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Minchanka [31]
3 years ago
10

10. During 4th period we put Klaudia in a box because she was talking too much. We still heard her voice through the box so we d

ecided to push her outside. The force of friction of the ground on the box was 68 N. If Mr.Whitmore can apply a force of 25 N and every other 7th grade student can apply a force of 6 N. How many students would Mr. Whitmore need to make the box start moving and go outside. (Think quickly, the faster we move the box out, the quicker she stops talking)
Physics
1 answer:
Alex787 [66]3 years ago
6 0

Answer: Mr. Whitmore would need 7 or more students ( 7.17) to make the box start moving and go outside

Explanation:

Given that;

friction force of ground box = 68 N

student of 7th grade = n

Whitmore can apply a force of 25 N

every other 7th grade student can apply a force of 6 N.

now

friction force = forced applied by whitmore + total force ny 7th grade student

we substitute

68 = 25 + 6n

6n = 68 - 25

6n = 43

n = 43/6

n = 7.17

Therefore Mr. Whitmore would need 7 or more students ( 7.17) to make the box start moving and go outside

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Explanation:

When water is boiled in the flask . Some portion of it is evaporated out . Now when cork is placed on it and is placed in the ice box . It cools down , by which the pressure inside decreases .

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3 years ago
Is it possible to have a net torque when all of the forces sum to zero? Explain.
nexus9112 [7]

Answer:

Yes it is possible

Explanation:

When two equal magnitude forces are acting on the rod in opposite direction

Then the net force on the system is always zero in that case

so we will have

F - F = 0

now for the system net torque due to these forces is given by

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here we know that

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3 years ago
A 46.8-g golf ball is driven from the tee with an initial speed of 58.8 m/s and rises to a height of 24.7 m. (a) Neglect air res
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Answer:

a) the kinetic energy of the ball at its highest point is 69.58 J

b) its speed when it is 8.11 m below its highest point is 55.97 m/s

Explanation:

Given that;

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initial speed of the ball v₁ = 58.8 m/s

height h = 24.7 m

acceleration due to gravity = 9.8 m/s²

the kinetic energy of the ball at its highest point = ?

from the conservation of energy;

Kinetic energy at the highest point will be;

K.Ei + P.Ei = KEf + PEf

now the Initial potential energy of the ball P.Ei = 0 J

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1/2mv² + 0 J = KEf + mgh

K.Ef = 1/2mv² - mgh

we substitute

K.Ef = [1/2 × 0.0468 × (58.8 )²] - [0.0468 × 9.8 × 24.7]

K.Ef  = 80.904 - 11.3284

K.Ef = 69.58 J

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so our velocity will be v₂

also using the principle of energy conservation;

K.Ei + P.Ei = KEh + PEh

1/2mv² + 0 J = 1/2mv₂² + mgh'

1/2mv₂² = 1/2mv² - mgh'

multiply through by 2/m

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v₂ = √( (58.8)² - 2×9.8×16.59 )

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Answer:

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