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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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A capacitor in a single-loop RC circuit is charged to 85% of its final potential difference in 2.4 s. What is the time constant
atroni [7]

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Explanation:

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The mathematically representation for voltage potential of a capacitor at different time is

        V  =  V_o  - e^{-\frac{t}{\tau} }

Where  \tau  is the time constant  

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     So  the capacitor potential will be  100%  when it is full thus  V_o  =100%  =  1  

and from the  question we are told that the  at the given time the potential of the capacitor is 85% = 0.85 of its final potential so

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Hence

     0.85 =  1 -  e^{-\frac{2.4}{\tau } }

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3 years ago
Vector ~A has a magnitude of 29 units and points in the positive y-direction. When vector ~B is added to ~A, the resultant vecto
timurjin [86]

Answer:

The magnitude of vector B is 43 units and it points in the negative y-direction.

Explanation:

Resultant of vectors = vector sum of all the vectors

Vector A = 29j

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B = -14j - 29j = - 43j

Hence, the magnitude of vector B is 43 units and it points in the negative y-direction.

4 0
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