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riadik2000 [5.3K]
3 years ago
15

200 mL of an ideal gas is placed in a piston and is held at a pressure of 500 torr. If the temperature is held constant and the

pressure is increased to 650 torr, what is the new volume of the gas?
Physics
1 answer:
galina1969 [7]3 years ago
7 0

Answer:

V₂=153.84 m L

Explanation:

Given that

Initial volume ,V₁ = 200 m L

Initial pressure ,P₁= 500 torr

Final pressure P₂ = 650 torr

Lets take the final volume =V₂

Given that process occurs at constant temperature and we know that for constant temperature process

P₁ V₁ = P₂ V₂

V_2=\dfrac{P_1V_1}{P_2}

Now by putting the values

V_2=\dfrac{500\times 200}{650}\ mL

V₂=153.84 m L

Therefore the new volume of the gas will be 153.8 4mL.

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What is a measure between the difference in start and end positions?
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Answer:

Displacement

General Formulas and Concepts:

<u>Kinematics</u>

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Explanation:

Displacement is the difference between the start position and end position.

Total Distance is the entire distance <em>traveled</em> between the start and end position.

Topic: AP Physics 1 Algebra-Based

Unit: Kinematics

5 0
3 years ago
Goldberg's sleigh currently runs at 203mph, but he needs it to reach 400mph with all the packages he has to deliver.
svp [43]

Answer:

c

Explanation:

6 0
3 years ago
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In the great shopping cart race, two students push on shopping carts. A having twice the mass of B, with the same force applied
Lesechka [4]

K = 1/2 m x v^2

m = mass on the cart

V = velocity imparted to the cart

KA = 1/2 mA x vA^2.......................(1)

KB = 1/2 mB x vB^2........................(2)

Diving equation 1 by equation 2, we get -

KA/KB = mA/mB

= 2

KA = 2 x KB

Option A is correct

6 0
4 years ago
Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at
PSYCHO15rus [73]

Answer:

|\Delta v |=\sqrt{\frac{4Gm}{d} }

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

\mu= \frac{m_1m_2}{m_1+m_2}

now, since both the masses have mass m

therefore,

\mu= \frac{m^2}{2m}

= m/2

The final K.E. of the particles is

KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2

Distance between two particles is d and the gravitational potential energy between them is given by

PE_{Gravitational}= \frac{Gmm}{d}

By law of conservation of energy we have

KE_{initial}+KE_{final}= PE_{gravitaional}

Now plugging the values we get

0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}

|\Delta v |=\sqrt{\frac{4Gm}{d} }

=\sqrt{\frac{Gm}{d} }

This the required relation between G,m and d

5 0
3 years ago
Two workers are sliding 290 kg crate across the floor. One worker pushes forward on the crate with a force of 430 N while the ot
m_a_m_a [10]

Answer:0.27

Explanation:

Given

One worker Pushes with force F_1=430 N

other Pulls it with a rope of rope F_2=360 N

mass of crate m=290 kg

both forces are horizontal and crate slides with a constant speed

Both forces are in the same direction so Friction will oppose the forces and will be equal in magnitude of sum of two forces because crate is moving with constant speed i.e. net force is zero on it

f_r=F_1+F_2

where f_r is the friction force

f_r=430+360

f_r=790 N

f_r=\mu N

where \muis the coefficient of static friction

N=mg

790=\mu 29\cdot 9.8

\mu =0.27

6 0
4 years ago
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