What will be the pressure exerted by the object if 5,000 N of force area of 200 cm²?

Physics

1 answer:

devlian [24]2 years ago

3 0

Answer:

$250000 \frac{n}{m {}^{2} }$

Explanation:

pressure is the force applied perpendicular to the surface area of an object per unit area.

for the question,

Force (F) = 5000N

$Area = 200cm {}^{2} = 0.02m {}^{2} \\ Pressure = \frac{force}{area} \\ Pressure = \frac{5000}{0.02} \\ Pressure = 250000 \frac{n}{m {}^{2} }$

You might be interested in

What is the

Len [333]

Answer:

$d=5\ g/cm^3$

Explanation:

Given that,

Mass of the object, m = 100 grams

Volume of the object, V = 20 cm³

We need to find the density of the object. We know that, density is equal to mass per unit volume. So,

$d=\dfrac{m}{V}\\\\d=\dfrac{100\ g}{20\ cm^3}\\\\d=5\ g/cm^3$

So, the density of the object is equal to $5\ g/cm^3$ .

6 0

2 years ago

the average time required for radioactive elements, whereby they decay to half their original mass is called the "average life"

VikaD [51]

I'm not sure what your question is. But, the half life is the amount of time required for half the material to decay. For U238 this is 4.5 billion years, whilst for Fr-223 (Francium) its about 22 minutes. To calculate the time for something to decay you need to use the equation:
Mass (after time t) = Mass (initial) * (0.5)^(time/half life)
Hope this helps

4 0

2 years ago

Read 2 more answers

The length of a rectangular sheet of metal decreases by 34.5 cm. Its width decreases proportionally. If the sheets original widt

devlian [24]

The original width was 94.71 cm
The area decreased 33.1% 

The equation for the final size is 
2X^2 = 1.2 m^2 
X^2 - 0.6 m^2 
X^2 = 10000 * .6 cm 
X = 77.46 cm (this is the width) 

The length is 2 * 77.46 = 154.92 cm 

The original length was 154.92 + 34.5 = 189.42 cm 
The original width was 189.42 / 2 = 94.71 cm 

The original area was 94.71 * 189.92 = 17939.9 cm^2 
The new area is 79.46 * 154.92 = 12000.1 cm^2 

The difference between the original and current area is 17939.9 - 12000.1 = 5939.86 cm^2 

The percentage the area decreased is 5939.86 ' 17939.9 = 33.1%

6 0

3 years ago

The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have st

irina [24]

Complete Question:

The elastic energy stored in your tendons can contribute up to 35 % of your energy needs when running. Sports scientists have studied the change in length of the knee extensor tendon in sprinters and nonathletes. They find (on average) that the sprinters' tendons stretch 43 mm , while nonathletes' stretch only 32 mm . The spring constant for the tendon is the same for both groups, $31 {\rm {N}/{mm}}$ . What is the difference in maximum stored energy between the sprinters and the nonathlethes?