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Westkost [7]
3 years ago
10

During a goal-line stand, a 112-kg fullback moving eastward with a speed of 6 m/s

Physics
1 answer:
11111nata11111 [884]3 years ago
7 0

Answer:

-1.24 m/s

Explanation:

Total momentum before collision = total momentum after collision

Total momentum before collision = (mass of full back * velocity of fullback) + (mass of lineman * velocity of line man).

Mass of full back = 112 kg, mass of line bag = 120 kg, velocity of full back  6 m/s (east), velocity of line back = -8 m/s (west). Hence:

Total momentum before collision = (112 * 6) + (120 * -8) = 672 - 960 = -288 kgm/s

The total momentum after collision = (mass of full back + mass of line back) * velocity after collision.

Let velocity after collision be v, hence:

The total momentum after collision = (112 + 120)v = 232v

Total momentum before collision = total momentum after collision

-288 = 232v

v = -288 / 232

v = -1.24 m/s

Therefore after collision, the two players would move at a velocity 1.24 m/s west (the same direction as the lineman).

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the sole of a tennis shoe has a surface area of 0.0290 m^2. if it is worn by a 65.0 kg person, what pressure does the shoe exert
AURORKA [14]

Answer: 21965.517 Pa

Explanation:

Pressure P is the force F exerted by a gas, a liquid or a solid on a surface (or area) A, its unit is Pascal Pa which is equal to N/m^{2} and its formula is:  

P=\frac{F}{A} (1)

In this case we have the surface of a sole of a tennis shoe:

A=0.0290 m^{2} (2)

And the mass m of the person who wears it:

m=65 kg

On the other hand, we know the weight is the force  F the Earth exerts on people and objects due gravity g :

F=m.g=(65 kg)(9.8m/^{2})

F=637N (3)

Substituting (2) and (3) in (1):

P=\frac{637N}{0.0290 m^{2}} (4)

Finally:

P=21965.517 Pa This is the pressure the shoe exert on the ground

5 0
3 years ago
What impulse occurs when an average force of 7.0 N is exerted on a cart for 2.5 s ?
makkiz [27]

Answer:1.7\times 10

Explanation:

Given

Force F=7 N

time interval t=2.5 s

Impulse is given by =Force \times time\ interval\ for\ applied\ force

Impulse=7\times 2.5=17.5

For two significant Figure

Impulse=1.7\times 10

                       

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The answer should be 88.56
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