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Julli [10]
2 years ago
11

A 50.0 g golf ball is struck by a driver whose head has a mass of 500.0 kg. The final speed of the golf ball after leaving the t

ee is 45.0 m/s. Find the final momentum of the golf ball.
A)2.5 kg·m/s
B)5.0 kg·m/s
C)7.5 kg·m/s
D)10.0 kg·m/s
Physics
1 answer:
Savatey [412]2 years ago
8 0

<h2>A. 2.5 kg.m/s</h2>

this is my answer sure po ako n tama iyan

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This is a science Question. Please help.
sattari [20]

Answer:

B

Explanation:

This is a physics question, know that force is equals to mass divided by acceleration (acc.), so if the same force is applied, say 10 Newton and the mass of A is 2 and the mass of B is 4, then the acceleration of A is 0.2 and that of B is 0.4 by equating, and this applies to all cases.

6 0
3 years ago
Put the following objects in order by their
maksim [4K]

Answer:

I Think its ABC

Explanation:

No air is the lightest

Helium is lighter than regular air that's why it goes up so then the regular air would be heaviest.

4 0
3 years ago
If the 50 kg boy were in a spacecraft 5.0r from the center of the earth, what would his weight be?
Lady_Fox [76]
·The acceleration of gravity is proportional to

                       1 / (the square of the distance from the center) .

When we're on the surface, we're 1 radius from the center of the Earth,
and the acceleration of gravity is  9.8 m/s² .

The boy's weight = (mass) · (gravity) = (50kg) · (9.8 m/s²)

                                                             =      490 newtons .

At the distance of 5 radii from the center (4 radii altitude from the surface),
the acceleration of gravity is

                 (9.8 m/s²) · (1/5)²  =  0.39 m/s² .

The boy's weight is    (mass) · (gravity)  =  (50kg) · (0.39 m/s²)

                                                                  =     19.6 newtons .

Just as we expected, his weight at that distance is

         (19.6 / 490) = 0.04  =  1/25  =  1/5²  of his weight on the surface.
7 0
3 years ago
Force F = − + ( 8.00 N i 6.00 N j ) ( ) acts on a particle with position vector r = + (3.00 m i 4.00 m j ) ( ) .
jek_recluse [69]

Explanation:

Given that,

Force, F=((-8i)+6j)\ N

Position of the particle, r=(3i+4j)\ m

(a) The toque on a particle about the origin is given by :

\tau=F\times r

\tau=((-8i)+6j) \times (3i+4j)

Taking the cross product of above two vectors, we get the value of torque as :

\tau=(0+0-50k)\ N-m

(b) Let \theta is the angle between r and F. The angle between two vectors is given by :

cos\theta=\dfrac{r.F}{|r|.|F|}

cos\theta=\dfrac{(3i+4j).((-8i)+6j)}{(\sqrt{3^2+4^2} ).(\sqrt{8^2+6^2}) }

cos\theta=\dfrac{0}{50}

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6 0
3 years ago
Find the heat energy is required to change 2Kg of ice at 0 C to water at 20 C ( specific latent heat of fusion of water = 336000
katrin2010 [14]

We want to find the energy that we need to transform 2kg of ice at 0°C to water at 20°C.

We will find that we must give 840,000 Joules.

First, we must change of phase from ice to water.

We use the specific latent heat of fusion to do this, this quantity tells us the amount of energy that we need to transform 1 kg of ice into water.

So we need 336,000 J of energy to transform 1kg of ice into water, and there are 2kg of ice, then we need twice that amount of energy:

2*336,000 J = 672,000 J

Now we have 2kg of water at 0°C, and we need to increase its temperature to 20°C.

Here we use the specific heat, it tell us the amount of energy that we need to increase the temperature per mass of water by 1°C.

We know that:

specific heat of capacity of water = 4200 J/kg°C

This means that we need to give 4,200 Joules of energy to increase the temperature by 1°C of 1kg of water.

Then to increase 1°C of 2kg of water we need twice that amount:

2*4,200 J = 8,400 J

And that is for 1°C, we need to give that amount 20 times (to increase 20°C) this is:

20*8,400 J = 168,000 J

Then the total amount of energy that we must give is:

E = 672,000 J + 168,000 J = 840,000 J

If you want to learn more, you can read:

brainly.com/question/12474790

5 0
3 years ago
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