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2 years ago

A system gains 767 kJ of heat, resulting in a change in internal energy of the system equal to +151 kJ. How much work is done?

Physics

1 answer:

Crazy boy [7]2 years ago

6 0

Answer:

The work done on the system is -616 kJ

Explanation:

Given;

Quantity of heat absorbed by the system, Q = 767 kJ

change in the internal energy of the system, ΔU = +151 kJ

Apply the first law of thermodynamics;

ΔU = W + Q

Where;

ΔU is the change in internal energy

W is the work done

Q is the heat gained

W = ΔU - Q

W = 151 - 767

W = -616 kJ (The negative sign indicates that the work is done on the system)

Therefore, the work done on the system is -616 kJ

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The angular momentum of a system remains constant when: A. the total kinetic energy is constant

Elenna [48]

Answer:

D. when no torque acts on the system

Explanation:

The angular momentum of a system is defined as:

$L=I\omega$

where

I is the moment of inertia of the system

$\omega$ is the angular velocity

This is the equivalent of the linear momentum, p, for rotational motion

$p=mv$

where m is the mass of the system and v is its velocity. From Newton' laws, we know that the change in linear momentum is equal to net force acting on the system, F:

$F=\frac{\Delta p}{\Delta t}$

The analogous relationship for a rotational motion is:

$\tau=\frac{\Delta L}{\Delta t}$

where $\tau$ is the net torque acting on the system. This is the law of conservation of angular momentum, which states that the rate of change of the angular momentum of an object is equal to the net torque on the system: therefore, when no torque acts on the system, the angular momentum is conserved.

8 0

3 years ago

A rifle of mass 2 kg is suspended by strings. The rifle fires a bullet of mass 0.01 kg at a speed of 200 m/s. The recoil velocit

ozzi

Answer:

option D

Explanation:

given,

mass of the rifle,M = 2 Kg

mass of the bullet , m = 0.01 Kg

speed of the bullet, v = 200 m/s

recoil velocity ,v'= ?

initial velocity of bullet and rifle is zero.

using conservation of energy

Mu + m u' = M v'+ m v

0 + 0 = 2 x v' + 0.01 x 200

2 v' = -2

v' = -1 m/s

negative sign represent the velocity of rifle is in opposite direction of bullet.

hence, the correct answer is option D

5 0

2 years ago

wo ships, one 200200 metres in length and the other 100100 metres in length, travel at constant but different speeds. When trave

shutvik [7]

Answer:

The speed of the faster ship is 22.5 m/s

Explanation:

The length of the ships are;

Ship 1 = 200 m

Ship 2 = 100 m

The time it takes for the ships to completely cross each other when travelling in opposite directions = 10 seconds

The time it takes both ships to cross each other when travelling in the same direction = 25 seconds

Let x represent the speed of the first ship, ship 1, and y represent the speed of the second ship, ship, 2, we have;

(x + y) × 10 = 200 + 100 = 300

10·x + 10·y = 300...(1)

(x - y) × 20 = 200 + 100 = 300

20·x - 20·y = 300...(2)

Multiply equation (1) by 2, to get;

(x + y) × 10 × 2 = 300 × 2

20·x + 20·y = 600...(3)

Adding equation (1) to equation (3) gives;

20·x + 20·y + 20·x - 20·y = 600 + 300

40·x = 900

x = 900/40 = 22.5

x = 22.5

The speed of the first ship, ship 1 = x = 22.5 m/s

From equation (1), we have;

10·x + 10·y = 300

y = (300 - 10·x)/10 = (300 - 10×22.5)/10 = 7.5

y = 7.5

The speed of the second ship, ship 2 = y = 7.5 m/s

Therefore, the faster ship is ship 1 with a speed of 22.5 m/s

7 0

2 years ago

Four identical capacitors are connected with a resistor in two different ways. When they are connected as in part a of the drawi

Nina [5.8K]

Answer:

$T_2 = 0.592$

Explanation:

Given

$T_1 = 1.48s$

See attachment for connection

Required

Determine the time constant in (b)

First, we calculate the total capacitance (C1) in (a):

The upper two connections are connected serially:

So, we have:

$\frac{1}{C_{up}} = \frac{1}{C} + \frac{1}{C}$

Take LCM

$\frac{1}{C_{up}} = \frac{1+1}{C}$

$\frac{1}{C_{up}}= \frac{2}{C}$

Cross Multiply

$C_{up} * 2 = C * 1$

$C_{up} * 2 = C$

Make $C_{up}$ the subject

$C_{up} = \frac{1}{2}C$

The bottom two are also connected serially.

In other words, the upper and the bottom have the same capacitance.

So, the total (C) is:

$C_1 = 2 * C_{up}$

$C_1 = 2 * \frac{1}{2}C$

$C_1 = C$

The total capacitance in (b) is calculated as:

First, we calculate the parallel capacitance (Cp) is:

$C_p = C+C$

$C_p = 2C$

So, the total capacitance (C2) is:

$\frac{1}{C_2} = \frac{1}{C_p} + \frac{1}{C} + \frac{1}{C}$

$\frac{1}{C_2} = \frac{1}{2C} + \frac{1}{C} + \frac{1}{C}$

Take LCM

$\frac{1}{C_2} = \frac{1 + 2 + 2}{2C}$

$\frac{1}{C_2} = \frac{5}{2C}$

Inverse both sides

$C_2 = \frac{2}{5}C$

Both (a) and (b) have the same resistance.

So:

We have:

Time constant is directional proportional to capacitance:

So:

$T\ \alpha\ C$

Convert to equation

$T\ =kC$

Make k the subject

$k = \frac{T}{C}$

$k = \frac{T_1}{C_1} = \frac{T_2}{C_2}$

$\frac{T_1}{C_1} = \frac{T_2}{C_2}$

Make T2 the subject

$T_2 = \frac{T_1 * C_2}{C_1}$

Substitute values for T1, C1 and C2

$T_2 = \frac{1.48 * \frac{2}{5}C}{C}$

$T_2 = \frac{1.48 * \frac{2}{5}}{1}$

$T_2 = \frac{0.592}{1}$

$T_2 = 0.592$

Hence, the time constance of (b) is 0.592 s

8 0

2 years ago

A hockey puck is struck so that it slides at a constant speed and strikes the far side of the rink, 58.2 m away. The shooter hea

DENIUS [597]

Answer:

$v = 33.66 m/s$

Explanation:

Let hockey puck is moving at constant speed v

so here we have

$d = vt$

so time taken by the puck to strike the wall is given as

$t = \frac{58.2}{v}$

now time taken by sound to come back at the position of shooter is given as

$t_2 = \frac{58.2}{340}$

$t_2 = 0.17s$

so we know that total time is 1.9 s

$1.9 = t + t_2$

$1.9 = t + 0.17$

$1.9 - 0.17 = t$

$t = 1.73 s$

now we have

$1.73 = \frac{58.2}{v}$

$v = 33.66 m/s$

7 0

3 years ago