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2 years ago

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A sample of gas is enclosed in a container of fixed volume. Identify which of the following statements are true. Check all that

apply.If the container is heated, the gas particles will lose kinetic energy and temperature will increase.

1 answer:

motikmotik2 years ago

7 0

Answer:

B. If the container is cooled, the gas particles will lose kinetic energy and temperature will decrease.

C. If the gas particles move more quickly, they will collide more frequently with the walls of the container and pressure will increase.

E. If the gas particles move more quickly, they will collide with the walls of the container more often and with more force, and pressure will increase.

#FreeMelvin

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About how long does it take ocean water to travel around a gyre?

mylen [45]

The Gulf Stream, for example, has an average speed of 4 mph, or 100 miles/day, or 3600 miles/year, in round numbers. A complete circuit around the Atlantic is 8 or 10 thousand miles, which would take around 3 years on a raft.

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3 years ago

An object is moving in a circle. If the radius of the object is doubled, and the period remains constant, the magnitude of the v

Anton [14]

Answer:

Twice

Explanation:

From the formula for velocity in a circle

V= 2πr/T

Where V is velocity

r is raduis

T is period

We see that as r increases V increases so if r is doubled V becomes doubled

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2 years ago

Oil having a density of 922kg/m^3 floats on water. A rectangular block of wood 3.97 cm high and with a density of 963 kg/m^3 flo

Blizzard [7]

Explanation:

For the equilibrium:

\rho_{wood}gh-\rho_{oil}g(h-x)-\rho_{water}gx=0ρ

wood

gh−ρ

oil

g(h−x)−ρ

water

gx=0

\rho_{wood}h-\rho_{oil}(h-x)-\rho_{water}x=0ρ

wood

h−ρ

oil

(h−x)−ρ

water

x=0

(974)(3.97)-928(3.97-x)-1000x=0(974)(3.97)−928(3.97−x)−1000x=0

x=2.54\ cmx=2.54 cm

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1 year ago

a 25 kg object is pushed with a horizontal force of 5 N esat across a table. If the force of friction is 2 N, what is the accele

Julli [10]

Answer:

Apply Newton's second law in the moving direction.

Explanation:

$\begin{aligned}F &= ma\\5N-2N &= 25kg\times a\\a & = \frac{3N}{25kg}\\&= 0.12ms^{-2}\end{aligned}$

Friction force applies in the opposite direction of motion; as a restriction.

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2 years ago

Pete Zaria works on weekends at Barnaby's Pizza Parlor. His primary responsibility is to fill drink orders for customers. He fil

laila [671]

Answer:

$W_n_e_t=7.648512 \approx 7.6J$

$K.E=0.8J$

$v=0.7844645406 \approx 0.78m/s$

Explanation:

From the question we are told that

Mass of pitcher $M= 2.6kg$

Force on pitcher $f=8.8N$

Distance traveled $48cm=>0.48m$

Coefficient of friction $\mu=0.28$

a)Generally frictional force is mathematically given by

$F=\mu N$

$F=0.28*2.6*9.8$

$F=7.1344N$

Generally work done on the pitcher is mathematically given as

$W_n_e_t=W_f+W_F$

$W_f=8.8*0.48=>4.224N\\W_F=7.1344*0.48=>3.424512N$

$W_n_e_t=4.224-3.424512$

$W_n_e_t=0.799488\approx 0.8J$

b)Generally K.E can be given mathematically as

$K.E= W_n_e_t$

Therefore

$K.E=0.8J$

c)Generally the equation for kinetic energy is mathematically represented by

$K.E=1/2mv^2$

$0.8=1/2mv^2$

Velocity as subject

$v=\sqrt{\frac{K.E*2}{m} }$

$v=\sqrt{\frac{0.8*2}{2.6} }$

$v=0.7844645406 \approx 0.78m/s$

6 0

2 years ago