Some amount of ideal gas with internal energy U and initial temperature 1000C was compressed to half of volume meanwhile absolute pressure inside of a container increased twice. We can say that internal energy of this gas after compression in terms of U is (20.2, 20.1, 19.4, 19.5) Group of answer choices

Answers

Answer 1

Answer:

Answer:

U. With no variation.

Explanation:

Note- since temperature remains constant when pressure becomes twice and volume becomes half, and internal energy of ideal gas is function of only temperature so it remains constant. The internal energy is independent of the variables stated in the exercise.

Related Questions

Carol is farsighted ( presbyopia) and cannot see objects clearly that are closer to her eyes than about meter. She sees objects clearly with a relaxed eye when they are distant. What is the refractive power of reading glasses that would allow her to read a book 50 cm away with a relaxed eye
Is the magnet in a compass a permanent magnet or an electromagnet?
The Richter scale is used to determine how strong the earthquake is (magnitude) of the earthquake. It touches the ground and feels the earth shaking. With the momentum of the earth shaking the device the needle on the device moves creating a wave looking line. According to the theory of plate tectonics, what happens at transform, divergent and convergent boundaries? On which of these boundary types would a volcano most likely take place, and why?
A 500-gram mass is attached to a spring and executes simple harmonic motion with a period of 0.25 second. If the total energy of the system is 4J, find the force constant of the spring?
Power P is the rate at which energy E is consumed per unit time. Ornithologists have found that the power consumed by a certain pigeon flying at velocity v m/s is described well by the function P(v)=16v−1+10−3v3 J/s. Assume that the pigeon can store 5×104 J of usable energy as body fat. Find the velocity vp min that minimizes power consumption. (Use decimal notation. Give your answer to two decimal places.)

In a collision between two objects having unequal masses, how does magnitude of the impulse imparted to the lighter object by the heavier one compare with the magnitude of the impulse imparted to the heavier object by the lighter one?a. the lighter object receives a larger impulse

b. the heavier object receives a larger impulse

c. both objects receive the same impulse

d. the answer depends on the ratio of the masses

e. the answer depends on the ratio of the speeds

Answers

Answer:

Explanation:

The impulse has as an expression

I = ∫ F dt

In addition, on impulse of isolated systems is

I =ΔP = m vf - m v₀

Let's replace

F dt = m (vf -vo)

The force in these shocks laqueunxcyerpoesjerce on the other, using the law of action and reaction state forces has the same magnitude and the time of the shock is equal for the two bodies, this implies that the impulse is equal for the two bodies

Let's check the answers

a) False as the forces are of action and reaction the impulse must be equal

b) False

c) True. Why do we have an action and reaction stop

d False

e) false

Final answer:

In collision between two objects, despite difference in their masses, both objects receive the same impulse, as per Newton's third law.

Explanation:

In a collision between two objects with unequal masses, according to Newton's third law of motion, 'For every action, there is an equal and opposite reaction', the magnitude of the impulse imparted to the lighter object by the heavy object is exactly the same as the magnitude of the impulse imparted to the heavier object by the lighter one.

This is regardless of the masses or speeds of the objects involved. The direction of these impulses will be opposite, but their magnitudes will be the equal. So, option c. 'both objects receive the same impulse' is correct.

Learn more about Impulse during Collision here:

brainly.com/question/31678431

#SPJ6

A typical laboratory centrifuge rotates at 3700 rpm . Test tubes have to be placed into a centrifuge very carefully because of the very large accelerations. Part A What is the acceleration at the end of a test tube that is 10 cm from the axis of rotation

Answers

Answer:

Explanation:

acceleration of test tube

= ω² R

= (2πn)² R

= 4π²n²R

n = no of rotation per second

= 3700 / 60

= 61.67

R = .10 m

acceleration

= 4π²n²R

= 4 x 3.14² x 61.67² x .10

= 14999 N Approx

Please use Gauss’s law to find the electric field strength E at a distance r from the center of a sphereof radius R with volume charge density ???? = cr 3 and total charge ????. Your answer should NOT contain c. Be sure to consider regions inside and outside the sphere.

Answers

Answer:

See the explaination for the details.

Explanation:

Gauss Law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux in an area is defined as the electric field multiplied by the area of the surface projected in a plane and perpendicular to the field.

According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.

Please kindly check attachment for the step by step explaination of the answer.

The erg is a unit of work in units of centimeters (cm), grams (g), and seconds (s), and 1 erg=1 g⋅cm^2/s^2 . Recall that the SI unit of force is the newton (N) and is equal to kg⋅m/s^2 . You push an object 0.032 m by exerting 0.010 N of force. The work is the force times the distance.How much work have you done, expressed in ergs?

A: 3,200 ergs

B: 32 ergs

C: 0.32 ergs

D: 0.00032 ergs

Answers

For the given problem, the amount of work done expressed in ergs is 3200 ergs.

Answer: Option A

Explanation:

The work done on an objects are the force acting on it to move the object to a particular distance. So, work done on the object will be directly proportional to the force acting on it and the displacement.

Here, the force acting on the object is given as 0.010 N and the displacement of the object is 0.032 m. So, the work done on the object is

$\text { Work done }=\text { Force } * \text { displacement }$

$\text { Work done }=0.010 \mathrm{N} * 0.032 \mathrm{m}=0.00032 \mathrm{Nm}$

It is known that $1 N=1 \mathrm{kg} \mathrm{ms}^(-2)$

So, the work done can be expressed in $k g m s^(-2)$ as,

$\text { Work done }=0.00032 \mathrm{kgm}^(2) \mathrm{s}^(-2)$

It is known that $1 \mathrm{erg}=1 \mathrm{g} \mathrm{cm}^(2) / \mathrm{s}^(2)$ , so the conversion of units from Nm to erg will be done as follows:

$\text { Work done }=0.00032 \mathrm{kgm}^(2) \mathrm{s}^(-2) * \frac{1000 \mathrm{g}}{1 \mathrm{kg}} * \frac{100 * 100 \mathrm{cm}^(2)}{m^(2)}=3200 \mathrm{g} \mathrm{cm}^(2) \mathrm{s}^(-2)$

Thus, work done in ergs is 3200 ergs.

While on a hike, a pair of friends get caught in a thunderstorm. Four seconds after seeing the flash of a distant lightning strike, they hear the thunder. How far away was this lightning strike in miles? Note: sound, in air, travels at 340 m/s.

Answers

Answer:

1360 m

Explanation:

Time taken for the thunder to travel the distance to the hikers = 4 seconds

Speed of the thunder = 340 m/s

Speed of light = 3×10⁸ m/s

It can be seen that the speed of light is substantially faster than the speed of sound. This is the reason why there is a delay in seeing the lightning and hearing the thunder.

Distance = Speed × Time

$\text{Distance}=340* 4\n\Rightarrow \text{Distance}=1360\ m$

Hence, the lightning strike was 1360 m away from the hikers

Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)

Answers

Answer:

1/4F

Explanation:

We already know thatThe electrostatic force is directly proportional to the product of the charge, from Coulomb's law.

So F α Qq

But if it is now half the initial charges, then

F α (1/2)Q *(1/2)q

F α (1/4)Qq

Thus the resultant charges are each halved is (1/4) and the first initial force experienced at full charge.

Thus the answer will be 1/4F

Other Questions

What is science? Plz help I cant find a answer thats not really complicated and i have to fit it within 3 sentences.

Two violin strings are tuned to the same frequency 294 H. The tension in one string is then decreased by 2.0%. What will be the beat frequency heard when the two strings are played together?

What is the speed of an ocean wave if it’s wavelength is 5.0 m and it’s frequency is 3/s?

A piccolo and a flute can be approximated as cylindrical tubes with both ends open. The lowest fundamental frequency produced by one kind of piccolo is 522.5 Hz, and that produced by one kind of flute is 256.9 Hz. What is the ratio of the piccolo's length to the flute's length?