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3 years ago

1) Do liquids and gases also exert pressure?

Physics

2 answers:

Marta_Voda [28]3 years ago

6 0

Explanation:

Liquids also exert pressure in all directions on the walls of the container they are stored in. We see water coming out from leaking pipes and taps. ... Gases (Air) also exert pressure in all directions

lions [1.4K]3 years ago

5 0

I’m pretty sure they do I’m not sure though

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A 1900kg car starts from rest and drives around a flat 65-m-diameter circular track. The forward force provided by the car's dri

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Answer:

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3 years ago

When a solid is in the process of changing into a liquid and extra energy is added to the system, the temperature does not chang

Alekssandra [29.7K]

Answer:

D. The temperature does not change during a phase change because the average kinetic energy does not change. Therefore, the potential energy in the bonds between molecules must change.

Explanation:

When there is a change of state (for example, from solid into a liquid, as in this example), when energy is added to the system, the temperature of the substance does not change.

The reason for this is that the energy supplied is no longer used to increase the average kinetic energy of the particle, but instead it is used to break the bonds between the different particles/molecules. For instance, since in this case the substance is changing from solid to liquid, all the energy supplied during the phase change is used to break the bonds between the molecules of the solid: when the process is done, all the molecules will be free to slide past each other, and the substance has turned completely into a liquid.

The bonds between molecules store potential energy: therefore, this means that the energy supplied during the phase change is not used to change the kinetic energy, but to change the potential energy in the bonds between the molecules.

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3 years ago

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A 7.5 kg cannon ball leaves a canon with a speed of 185 m/s. Find the average net force applied to the ball if the cannon muzzle

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3 years ago

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Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless

drek231 [11]

Answer:

$v_f = 15 \frac{m}{s}$

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum $\vec{L}$ is

$\vec{L} = \vec{r} \times \vec{p}$

where $\vec{r}$ is the position and $\vec{p}$ the linear momentum.

We also know that the torque is

$\vec{\tau} = \frac{d\vec{L}}{dt} = \frac{d}{dt} ( \vec{r} \times \vec{p} )$

$\vec{\tau} = \frac{d}{dt} \vec{r} \times \vec{p} + \vec{r} \times \frac{d}{dt} \vec{p}$

$\vec{\tau} = \vec{v} \times \vec{p} + \vec{r} \times \vec{F}$

but, as the linear momentum is $\vec{p} = m \vec{v}$ this means that is parallel to the velocity, and the first term must equal zero

$\vec{v} \times \vec{p}=0$

$\vec{\tau} = \vec{r} \times \vec{F}$

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

$\vec{\tau}_{rod} = 0$