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irga5000 [103]
3 years ago
7

A box exerts a force of 540 N onto the floor and the bottom of the box is 0.5 m2. What is the pressure the box exerts on the flo

or?
Physics
2 answers:
Sever21 [200]3 years ago
7 0

Answer:

pressure = force/Area

pressure = 540/0.5

pressure = 1080 n/m2

krek1111 [17]3 years ago
7 0

Answer:

I hope the above pic will help you

I will always help you understanding your assingments

have a great day

#Captainpower:)

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Can the boiling point of water exceed 100c
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Superheated water is liquid water under pressure at temperatures between the usual boiling point, 100c (212 F ) and the critical temperature , 374 C (705F) . It is also known as “subcritical water”
Or “pressurized hot water”.
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Why is plate tectonics a widely accepted theory?
MariettaO [177]

Answer: The theory of Plate Tectonics is now widely accepted because there is sufficient proof to support it, and it is an important aspect of geology, oceanography, geophysics and even paleontology.

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6 0
3 years ago
If the mass of both objects are doubled while the
lutik1710 [3]

Answer:

The force becomes 16 times what it is now.

Explanation:

The formula for gravitational force is

F = G * m1 * m2 / r^2

When you do what you have described, you are setting a stage that not even the USS Enterprise (Star Trek) can get out of. The increase is huge.

If you double m1 and m2 and don't do anything to r, you've already increased the force by 4 times. (2m1 * 2m2 = 4 * m1 * m2)

But you are not finished. If you 1/2 the distance, you are again increasing the Force by 4 times. 1 / (2r) ^2 = 1/ 4* r^2

Because this is in the denominator, the 1/4 is going to flip to the numerator.

So the total increase is going to be 4 * (4 * m1 * m2) = 16 * m1 * m2.

Think about what that means. If you were out golfing, your drives would be roughly 1/16 times as far as they are now. Also you would be lugging around 16 times your weight around the golf course. My feeling is that you would never finish 5 holes at that rate.

3 0
2 years ago
A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar
notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2>\omega = 0.23 rad/s</h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

L_i = L_f

now we can say

m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega

so we will have

0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega

\omega = 0.23 rad/s

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

6 0
3 years ago
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