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3 years ago

a body of mass 5kg falls from height of 10m above the ground what kinetic energy of the body before it strike the ground

1 answer:

4 0

Gravitational Potential Energy (GPE) before fall = Kinetic energy on impact
GPE = mgh
GPE = 5kg x 9.8m/s^2 x 10m
GPE = 490 J
Kinetic Energy = 490 J

(This is assuming that gravitation field strength (g) is 9.8m/s^2, sometimes you may round this to 10m/s^2, therefore making the final result:
Kinetic energy = 500 J)

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A 250 mL sample of nitrogen(N2) has a pressure of 745 mm Hg at 30 C.What is the mass of nitrogen?

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We assume that nitrogen gas sample here is an ideal gas. So that, we can use the ideal gas equation which is expressed as follows:

PV = nRT

First, we calculate the number of moles from the equation above. Then, use molar mass to calculate mass.

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n = 0.01 mol

m = 0.01 ( 28) = 0.28 g

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The specific heat of water is 4.184 J g—1 K—1. A piece of iron (Fe) weighing 40 g is heated to 80EC and dropped into 100 g of wa

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2 years ago

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3 years ago

A particular baseball pitcher throws a baseball at a speed of 39.1 m/s (about 87.5 mi/hr) toward home plate. We use g = 9.8 m/s2

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2 years ago

Refer to the first diagram. What is the weight of the person hanging on the end of the seesaw in Newtons?

irina1246 [14]

Due to equilibrium of moments:

1) The weight of the person hanging on the left is 250 N

2) The 400 N person is 3 m from the fulcrum

3) The weight of the board is 200 N

Explanation:

To solve the problem, we use the principle of equilibrium of moments.

In fact, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

The moment of a force is defined as:

$M=Fd$

where

F is the magnitude of the force

d is the perpendicular distance of the force from the fulcrum

In the first diagram:

- The clockwise moment is due to the person on the right is

$M_c = W_2 d_2$

where $W_2 = 500 N$ is the weight of the person and $d_2 = 2 m$ is its distance from the fulcrum

- The anticlockwise moment due to the person hanging on the left is

$M_a = W_1 d_1$

where $W_1$ is his weight and $d_1 = 4 m$ is the distance from the fulcrum

Since the seesaw is in equilibrium,

$M_c = M_a$

So we can find the weight of the person on the left:

$W_1 d_1 = W_2 d_2\\W_1 = \frac{W_2 d_2}{d_1}=\frac{(500)(2)}{4}=250 N$

Again, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

- The clockwise moment due to the person on the right is

$M_c = W_2 d_2$

where $W_2 = 400 N$ is the weight of the person and $d_2$ is its distance from the fulcrum

- The anticlockwise moment due to the person on the left is

$M_a = W_1 d_1$

where $W_1 = 300 N$ is his weight and $d_1 = 4 m$ is the distance from the fulcrum.

Since the seesaw is in equilibrium,

$M_c = M_a$

So we can find the distance of the person on the right:

$W_1 d_1 = W_2 d_2\\d_2 = \frac{W_1 d_1}{W_2}=\frac{(300)(4)}{400}=3 m$

As before, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

- The clockwise moment around the fulcrum this time is due to the weight of the seesaw:

$M_c = W_2 d_2$

where $W_2$ is the weight of the seesaw and $d_2 = 3 m$ is the distance of its centre of mass from the fulcrum

- The anticlockwise moment due to the person on the left is

$M_a = W_1 d_1$

where $W_1 = 600 N$ is his weight and $d_1 = 1 m$ is the distance from the fulcrum

Since the seesaw is in equilibrium,

$M_c = M_a$

So we can find the weight of the seesaw:

$W_1 d_1 = W_2 d_2\\W_2 =\frac{W_1 d_1}{d_2}= \frac{(600)(1)}{3}=200 N$

#LearnwithBrainly

8 0

2 years ago