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3 years ago

Explain how fluids exert pressure

1 answer:

4 0

In a fluid, all the forces exerted by the individual particles combine to make up the pressure exerted by the fluid
Due to fundamental nature of fluids, a fluid cannot remain at rest under the presence of shear stress. However, fluids can exert pressure normal to any contacting surface. If a point in the fluid is thought of as a small cube, then it follows from the principles of equilibrium that the pressure on every side of this unit of fluid must be equal. but if this were not a case, the fluid would move in the directions of the resulting force, So the pressure on a fluid at rest is isotropic.
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Force due to gravity. While JWST is in orbit, the Earth will be at a distance of 1.494 x 109 m from the telescope, and the Sun w

Alona [7]

Answer:

the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

Explanation:

Given that;

distance of earth from the telescope r1 = 1.494 × 10⁹ m

and the Sun will be 1.49598 × 10¹¹ m further away

so r2 = r1 + 1.49598 × 10¹¹

r2 = 1.494 × 10⁹ m + 1.49598 × 10¹¹ m = 1.51092 × 10¹¹ m

mass of sun Ms = 1.9884 × 10³⁰ kg

mass of earth Me = 5.945× 10²⁴ kg

mass of JWST Mj = 6500 kg

What is the gravitational force JWST will feel from the Sun (strength and direction)?

the gravitational force of the sun will be attractive based on Newton law of gravitational force; so

Fjs = GMjMs / r2²

constant G = 6.674 × 10⁻¹¹ Nm²/kg²

Force on the JWST by the sun will be;

Fjs = GMjMs / r2² { leftward}

we substitute

Fjs = [(6.674 × 10⁻¹¹ Nm²/kg²)(6500 kg )(1.9884 × 10³⁰ kg)] / (1.51092 × 10¹¹ m)²

= (8.62587804 × 10²³) / ( 2.28287925 × 10²² )

= 37.785 Newton { leftward }

Therefore, the gravitational force JWST will feel from the Sun is 37.785 Newton { leftward }

7 0

3 years ago

At a point 1.2 m out from the hinge, 14.0 N force is exerted at an angle of 27 degrees to the moment arm in a plane which is per

geniusboy [140]

Answer:

$\tau = 7.63 Nm$

Explanation:

As we know that moment of force is given as

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

now we have

$\vec r = 1.2 m$

$\vec F = 14 N$

now from above formula we have

$\tau = r F sin\theta$

here we know that

$\theta = 27 degree$

so we have

$\tau = (1.2)(14) sin27$

$\tau = 7.63 Nm$

3 0

3 years ago

a car with a mass of 100 kg is stopped on the side of the road after getting a flat tire. the two people that were riding in the

notsponge [240]

Answer:

Please mark me brainliest and thank me and rate me

3 0

3 years ago

A ball rolls down the hill which has a vertical height of 15 m. Ignoring friction what would be the gravitational potential ener

trasher [3.6K]

a) Potential energy: 147 m [J]

The gravitational potential energy of an object is given by

$U=mgh$

where

m is its mass

$g=9.8 m/s^2$ is the acceleration of gravity

h is the height of the object above the ground

In this problem,

h = 15 m

We call 'm' the mass of the ball, since we don't know it

So, the potential energy of the ball at the top of the hill is

$U=(m)(9.8)(15)=147 m$ (J)

b) Velocity of the ball at the bottom of the hill: 17.1 m/s

According to the law of conservation of energy, in absence of friction all the potential energy of the ball is converted into kinetic energy as the ball reaches the bottom of the hill. Therefore we can write:

$U=K=\frac{1}{2}mv^2$

where

v is the final velocity of the ball

We know from part a) that

U = 147 m

Substituting into the equation above,

$147 m = \frac{1}{2}mv^2$

And re-arranging for v, we find the velocity:

$v=\sqrt{2\cdot 147}=17.1 m/s$

5 0

3 years ago

.hjvhhchhvvgvfggghhjidaaawryui<br>

Naddik [55]

Answer:

yes

Explanation:

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5 0

3 years ago