All categories

3 years ago

How does heating a gas rigid container change its pressure

1 answer:

6 0

The particles are heating up so they are starting to move faster and closer together which causes them to change the pressure of the container

You might be interested in

Consider Earth and the Moon. As you should now realize, thegravitational force that Earth exerts on the Moon is equal andopposit

dolphi86 [110]

Answer:

A. the Moon has a larger acceleration than Earth, because it has a smaller mass.

Explanation:

A. the Moon has a larger acceleration than Earth, because it has a smaller mass.

According to Newton's second law F=ma, if we solve for a we have that

a=F/m, the smaller the mass the larger the acceleration given the force is the same, in this case the moon has a smaller mass and therefore a larger acceleration

3 0

3 years ago

Can someone please help me with this

Mekhanik [1.2K]

Answer:

B because the earth rotational axis tilt away from the sun.

3 0

3 years ago

A 45 kg object has a momentum of 225 kg-m/s northward. What is the object's velocity?

melomori [17]

Answer: 5 m/s North

Explanation:

Velocity = Momentum/Mass

Velocity = 225 kg*m/s/45kg

Velocity = 5 m/s North

3 0

2 years ago

After a trip to Mars, the human heart would be much stronger than it was on Earth

Mariulka [41]

My answer to this question honestly is no

7 0

3 years ago

A circular rod has a radius of curvature R = 9.09 cm and a uniformly distributed positive charge Q = 6.49 pC and subtends an ang

Digiron [165]

Answer:

E = 1.19 N/C

Explanation:

Let's first determine the length of the arc which can be given as:

L= Rθ

where:

L = length of the arc

R = radius of curvature

θ = angle in radius

L = (9.09×10⁻²m)(2.59)

L = (0.0909)(2.59)

L = 0.235431 m

Then, the magnitude of electric field that Q produces at the center of curvature can be calculated by using the formula:

$E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}-sin(-\frac{\theta}{2})]$

$E= \frac{\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}+sin(\frac{\theta}{2})]$

$E= \frac{2\lambda}{4 \pi E_oR}[sin\frac{\theta}{2}]$

Since $\lambda = \frac{Q}{L}$

where;

L = length

Q = charge

λ = density of the charge;

then substituting $\frac{Q}{L}$ for λ, we have :

$E= \frac{2(\frac{Q}{L})}{4 \pi E_oR}[sin\frac{\theta}{2}]$

$E= \frac{2Q[sin\frac{\theta}{2}]}{4 \pi E_oLR}$

substituting our given parameter; we have:

$E= \frac{2(6.26*10^{-12}C)[sin\frac{2.59rad}{2}]}{4 \pi (8.85*10^{-12}C^2/N.m^2)(0.235431)(0.0909)}$

E = 1.1889 N/C

E = 1.19 N/C

∴ the magnitude of the electric field that Q produces at the center of curvature = 1.19 N/C

4 0

3 years ago