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natulia [17]
2 years ago
9

A 120 kg quarterback is stationary in the "pocket" and hit by a defensive tackle with 4000 Newtons of force. What acceleration d

id the quarterback experience after being hit
Physics
1 answer:
Leni [432]2 years ago
6 0

Answer: 33.3 m/s^2

Explanation:

Here we need to use the second Newton's law.

F = m*a

Force equals mass times the acceleration.

Here we know that the force is F = 4000 N, and the mass is 120kg, then the acceleration must be:

4000 N = 120kg* a

a = (4000/120) m/s^2 = 33.3 m/s^2

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You throw a tennis ball (mass 0.0570 kg) vertically upward. It leaves your hand moving at 15.0 m/s. Air resistance cannot be neg
Deffense [45]

Answer:195 J

Explanation:

Given

mass of ball m=0.0570\ kg

ball leaves the hand with u=15\ m/s

maximum height reached by ball h=8\ m

Initial Mechanical energy when ball just leaves the hand

M.E._1=(P.E.+K.E.)_1

M.E._1=(mgh)_1+(\frac{1}{2}mv^2)_1

considering hand to be datum so h_1=0[/tex]

so Potential energy at ground is zero

M.E._1=\frac{1}{2}\times m\times (15)^2

M.E._1=6.41\ J

Mechanical Energy at highest point

(M.E.)_2=(P.E.+K.E.)_2

at highest Point velocity is zero

(M.E.)_2=mgh_2+0

(M.E.)_2=0.0570\times 9.8\times 8

(M.E.)_2=4.46\ J

Decrease in Mechanical energy

(M.E.)_1-(M.E.)_2=6.41-4.46

(M.E.)_1-(M.E.)_2=1.95\ J

3 0
2 years ago
To what potential should you charge a 2.0 μF capacitor to store 1.0 J of energy?
Bess [88]
E = (1/2)CV²
1 = (1/2)*(2*10⁻⁶)V²
10⁶ = V²
1000 = V

You should charge it to 1000 volts to store 1.0 J of energy.
6 0
2 years ago
PLEASE HELP!!!!!! WILL GIVE BRAINLIEST TO WHOEVER ANSWERS WITH THE RIGHT ANSWER !!!!!!!! 
Solnce55 [7]

It would be B and D your welcome


7 0
2 years ago
Read 2 more answers
Physics in motion unit 6a the nature of waves
mylen [45]
What’s the question?
7 0
2 years ago
An 80-g particle moving with an initial speed of 50 m/s in the positive x direction strikes and sticks to a 60-g particle moving
liubo4ka [24]

The collision is a form of inelastic collision because the it forms a single mass after is collides. So it can be solve by momentum balance

( 0.08 kg * 50 m/s ) + ( 0.06 kg * 50 m/s) = ( 0.08 + 0.06 kg ) v

V = 50 m/s

So the kinetic energy lost is

KE = 0.5 (50 m/s)^2) *( 0.14 – 0.08kg )

KE = 75 J

8 0
2 years ago
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