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liraira [26]
3 years ago
10

What is the force of a lead ball that starts with a momentum of 7 kg*m/s and end with a momentum of 22 kg*m/s over a time of 31

seconds?
Explain how you got your answer.
Physics
1 answer:
Gwar [14]3 years ago
6 0

Answer:

F = 0.483 N

Explanation:

Initial momentum, P_i= 7\ kg-m/s

Final momentum, P_f= 22\ kg-m/s

Time, t = 31 s

We need to find the force of a lead ball. We can use here the impulse momentum theorem.

\text{Change in momentum}=F\times t

F is force

P_f-P_i=Ft\\\\F=\dfrac{P_f-P_i}{t}\\\\F=\dfrac{22-7}{31}\\\\F=0.483\ N

So, the force is 0.483 N.

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An action force is 40 N to the right. The reaction force must be: A. 20 N left B. 40 N left C. 20 N right D. 40 N right
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Please help me with this question​
Alexandra [31]

Answer:

(4) A = 3 A, A₂ = 11 A

(5) 7 A

Explanation:

(4)

From the diagram,

A = 3+6+2

A = 11 A

V = A₂R

A₂ = V/R₂............ Equation 1

Given: V = 12 V,  R₂ = 4 Ω

Substitute these values into equation 1

A₂ = 12/4

A₂ = 3 A

(5) Applying,

V = IR'

I = V/R'............ Equation 1

Where V = Voltage, I = cuurent, R' = total resistance.

But,

1/R' = (1/3)+(1/4)

1/R' = (3+4)/12

1/R' = 7/12

R' = 12/7 Ω

Given: V = 12 V

Substitute these values into equation 1

I = 12/(12/7)

I = 7 A

Therefore

A = 7 A

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