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3 years ago

A person weighs 60 kg. The area under the foot of the person is 150 cm2. Find the pressure exerted on the ground by the person.

Physics

1 answer:

zubka84 [21]3 years ago

7 0

Answer:

Pressure = 39200 N/m²

Explanation:

Given the following data;

Mass = 60 kg

Area = 150 cm²

Conversion:

1 centimeter square = 0.0001 meter square

150 centimeter square = 150 * 0.0001 = 0.015 m²

To find the pressure exerted on the ground by the person;

Mathematically, pressure is given by the formula;

$Pressure = \frac {Force}{area}$

First of all, we would determine the force exerted by the person.

Force = mass * acceleration due to gravity.

We know that acceleration due to gravity is equal to 9.8 m/s².

Substituting into the formula, we have;

Force = 60 * 9.8

Force = 588 Newton

Next, find the pressure;

$Pressure = \frac {Force}{area}$

Substituting into the formula, we have;

$Pressure = \frac {588}{0.015}$

Pressure = 39200 N/m²

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Answer:

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Explanation:

Drawing a free body diagram on the two boxes we can analyze the system more easily.

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because the two boxes has the same acceleration because they are attached together:

$m_B*g-m_B*a-m_a*g=m_a*a\\(m_B-m_A)*g=(m_a+m_B)*a\\a=\frac{(7.05kg-3.65kg)*9.8m/s^2}{3.65kg+7.05kg}\\\\a=3.11m/s^2$

So the magintude of the acceleration for both objects is $3.11m/s^2$

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3 years ago

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gamma radiation and heat flares from the sun, they use refelective gold sheets

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3 years ago

A roller coaster car may be approximated by a block of mass m. Thecar, which starts from rest, is released at a height h above t

elena55 [62]

Answer:

The first part can be solved via conservation of energy.

$mgh = mg2R + K\\K = mg(h-2R)$

For the second part,

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$F = ma = \frac{mv^2}{R}\\mg + N = \frac{mv^2}{R}$

where $N = 0$ because we are looking for the case where the car loses contact.

$mg = \frac{mv^2}{R}\\v^2 = gR\\v = \sqrt{gR}$

Now we know the minimum velocity that the car should have. Using the energy conservation found in the first part, we can calculate the minimum height.

$mgh = mg2R + \frac{1}{2}mv^2\\mgh = mg2R + \frac{1}{2}m(gR)\\gh = g2R + \frac{1}{2}gR\\h = 2R + \frac{R}{2}\\h = \frac{5R}{2}$

Explanation:

The point that might confuse you in this question is the direction of the normal force at the top of the loop.

We usually use the normal force opposite to the weight. However, normal force is the force that the road exerts on us. Imagine that the car goes through the loop very very fast. Its tires will feel a great amount of normal force, if its velocity is quite high. By the same logic, if its velocity is too low, it might not feel a normal force at all, which means losing contact with the track.

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3 years ago

A particle has a charge of -4.25 nC.

SpyIntel [72]

Answer:

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Explanation:

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The magnitude is 611.32 N/C

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The distance from the electric field is 1.71436 m

4 0

3 years ago

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