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4 years ago

A motorcycle travels 889 meters in 15 seconds. What is its average speed?______m/s

2 answers:

8 0

So here you would have to do
distance = time x speed.
distance is 889 meters
time is 15 seconds.
so 889 = 15 x speed
divide 15 on both sides.
889/15
Average speed is 59.2666667 m/s
if it says in the question to round it then just round it.
But whilst checking to see if this answer is correct, i did 59.2666667 multiplied by 15, which equals 889
hope this helped.

german4 years ago

6 0

The answer is about 59.3 m/s

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Given

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3 years ago

023 (part 1 of 2) 10.0 points

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Answer:

Part 1

The angular speed is approximately 1.31947 rad/s

Part 2

The change in kinetic energy due to the movement is approximately 675.65 J

Explanation:

The given parameters are;

The rotation rate of the merry-go-round, n = 0.21 rev/s

The mass of the man on the merry-go-round = 99 kg

The distance of the point the man stands from the axis of rotation = 2.8 m

Part 1

The angular speed, ω = 2·π·n = 2·π × 0.21 rev/s ≈ 1.31947 rad/s

The angular speed is constant through out the axis of rotation

Therefore, when the man walks to a point 0 m from the center, the angular speed ≈ 1.31947 rad/s

Part 2

Given that the kinetic energy of the merry-go-round is constant, the change in kinetic energy, for a change from a radius of of the man from 2.8 m to 0 m, is given as follows;

$\Delta KE_{rotational} = \dfrac{1}{2} \cdot I \cdot \omega ^2 = \dfrac{1}{2} \cdot m \cdot v ^2$

I = m·r²

Where;

m = The mass of the man alone = 99 kg

r = The distance of the point the man stands from the axis, r = 2.8 m

v = The tangential velocity = ω/r

ω ≈ 1.31947 rad/s

Therefore, we have;

I = 99 × 2.8² = 776.16 kg·m²

$\Delta KE_{rotational}$ = 1/2 × 776.16 kg·m² × (1.31947)² ≈ 675.65 J

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