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zysi [14]
3 years ago
7

A man weighing 700 NN and a woman weighing 440 NN have the same momentum. What is the ratio of the man's kinetic energy KmKmK_m

to that of the woman K
Physics
1 answer:
Eduardwww [97]3 years ago
3 0

Answer:

The ratio of the man's kinetic energy to that of the woman's kinetic energy is 0.629.

Explanation:

Given;

weight of the man, W = 700 N

Weight of the woman, W = 440 N

momentum is given by;

P = mv\\\\v = \frac{P}{m}

Kinetic energy of the man;

K_m = \frac{1}{2}m_m(\frac{P_m}{m_m})^2  \\\\K_m = \frac{P_m^2}{2m_m}

Momentum of the man is calculated as;

P_m^2 = 2m_mK_m

The kinetic energy of the woman is given by;

K_w = \frac{P_w^2}{2m_w}

The momentum of the woman is given;

P_w^2 = 2m_wK_w

Since, momentum of the man = momentum of the woman

P_m^2 = P_w^2

2m_mK_m = 2m_wK_w\\\\\frac{K_m}{K_w} = \frac{2m_w}{2m_m}\\\\\frac{K_m}{K_w} = \frac{m_w}{m_m}

mass of the mas = 700 / 9.8 = 71.429

mass of the woman is = 440 / 9.8 = 44.898

\frac{K_m}{K_w} = \frac{44.898}{71.429}\\\\\frac{K_m}{K_w} =0.629

Therefore, the ratio of the man's kinetic energy to that of the woman's kinetic energy is 0.629.

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An applied force of 50 N is used to accelerate an object, that weighs 73 N, to the right across a frictional surface. The object
Hunter-Best [27]

Answer:

5.38 m/s^2

Explanation:

NET force causing the object to accelerate  =  50 -10 = 40 N

Mass of the object =  73 N / 9.81 m/s^2 = 7.44 kg

F = ma

40 = 7.44 * a         a = 5.38 m/s^2

6 0
1 year ago
An object is placed at zero zero on a Number line. It moves three units to the right, then four units to the left, and then 60 u
STatiana [176]

Answer:

the displacement of the object is 5 units

Explanation:

The computation of the displacement of the object is shown below:

= Move to the right + move to the right - move to the left

= 6 units + 3 units - 4 units

= 9 units - 4 units

= 5 units

Hence, the displacement of the object is 5 units

7 0
2 years ago
A horizontal force of 750 N is needed to overcome the force of static friction between a level floor and a 250-kg crate. What is
Aleksandr [31]

Answer:

The acceleration of the crate is 1.8 m/s² so the answer is a.

Explanation:

The very first thing you must do when solving this problem is to draw a free body diagram. (The body diagram is attached to this answer)

So once we got the free body diagram, we can analyze it and build our sum of forces in the x and y directions. Notice that according to the diagram, there are 4 forces to this problem, Normal (N), Weight (W), kinetic friction (fk) and the 750N force.

As one may see in the free body diagram, two of the forces are vertical forces: N and W, so we can use them to build a sum of forces:

Starting with the sum of forces in the y-direction, we get:

ΣF_{y}=0

We set the sum equal to zero because there is no movement in the y-direction, so the system is in vertical equilibrium.

so the sum will be:

N-W=0

when solving for N we get that:

N=W

where W is found by multiplying the mass of the crate by the acceleration of gravity:

N=250kg*9.8m/s²

N=2450N

Once we found the normal force, we can use it to find the kinetic friction which is given by the following formula:

f_{k}=Nμ

where μ is the kinetic friction coefficient.

So we get that the kinetic friction is:

f_{k}=2450N*0.12

so

f_{k}=294

With this information we can go ahead and find the sum of horizontal forces:

ΣF_{x}=ma

In this case the sum is equal to mass times acceleration because the crate is moving horizontally due to the action of a force, so it will have an acceleration.

so the sum of forces look like this:

750N-f_{k}=ma

so

750N-294N=(250kg)a

when solving for a we get:

a=\frac{759N-294N}{250kg}\\ \\a=1.8m/s^{2}

so the crate's acceleration is 1.82m/s².

5 0
3 years ago
How does this diagram demonstrate the law of superposition?
Mandarinka [93]

Answer:

well, as u can tell the top layer will always be the youngest layer aka the newest layer. The farther u go down the older the layers get. So the deeper u dig the farther back in time we see.

Explanation:

8 0
2 years ago
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If a rock sample has a mass of 1.17 g and a volume of 0.33 cm3, what type of rock is it?
Marat540 [252]

The correct answer is diamond

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3 years ago
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