All categories

3 years ago

4) A force of 500 N acts on an area of 0.05m2. Find the pressure in Pascal.

Physics

1 answer:

snow_tiger [21]3 years ago

6 0

Answer:

pressure = force ie 500 N divided by area ie 0.05m².

p=f by a

p= 500n divided by 0.05 m²

p= 10,000 pascal

You might be interested in

A body is traveling at 5.0 m/s along the positive direction of an x axis; no net force acts on the body. An internal explosion s

loris [4]

To solve this problem it is necessary to apply the concepts related to the conservation of momentum and conservation of kinetic energy.

By definition kinetic energy is defined as

$KE = \frac{1}{2} mv^2$

Where,

m = Mass

v = Velocity

On the other hand we have the conservation of the moment, which for this case would be defined as

$m*V_i = m_1V_1+m_2V_2$

Here,

m = Total mass (8Kg at this case)

$m_1=m_2 =$ Mass each part

$V_i =$ Initial velocity

$V_2 =$ Final velocity particle 2

$V_1 =$ Final velocity particle 1

The initial kinetic energy would be given by,

$KE_i=\frac{1}{2}mv^2$

$KE_i = \frac{1}{2}8*5^2$

$KE_i = 100J$

In the end the energy increased 100J, that is,

$KE_f = KE_i KE_{increased}$

$KE_f = 100+100 = 200J$

By conservation of the moment then,

$m*V_i = m_1V_1+m_2V_2$

Replacing we have,

$(8)*5 = 4*V_1+4*V_2$

$40 = 4(V_1+V_2)$

$V_1+V_2 = 10$

$V_2 = 10-V_1$ (1)

In the final point the cinematic energy of EACH particle would be given by

$KE_f = \frac{1}{2}mv^2$

$KE_f = \frac{1}{2}4*(V_1^2+V_2^2)$

$200J=\frac{1}{2}4*(V_1^2+V_2^2)$ (2)

So we have a system of 2x2 equations

$V_2 = 10-V_1$

$200J=\frac{1}{2}4*(V_1^2+V_2^2)$

Replacing (1) in (2) and solving we have to,

200J=\frac{1}{2}4*(V_1^2+(10-V_1)^2)

PART A: $V_1 = 10m/s$

Then replacing in (1) we have that

PART B: $V_2 = 0m/s$

8 0

3 years ago

Changes in motion come from;

almond37 [142]

Answer:

C-Mass

Explanation:

3 0

3 years ago

True or false if two forces are in the same direction they cancel each other out

Hitman42 [59]

I believe the correct answer is true.

Hope this helps! ;)

6 0

3 years ago

17. A 60.-kilogram student climbs a ladder a vertical distance of 4.0 meters in 8.0 seconds. Approximately how much total work i

Margarita [4]

Work done = change in PE = mgh = 60*10*4 = 2400 J

8 0

3 years ago

What is the potential energy of a 30 kg object that is 10 meters up on a ladder?

scZoUnD [109]

Answer:

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 30 × 10 × 10

We have the final answer as

Hope this helps you

3 0

3 years ago