Two bodies, one hot and the other cold kept in vacuum.what will happen to the tempreture of bodies after some time.


Answer 1
Answer: Hot body will lose heat from it, and that heat will goes out from it through radiation, so it's temperature will decrease after some time.

In same manner, cold body will take the heat, and it's temperature will increase

Hope this helps!

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A rock is thrown vertically upward from some height above the ground. It rises to some maximum height and falls back to the ground. What one of the following statements is true if air resistance is neglected? The acceleration of the rock is zero when it is at the highest point. The speed of the rock is negative while it falls toward the ground. As the rock rises, its acceleration vector points upward. At the highest point the velocity is zero, the acceleration is directed downward. The velocity and acceleration of the rock always point in the same direction.

Garza travels at a speed of 5 m/s. How long will it take him to travel 640 m?



128 s


The distance, speed and time are related as;

Distance=Speed* Time

Given that the speed = 5 m/s

Distance = 640 m

Time = ?


Distance=Speed* Time

640\ m=5\ m/s* Time

Time=\frac {640\ m}{5\ m/s}=128\ s

Thus, Garza takes 128 s to travel 640 m at 5 m/s speed.

A uniform 190 g rod with length 43 cm rotates in a horizontal plane about a fixed, vertical, frictionless pin through its center. Two small 38 g beads are mounted on the rod such that they are able to slide without friction along its length. Initially the beads are held by catches at positions 10 cm on each sides of the center, at which time the system rotates at an angular speed of 12 rad/s. Suddenly, the catches are released and the small beads slide outward along the rod. Find the angular speed of the system at the instant the beads reach the ends of the rod. Answer in units of rad/s.



The angular speed of the system at the instant the beads reach the ends of the rod is 14.87 rad/s


Moment of inertia is given as;

I = ¹/₁₂×ML² + 2mr²


I is the moment of inertia

M is the mass of the rod = 0.19 kg

L is the length of the rod = 0.43 m

m is the mass of the bead = 0.038 kg

r is the distance of one bead

Initial moment of inertial is given as;

I_i = (1)/(12)ML^2 +2mr_1^2

Final moment of inertia is also given as

I_f= (1)/(12)ML^2 +2mr_2^2

Angular momentum is the product of angular speed and moment of inertia;

= Iω

From the principle of conservation of angular momentum;

I_i \omega_i = I__f } \omega_f

((1)/(12)ML^2 +2mr_1^2) \omega_i =  ((1)/(12)ML^2 +2mr_2^2) \omega_f


ωi = 12 rad/s

r₁ = 10.0 cm = 0.1 m

r₂ = 10.0cm/4 = 2.5 cm = 0.025 m

Substitute these values in the above equation, we will have;

((1)/(12)*0.19*(0.43)^2 +2*0.038(0.1)^2) 12 =  ((1)/(12)*0.19*(0.43)^2 +2*0.038*(0.025)^2) \omega_f\n\n0.04425 =0.002975\ \omega_f\n\n\omega_f = (0.04425)/(0.002975) = 14.87\ rad/s

Therefore, the angular speed of the system at the instant the beads reach the ends of the rod is 14.87 rad/s

Four +2 µC point charges are at the corners of a square of side 2 m. Find the potential at the center of the square (relative to zero potential at infinity) for each of the following conditions.(a) All the charges are positive(b) Three of the charges are positive and one is negative(c) Two are positive and two are negative



(a) 51428.59 J/C

(b) 25714.29 J/C

(c) 0 J/C


Parameters given:

Q1 = 2 * 10^-6 C

Q2 = 2 * 10^-6 C

Q3 = 2 * 10^-6 C

Q4 = 2 * 10^-6 C

=> Q1 = Q2 = Q3 = Q4 = Q

Side of the square = 2m

The center of the square is the midpoint of the diagonals, i.e. Using Pythagoras theorem:

BD² = 2² + 2²

BD² = 8

BD = √(8) = 2.8m

OD = 1.4m

(The attached diagram explains better)

Hence, the distance between the center and each point charge, r, is 1.4m.

Electric Potential, V = kQ/r

k = Coulombs constant

(a) If all charges are positive:

V(Total) = V1 + V2 + V3 + V4

V1 = Potential due to Q1

V2 = Potential due to Q2

V3 = Potential due to Q3

V4 =Potential due to Q4

Since Q1 = Q2 = Q3 = Q4 = Q

=> V1 = V2 = V3 = V4

=> V(Total) = 4V1

V  = (4 * 9 * 10^9 * 2 * 10^-6)/1.4

V = 51428.59J/C

(b) If 3 charges are positive and 1 is negative:

Since Q1 = Q2 = Q3 = Q

and Q4 = -Q

The total potential becomes:

V(Total) = V1 + V2 + V3 - V4

Since V1, V2, V3 and V4 have the same value,

V(Total) = V1 + V2

V(Total) = 2V1

V(Total) = (2 * 9 * 10^9 * 2 * 10^-6)/1.4

V(Total) = 25174.29 J/C

(c) Two charges are positive and two are negative:

Since Q1 = Q2  = Q

and Q3 = Q4 = -Q

The total potential becomes:

V(Total) = V1 + V2 - V3 - V4

Since V1, V2, V3 and V4 have the same value,

V(Total)  = 0 J/C

Ezra (m = 20.0 kg) has a tire swing and wants to swing as high as possible. He thinks that his best option is to run as fast as he can and jump onto the tire at full speed. The tire has a mass of 10.0 kg and hangs 3.50 m straight down from a tree branch. Ezra stands back 10.0 m and accelerates to a speed of 3.62 m/s before jumping onto the tire swing. (a) How fast are Ezra and the tire moving immediately after he jumps onto the swing? m/s (b) How high does the tire travel above its initial height?



a) v=5.6725\,m.s^(-1)

b) h= 1.6420\,m



  • mass of the body, M=20\,kg
  • mass of the tyre,m=10\,kg
  • length of hanging of tyre, l=3.5m
  • distance run by the body, d=10m
  • acceleration of the body, a=3.62m.s^(-2)


Using the equation of motion :



v=final velocity of the body

u=initial velocity of the body

here, since the body starts from rest state:


putting the values in eq. (1)

v^2=0^2+2* 3.62 * 10


Now, the momentum of the body just before the jump onto the tyre will be:


p=20* 8.5088


Now using the conservation on momentum, the momentum just before climbing on the tyre will be equal to the momentum just after climbing on it.

(M+m)* v'=p

(20+10)* v'=170.1764



Now, from the case of a swinging pendulum we know that the kinetic energy which is maximum at the vertical position of the pendulum gets completely converted into the potential energy at the maximum height.


(1)/(2) (M+m).v'^2=(M+m).g.h

(1)/(2) (20+10)* 5.6725^2=(20+10)* 9.8* h

h\approx 1.6420\,m

above the normal hanging position.

How much heat is required to convert 0.3 kilogram of ice at 0°C to water at the same temperature? A. 334,584 J B. 167,292 J C. 100,375 J D. 450,759 J



Option C is the correct answer.


Heat required to melt solid in to liquid is calculated using the formula

            H = mL, where m is the mass and L is the latent heat of fusion.

Latent heat of fusion for water = 333.55 J/g

Mass of ice = 0.3 kg = 300 g

Heat required to convert 0.3 kilogram of ice at 0°C to water at the same temperature

          H = mL = 300 x 333.55 = 100,375 J

Option C is the correct answer.

Convert 56km/h to m/s.​



15.556 metres per second