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Kisachek [45]
2 years ago
13

PLEASE HELP!! When a box rests on a round sheet of wood on the ground, it exerts an average pressure p on the wood. If the wood

is replaced by a sheet that has half the diameter of the original piece, what is the new average pressure? I think the answer is 4p but I don't know why.
Physics
1 answer:
valkas [14]2 years ago
4 0
Pressure = Force ÷ Area

In this case, since the area is circular, we can rewrite the equation as:

Pressure = Force ÷ pi*r^2

Decreasing the diameter by half is the same as decreasing the radius by half.

We can replace all constants with 1 to compare the relationship between the two scenarios.

Original scenario with original area wood:

Pressure = 1 ÷ 1^2
Pressure = 1

Scenario with 0.5 radius of original scenario:
Pressure = 1 ÷ (1/2)^2
Pressure = 1 ÷ 1/4
Pressure = 4

This confirms that your guess of 4x the pressure is correct.

So as can be seen by the relationship, pressure for circular area is inversely proportional to the radius squared.


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Answer:

The value of tension on the cable T = 1065.6 N

Explanation:

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Initial velocity ( u )= 0.8 \frac{m}{sec}

Final velocity ( V ) = 0

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Now use third law of motion V^{2} = u^{2} - 2 a s

Put all the values in above formula we get,

⇒ 0 = 0.8^{2} - 2 × a ×0.2667

⇒ a = 1.2 \frac{m}{sec^{2} }

This is the deceleration of the box.

Tension in the cable is given by T = F = m × a

Put all the values in above formula we get,

T = 888 × 1.2

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This is the value of tension on the cable.

5 0
2 years ago
A child is riding a merry-go-round that has an instantaneous angular speed of 1.25 rad/s and an angular acceleration of 0.745 ra
skelet666 [1.2K]

Answer:

So the acceleration of the child will be 8.05m/sec^2

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We have given angular speed of the child \omega =1.25rad/sec

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Angular acceleration \alpha =0.745rad/sec^2

We know that linear velocity is given by v=\omega r=1.25\times 4.65=5.815m/sec

We know that radial acceleration is given by a=\frac{v^2}{r}=\frac{5.815^2}{4.65}=7.2718m/sec^2

Tangential acceleration is given by

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F=(0.850) sin (\frac{x}{2.00}) [N]

where x is the position in metres.

The acceleration can be found by using Newton's second law:

a=\frac{F}{m}

where

m = 150 g = 0.150 kg is the mass of the particle. Substituting into the equation,

a=\frac{0.850}{0.150}sin (\frac{x}{2.00})=5.67 sin(\frac{x}{2.00}) [m/s^2]

When x = 3.14 m, the acceleration is:

a=5.67 sin(\frac{3.14}{2.00})=5.67 m/s^2

Now we can find the final speed of the particle by using the suvat equation:

v^2-u^2=2ax

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u = 8.00 m/s is the initial velocity

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Solving for v,

v=\sqrt{u^2+2ax}=\sqrt{8.00^2+2(5.67)(3.14)}=9.98 m/s

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Here,only those radiations will be capable of emitting electrons irrespective of surface barrier of metals whose energy is greater than the work function.

We know that the radiation having long wavelength has least energy as energy and wavelength are inversely proportional to each other.

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