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Helen [10]
2 years ago
12

An iron bal of mass 20kg is rolling on a flat surface. On applying force, the velocity change from 17ms to 27m/s in 5s. Calculat

e the magnitude of the force. (Ans:40N)
Physics
1 answer:
pentagon [3]2 years ago
4 0

Answer:

40 N

Explanation:

We first need to calculate the acceleration of the tron ball.

Since acceleration, a = (v - u)/t where u = initial velocity of iron ball = 17m/s, v = final velocity of iron ball = 27m/s and t = time taken for the change in velocity = 5 s.

So, a = (v - u)/t

= (27 m/s - 17 m/s)/5 s

= 10 m/s ÷ 5 s

= 2 m/s²

We know force on iron ball, F = ma where m = mass of iron ball = 20 kg and a = acceleration = 2 m/s²

So, F = ma

= 20 kg × 2 m/s²

= 40 kgm/s²

= 40 N

So, the magnitude of the force on the iron ball is 40 N.

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If a 25 kg lawnmower produces 347 w and does 9514 J of work, for
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Steps 1 and 2)

The variables are W = work, P = power, and t = time. In this case, W = 9514 joules and P = 347 watts.

The goal is to solve for the unknown time t.

-----------------------

Step 3)

Since we want to solve for the time, and we have known W and P values, we use the equation t = W/P

-----------------------

Step 4)

t = W/P

t = 9514/347

t = 27.4178674351586

t = 27.4 seconds

-----------------------

Step 5)

The lawn mower ran for about 27.4 seconds. I rounded to three sig figs because this was the lower amount of sig figs when comparing 9514 and 347.

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Note: we don't use the mass at all

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In a pool game, the cue ball, which has an initial speed of 3.0 m/s, make an elastic collision with the eight ball, which is ini
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Explanation:

Given

initial speed(u)=3 m/s

mass of each ball is m

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In elastic collision Energy and momentum is conserved

Let u be the initial velocity and v_1 , v_2 be the final velocity of 8 ball and cue ball respectively

\frac{mu^2}{2}+0=\frac{mv^2_1}{2}+\frac{mv^2_2}{2}

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3=v_1cos40+v_2cos50

Along Y axis

0+0=v_1sin40-v_2sin50

v_1sin40=v_2sin50

substitute the value of v_1

we get v_2=1.912 m/s

v_1=2.27 m/s

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