states that the force between two charges is given by: F = (k*q1*q2)/r^2, where F is the force, k is the electrostatic constant, q1 and q2 are the charges, and r is the distance between them. The force experienced by each raindrop is equal in magnitude but opposite in direction.

The acceleration of each raindrop can be found using Newton's second law: F = m*a, where F is the net force, m is the mass of the raindrop, and a is the acceleration. Rearranging the equation, we have: a = F/m. Substituting the values into the equations and solving for acceleration, we find: a = (k*q^2)/(m*r^2). Plugging in the values, we have: a = (9 x 10^9 N*m^2/C^2 * (1 x 10^-3 C)^2) / (0.500 g * 10^-3 kg/g * (1 x 10^-2 m)^2). After calculating the values, we find that the acceleration of the two raindrops due to their charges is 1.8 x 10^13 m/s^2.

(b) What is unreasonable about this result? The result seems unreasonable because the acceleration is extremely high. Normally, objects with masses like raindrops would experience much lower accelerations. (c) The premise or assumption responsible for this result is that the raindrops have acquired charges of 1.00 mC, which is an extremely high charge for raindrops and likely not possible in reality.

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Answer:

c. No. An equation may have consistent units but still be numerically invaid.

Explanation:

For an equation to be corrected, it should have consistent units and also be numerically correct.

Most equation are of the form;

(Actual quantity) = (dimensionless constant) × (dimensionally correct quantity)

From the above, without the dimensionless constant the equation would be numerically wrong.

For example; Kinetic energy equation.

KE = 0.5(mv^2)

Without the dimensionless constant '0.5' the equation would be dimensionally correct but numerically wrong.

Answer:

No. An equation may have consistent units but still be numerically invalid.

Explanation:

Final answer:

The minimum possible temperature of the hot reservoir for an engine that does 25 J of work and exhausts 20 J of waste heat in each cycle, with a cold-reservoir temperature of 30°C is 408.87°C.

Explanation:

To find the minimum possible temperature of the hot reservoir for an engine that does 25 J of work and exhausts 20 J of waste heat during each cycle, with a cold-reservoir temperature of 30 °C, we have to use the concept of efficiency and the Carnot engine.

Firstly, let's convert the cold-reservoir temperature from Celsius to Kelvin:

Temperature in Kelvin (K) = Temperature in Celsius (°C) + 273.15

Tc (cold reservoir) = 30°C + 273.15 = 303.15 K

Next, we can calculate the efficiency (ε) of the engine using the formula:

ε = Work done (W) / Heat absorbed (Qh)

Qh (heat absorbed) = W (work done) + Qc (waste heat) = 25 J + 20 J = 45 J

So, ε = 25 J / 45 J = 0.555... (repeating)

The efficiency of a Carnot engine is also given by:

ε = 1 - (Tc/Th)

Now, we solve for Th (the hot reservoir temperature in Kelvin):

Th = Tc / (1 - ε)

Th = 303.15 K / (1 - 0.555...) = 303.15 K / 0.444... = 682.02 K

Finally, we convert the hot reservoir temperature back to Celsius:

Temperature in Celsius (°C) = Temperature in Kelvin (K) - 273.15

Th (hot reservoir) in °C = 682.02 K - 273.15 = 408.87°C

Thus, the minimum possible temperature of the hot reservoir is 408.87°C.

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.

Answer:

binding energy will be [tex]0.1633\times 10^{-15}J[/tex]

Explanation:

We have given wavelength of photon [tex]\lambda =121pm=121\times 10^{-12}m[/tex]

Velocity of light [tex]c=3\times 10^8m/sec[/tex]

Plank's constant [tex]h=6.6\times 10^{-34}Js[/tex]

So energy of photon [tex]E=\frac{hc}{\lambda }=\frac{6.6\times 10^{-34}\times 3\times 10^8}{121\times 10^{-12}}=1.636\times 10^{-15}J[/tex]

Mass of electron [tex]m=9.1\times 10^{-31}kg[/tex]

Velocity of electron is given [tex]v=56.9\times 10^6m/sec[/tex]

So kinetic energy of electron [tex]KE=\frac{1}{2}mv^2=\frac{1}{2}\times 9.1\times 10^{-31}\times (56.9\times 10^6)^2=1.473\times 10^{-15}J[/tex]

So binding energy = plank's energy - kinetic energy

[tex]=1.636\times 10^{-15}-1.473\times 10^{-15}=0.1633\times 10^{-15}J[/tex]

So binding energy will be [tex]0.1633\times 10^{-15}J[/tex]

The binding energy of an electron in an X-ray photoelectron experiment can be calculated by first determining the energy of the incoming photon and the kinetic energy of the ejected electron. The binding energy is then the difference between these two energies.

To calculate the binding energy of an electron in an X-ray photoelectron experiment, we'll require the principles of quantum mechanics and the concept of photoelectric effect. The energy of the incoming photon can be calculated using the equation:

E_photon = h*c/λ

where h is Planck's constant, c is the speed of light, and λ is the photon's wavelength. Converting the wavelength to the correct unit (meters), we get the energy of the photon.

Next, we calculate the kinetic energy of the ejected electron, using the equation:

E_kinetic = 0.5*m*v^2

where m is the mass of the electron, and v is its speed. Again, ensure that the speed is in the correct unit (meters per second).

The binding energy E_b of the electron then is the energy of the photon minus the kinetic energy of the electron:

E_b = E_photon - E_kinetic

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To solve this problem we will make a free body diagram to better understand the displacement measurements made by the body. From there we will apply the linear motion kinematic equations that describe the position of the body in reference to its vertical displacement, acceleration and speed. With this speed found we will apply the energy conservation theorem that will allow us to find the Force.

Equation of trajectory of a projectile is

[tex]y = xtan\theta - x^2 \frac{g}{2u^2cos^2\theta}[/tex]

Here

u = Initial velocity

x = Horizontal displacement

g = Acceleration due to gravity

y = Vertical displacement

We have that

[tex]x = 22m[/tex]

[tex]y = -2\sqrt{2m}[/tex]

Replacing we have that,

[tex]-2\sqrt{2} = 22tan45\° -\frac{9.8}{2u^2cos^2 45}(22)^2[/tex]

[tex]-2\sqrt{2} =22 -\frac{9.8*484}{2u^2(1/2)}[/tex]

[tex]-2\sqrt{2} =22 -\frac{4743.2}{u^2}[/tex]

[tex]u^2 = \frac{4743.2}{22+2\sqrt{2}}[/tex]

[tex]u = 191.03m/s[/tex]

From the work energy theorem

[tex]W_{net} = K_f +K_i[/tex]

Here,

[tex]K_f = \frac{1}{2} mu^2[/tex]

[tex]K_i = \frac{1}{2} m(0)^2 = 0[/tex]

[tex]W_{net} = W_m+W_g[/tex]

Where,

[tex]W_m =\text{Work by man} = F_s[/tex]

[tex]W_{g} = \text{Work by gravity} = -mgh[/tex]

Therefore

[tex]F_s -mgh = \frac{1}{2} mu^2[/tex]

[tex]F_s = \frac{m}{s} (\frac{u^2}{2}+gh)[/tex]

[tex]F_s = \frac{7.26}{2\sqrt{2}}(\frac{191.03}{2}+9.8*2)[/tex]

[tex]F_s = 295.477N[/tex]

The density of a fluid in which an object is submerged can be determined using the original mass of the object, the apparent submerged mass of the object, and the density of the object, using the formula provided above. Applying this formula to a 4.00-kg aluminum ball submersed in a liquid and appearing to be 2.10 kg yields a liquid density of 1247 kg/m^3.

According to Archimedes' Principle, the buoyant force on a submerged object is equal to the weight of the fluid displaced by the object. From this principle, if an object is fully submerged in a fluid then the volume of the fluid displaced is equal to the volume of the object.

To calculate the density of a liquid, using the mass of the object, the apparent mass, and the density of the object, the formula commonly used is:
ρliquid = (mobject - mapparent) / (mobject / ρobject - mapparent / ρobject).

Let's apply this formula to the situation you described. An aluminum ball of mass 4.00 kg is submerged in a liquid and has an apparent mass of 2.10 kg. The density of aluminum is approximately 2700 kg/m3. Therefore, the density of the liquid is (4.00kg - 2.10kg) / (4.00kg / 2700kg/m3 - 2.10kg / 2700kg/m3) = 1247 kg/m3.

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Archimedes' principle states that when a body is submerged in a liquid, it experiences an upward force equal to the weight of the liquid displaced by the object. The magnitude of the buoyant force is equal to the weight of the displaced liquid and is directed upward. The buoyant force can be used to determine the density of an object as well as the density of a liquid. The formula for Archimedes' principle is as follows:

Buoyant force = Weight of displaced fluid = density of fluid x volume of fluid displaced x g

where g is the acceleration due to gravity.

In this case, we can use the formula to determine the density of the liquid as follows:

Buoyant force = Weight of the object - Apparent weight of the object

Density of fluid x Volume of fluid displaced x g = m_object x g - m_apparent x g

Density of fluid = (m_object - m_apparent) / volume of fluid displaced

Now, substituting the given values, we get:

Density of fluid = (4.00 kg - 2.10 kg) / [(4.00 kg - 2.10 kg)/1000 kg/m³]

Density of fluid = 1900 kg/m³

Therefore, the density of the liquid is 1900 kg/m³.

Answer:

[tex]\Delta t =1.31\ s[/tex]

Explanation:

given,

coefficient of kinetic friction, μ = 0.25

Speed of sled at point A = 8.6 m/s

Speed of sled at point B = 5.4 m/s

time taken to travel from point A to B.

we know,

J = F Δ t

J is the impulse

where  F is the frictional force.

t is the time.

we also know that impulse is equal to change in momentum.

[tex]J = m(v_f - v_i)[/tex]

frictional force

F = μ N

where as N is the normal force

now,

[tex]F\Delta t = m(v_f -v_i)[/tex]

[tex]\mu m g \times \Delta t = m(v_f-v_i)[/tex]

[tex]\mu g \times \Delta t = v_f-v_i[/tex]

[tex]\Delta t =\dfrac{v_f-v_i}{\mu g}[/tex]

[tex]\Delta t =\dfrac{8.6-5.4}{0.25\times 9.8}[/tex]

[tex]\Delta t =1.31\ s[/tex]

time taken to move from A to B is equal to 1.31 s

Answer:

Time taken by the sled is 1.31 s

Solution:

As per the question:

Coefficient of kinetic friction, [tex]\mu_{k} = 0.25[/tex]

Velocity at point A, [tex]v_{A} = 8.6\ m/s[/tex]

Velocity at point A, [tex]v_{B} = 5.4\ m/s[/tex]

Now,

To calculate the time taken by the sled to travel from A to B:

According to the impulse-momentum theorem, impulse and the change in the momentum of an object are equal:

Impulse, I = Change in momentum of the sled, [tex]\Delta p[/tex]        (1)

[tex]I = Ft[/tex]                                 (2)

where,

F = Force

t = time

p = momentum of the sled

Force on the sled is given by:

[tex]F = \mu_{k}N[/tex]

where

N = normal reaction force = mg

where

m = mass of the sled

g = acceleration due to gravity

Thus

[tex]F = \mu_{k}mg[/tex]                     (3)

Using eqn (1), (2) and (3):

[tex]\mu_{k}mgt = m\Delta v[/tex]

[tex]\mu_{k}gt = v_{A} - v_{B}[/tex]

[tex]t = \frac{v_{A} - v_{B}}{\mu_{k}g}[/tex]

[tex]t = \frac{8.6 - 5.4}{0.25\times 9.8}[/tex]

t = 1.31 s

To solve this problem we will apply the linear motion kinematic equations, for this purpose we will define the time of each of the sections in which the speed is different. After determining the segments of these speeds we can calculate the average distance and the average speed. Our values are given as

x = 180 km

v = 95 km / h

Speed can be described as the displacement of a body per unit of time, and from that definition clearing the time we would have

[tex]v = \frac{x}{t} \rightarrow t = \frac{x}{v}[/tex]

[tex]v = \frac{180}{95}[/tex]

[tex]v = 1.89473 hrs[/tex]

From the statement we have that the total time is 4.5, then he remaining time is

[tex]t' = T-t =4.5-1.89473 = 2.60526 hrs[/tex]

When it starts to rain there is a phase change in the speed which is given by

[tex]v' = 65km/h[/tex]

Then the distance travel in this velocity

[tex]x' = v ' t '[/tex]

[tex]x' = 65*2.60526[/tex]

[tex]x = 169.34 km[/tex]

(a).  Distance of your hometown from school ,

[tex]x " = x + x '[/tex]

[tex]x'' = 180+169.34[/tex]

[tex]x= 349.34 km[/tex]

(b) The  average speed

[tex]V = \frac{x "}{ T}[/tex]

[tex]v = \frac{349.34 km}{4.5 h}[/tex]

[tex]v = 77.63 km/ h[/tex]

Final answer:

The student's hometown is 350 km away from school, and their average speed for the trip was 77.78 km/h.

Explanation:

To determine how far the student's hometown is from school and their average speed, we need to use their driving speeds and times. Firstly, let's calculate the time taken to travel the first 180 km at 95 km/h.

Time = Distance \/ Speed = 180 km \/ 95 km/h = 1.8947 hours (approximately 1.89 hours).

Since the total travel time is 4.5 hours, the time spent driving in the rain at 65 km/h is 4.5 hours - 1.89 hours = 2.61 hours.

Distance driven in the rain = Speed × Time = 65 km/h × 2.61 hours = 169.65 km (approximately 170 km).

Therefore, the total distance from school to the student's hometown is the sum of both distances: 180 km (before rain) + 170 km (in rain) = 350 km.

Now, to find the average speed, we use the total distance and total time.

Average speed = Total distance \/ Total time = 350 km \/ 4.5 hours = 77.78 km/h.

Thus, the hometown is 350 km away from school, and the student's average speed for the trip was 77.78 km/h.

Answer:

Incomplete question.

Complete question given below

a) 2.7933 C/s

b) 714.45 s

Explanation:

Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150 MW by the radioactive decay of fission products. This heat transfer causes a rapid increase in temperature if the cooling system fails {1 watt = 1 joule/second or 1 W = 1 J/s and 1 MW = 1 megawatt.

(a) Calculate the rate of temperature increase in degrees Celsius per second  if the mass of the reactor core is  1.60105*10^6 kg and it has an average specific heat of  0.3349 kJ/kgº.

(b) How long would it take to obtain a temperature increase of  which could cause some metals holding the radioactive materials to melt?

Part a)

[tex]Q_{rate} = m*c*\frac{dT}{dt}\\\frac{dT}{dt} = \frac{Q_{rate}}{m*c} \\\\\frac{dT}{dt} = \frac{150*10^6}{1.6 * 10^5*334.9} \\\\\frac{dT}{dt} = 2.7993 C/s[/tex]

Part b)

[tex]dt = \frac{dT}{2.7993} \\\\dt = \frac{2000}{2.7993} \\\\dt = 714.45 s[/tex]