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2 years ago

A body of mass 100kg exerts pressure of 50.000N\M².calculate the area over which the force acts

Physics

2 answers:

yanalaym [24]2 years ago

6 0

Answer:

the area is 0.0196 m^2

Explanation:

Recall that pressure is defined as the force over the surface area:

Pressure = Force / Area

Therefore, Area = Force / Pressure

in our case, the acting force is the gravitational force Fg = m * g = 100 * 9.8 N = 980 N

and the pressure is 50000 N/m^2

Then the acting area is:

980 / 50000 m^2 = 0.0196 m^2

hoa [83]2 years ago

4 0

The area is 0.0196m^2

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melisa1 [442]

Answer:

m = 1.26*10²⁵ kg.

Explanation:

Assuming that the mass of the stone is much smaller than the mass of the planet, we can get the mass, applying the Universal Law of Gravitation to both masses, as follows:

Fg = G* ms* mp / rp²

Now, if we apply Newton's 2nd Law to the mass of the stone, we can get the gravitational acceleration, as follows:

Fg = ms*a = ms*g ⇒ g = G*mp / rp²

First of all, we need to get the value of g.

Assuming that this acceleration is constant, we can appy the kinematic equations to this situation.

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At the highest point in the trajectory, just before of changing direction, the stone comes momentarily to a stop.

At this point, applying the definition of acceleration, we can write:

vf = vo -g*t ⇒ 0 = vo -gt ⇒ g = vo/t (1)

We have the total time since the stone was thrown upwards, not the one used for the upward trajectory.

It can be showed, using the expression for the displacement (which is the same in both directions) that the time used for going up, it's the same used to go down, so the time that we need to put in (1). is just the half of the total time.

So, replacing in (1) we get the value of g, as follows:

g = 15 m/s / 4.5 s = 3.33 m/s²

Now, we can replace this value in the equation that gives us g based in the Universal Law of Gravitation, as follows:

g=G*mp / rp² (2)

Before solving for mp, however, we need to get the value of the radius of the planet.

Assuming that it's a perfect sphere, we can get this value from the value of the circumference at the planet's equator:

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With this value for rp, we can solve (2) for mp, as follows:

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mp = 1.26*10²⁵ kg.

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3 years ago

an object with a mass of 5kg is moving with an initial velocity of 20m/s.the object accelerates at a rate of 15m/s/s for 8s.what

Alinara [238K]

It doesn't matter what the object's initial velocity is, or how long
the acceleration lasts. All that matters is the object's mass and
acceleration.

Force = (mass) x (acceleration) =

(5kg) x (15 m/s²) =

75 kg-m/s² = <em>75 newtons .</em>

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3 years ago

A block of mass m slides on a horizontal frictionless surface. The block is attached to a spring with a spring constant K. At th

Fittoniya [83]

Answer:

b) a = -k / m x , c) d²x / dt² = - A w² cos (wt+Ф) , d) and e) T = 2π √m / k

h) a = - A w² cos (wt+Ф)

Explanation:

a) see free body diagram in the attachment

b) We write Newton's second law

Fe = m a

-k x = ma

a = -k / m x

c) the acceleration is

a = d²x / dt²

If x = A cos wt

v = dx / dt = -A w sin (wt +Ф)

a = d²x / dt² = - A w² cos (wt+Ф)

d) we substitute in Newton's second law

d²x / dt² = -k / m x

We call

w² = k / m

e) substitute to find w

-A w² cos (wt+Ф) = -k / m A cos (wt+Ф)

w² = k / m

Angular velocity and frequency are related

w = 2π f

f = 1 / T

We substitute

T = 2π / w

T = 2π √m / k

g) v= - A w sin (wt+Ф)

h) acceleration is

a = - A w² cos (wt+Ф)

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3 years ago

Which of the following weighing balances performs measurement in a closed compartment with no air currents to disturb measuremen

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3 years ago

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A block of ice at 0 degrees C, whose mass is initially 62 kg, slides along a horizontal surface, starting at a speed of 5.48 m/s

Kryger [21]

The mass of ice melted as a result of friction between the ice and the horizontal surface is 2.78g

<u>Explanation:</u>

Given,

Temperature, T = 0°C

Initial mass, Mi = 62kg

Speed, s = 5.48m/s

Distance, x = 26.8m

Friction is present.

Mass of ice melted = ?

We know,

The amount of energy required for the melting of ice is exactly equal to the initial kinetic energy of the block of ice

and

$Kinetic Energy, KE = \frac{1}{2} mv^2$

Therefore, $KE = \frac{62 X 5.48 X 5.48}{2}$

KE = 930.94 Joules

Ice melting lateral heat is 334 kJ/kg = 334000 J/kg.

Therefore, the melted mass of the ice = 930.94 / 334000 = 0.00278 kg = 2.78 g.

Thus, The mass of ice melted as a result of friction between the ice and the horizontal surface is 2.78g

4 0

3 years ago

A body of mass 100kg exerts pressure of 50.000N\M².calculate the area over which the force acts​

A body of mass 100kg exerts pressure of 50.000N\M².calculate the area over which the force acts