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STALIN [3.7K]
3 years ago
14

A football player throws and accelerates a ball (m = 0.150 kg) from rest to a velocity of 50 m/s in 0.0121 sec. Determine the ac

celeration of the ball first (this is not the answer) and then the force applied by the player's arm to the ball (enter this value). Neglect resistance forces.
Physics
1 answer:
Rasek [7]3 years ago
5 0

Answer:

Acceleration=4m/s²

Force applied =619.8N

Explanation:

Using equation of motion

V=u+at we have: u=o, v=50m/s

50= 0 + a×0.0121

a = 50/0.0121

a= 4m/s²

Neglecting resistance forces

F= ma, where a = v-u/t

F=m×(v-u)/t

F= 0.150 ×(50-0)/0.0121

F=7.5/0.0121

F= 619.8N

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the velocity of a car traveling in the positive direction decreases from 32 m/s to 24 m/s in 4 seconds. what is the average acce
anastassius [24]

Answer:

a=-2m/sec^2

Negative sign shows that velocity of the car is decreases at a constant rate

Explanation:

We have given velocity of the car is decreases from 32 m /sec to 24 m/sec in 4 second

So initial velocity of the car u = 32 m /sec

And finally car reaches to a velocity of 24 m/sec

Time taken to change in velocity = 4 sec

So final velocity v = 24 m/sec

From first equation of motion v = u+at

So 24=32+a\times 4

a=-2m/sec^2

Negative sign shows that velocity of the car is decreases at a constant rate

6 0
3 years ago
What is the length of AR?​
AlladinOne [14]

Answer:Assault rifle?

Explanation:

20 m16 size and a 14.5 barrel and get a pump shot and get the blue version and go to tilted and one pump even tho forkknife is garbage.:)

6 0
3 years ago
What is the strength of the electric field ep 0.90 mm from a proton?
dem82 [27]

The electric field strength is inversely related to the square of the distance.so  the strength of the electric field is given by

E=\frac{1}{4\pi \epsilon _{0}  } \frac{q}{r^{2} } = \frac{k q}{r^{2} }

Here,  \frac{1}{4\pi \epsilon _{0}  } = k  is constant depend upon medium and its value is 9.0 \times10^{9} \ N m^2/C^2 and q is charge and  r is the distance.

Given  r = 0.90 mm = 9.0 \times 10^{-4} m and we know the charge of proton, q = 1.6\times 10^{-19} \ C.

Therefore,

E=\frac{9.0 \times10^{9} \ N m^2/C^2 \times1.6\times 10^{-19} \ C  }{(9.0 \times 10^{-4} m)^2} = 0.177 \times 10^{-2} \ N/C \\\\  E= 1.77 \times 10^{-3}  N/C

4 0
3 years ago
Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is th
katrin [286]

Complete Question

Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is the nail compressed if it is 2.50 mm in diameter and 6.00-cm long.What Average force is excreted on the Nail

Answer:

F=2*10^{4}N

Explanation:

From the question we are told that:

Mass m=0.500kg

Initial Velocity V=15.0m/s

Distance x=2.80mm=>0.00280m

Diameter d=2.50mm=>0.00250m

Length l=6.00cm=>0.6m

Generally the equation for Force is mathematically given by

 F=\frac{mv^2}{2d}

 F=\frac{0.500*15^2}{2.80*10^{-3}}

 F=2*10^{4}N

6 0
3 years ago
Power is the product of_____.
tia_tia [17]
Heya user☺☺

All options are wrong here.

The correct answer is..

Work/Time.

Hope this will help☺☺
3 0
3 years ago
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