Questions 15 out of 20

During an isothermal process, 10 j of heat is removed from an ideal gas. What is the work done by the gas in the process?

Black_prince [1.1K]

The work done in the isothermal process is 10 joule.

We need to know about the isotherm process to solve this problem. The isotherm process can be described as a process where the initial temperature system will be the same as the final temperature. Hence, the internal energy change will be zero.

ΔU = 0

Hence,

ΔU = Q - W

0 = Q - W

Q = W

It means that the heat transferred is the same as the work done.

From the question above, we know that the heat transferred is 10 joule. Thus, the work done in the isothermal process is 10 joule.

Find more on isothermal at: brainly.com/question/17097259

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8 0

2 years ago

Please do all of i will give you brainlest and thanks to best answer

vlabodo [156]

Answer:the answer is A

Explanation:

3 0

3 years ago

In February 1955, a paratrooper fell 370 m from an airplane without being able to open his chute but happened to land in snow, s

nevsk [136]

a) 0.94 m

The work done by the snow to decelerate the paratrooper is equal to the change in kinetic energy of the man:

$W=\Delta K\\-F d = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

where:

$F=1.1 \cdot 10^5 N$ is the force applied by the snow

d is the displacement of the man in the snow, so it is the depth of the snow that stopped him

m = 68 kg is the man's mass

v = 0 is the final speed of the man

u = 55 m/s is the initial speed of the man (when it touches the ground)

and where the negative sign in the work is due to the fact that the force exerted by the snow on the man (upward) is opposite to the displacement of the man (downward)

Solving the equation for d, we find:

$d=\frac{1}{2F}mu^2 = \frac{(68 kg)(55 m/s)^2}{2(1.1\cdot 10^5 N)}=0.94 m$

b) -3740 kg m/s

The magnitude of the impulse exerted by the snow on the man is equal to the variation of momentum of the man:

$I=\Delta p = m \Delta v$

where

m = 68 kg is the mass of the man

$\Delta v = 0-55 m/s = -55 m/s$ is the change in velocity of the man

Substituting,

$I=(68 kg)(-55 m/s)=-3740 kg m/s$

7 0

3 years ago

When a guitar string plays the note "a," the string vibrates at 440 hz ?

Rama09 [41]

Yes, that's correct. The note "A" (which is used to tune the other strings of the guitar) corresponds to a frequency of 440 Hz.

8 0

3 years ago

A ball is thrown horizontally from the top of a building 21.8 m high. The ball strikes the ground at a point 101 m from the base

riadik2000 [5.3K]

Answer:

t=2.10 s

u= 47.40 m/s

Explanation:

given that

h= 21.8 m

x= 101 m

g=9.8 m/s²

Lets take horizontal speed of ball = u m/s

The vertical speed of the car at initial condition is zero ( v= 0).

We know that

$h=vt+\dfrac{1}{2}gt^2$

v= 0 m/s

$h=\dfrac{1}{2}gt^2$

now by putting the values

21.8 = 1/2 x 9.8 x t²

t=2.10 s

This is time when ball was in motion.

Now in horizontal direction

x = u .t

101 = u x 2.1

u= 47.40 m/s

6 0

3 years ago