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Ray Of Light [21]
2 years ago
7

A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?

Physics
1 answer:
dexar [7]2 years ago
3 0

Answer:

1.08 s

Explanation:

From the question given above, the following data were obtained:

Height (h) reached = 1.45 m

Time of flight (T) =?

Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:

Height (h) = 1.45 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1.45 = ½ × 9.8 × t²

1.45 = 4.9 × t²

Divide both side by 4.9

t² = 1.45/4.9

Take the square root of both side

t = √(1.45/4.9)

t = 0.54 s

Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).

Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:

Time (t) taken to reach the height = 0.54 s

Time of flight (T) =?

T = 2t

T = 2 × 0.54

T = 1.08 s

Therefore, it will take the kangaroo 1.08 s to return to the earth.

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Answer:

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f = 662 Hz

Explanation:

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x =  52 cm * 1 / 100m = 0.52m

The wave length is the distance between the peak and peak so

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d = 2*0.52 m

d = 1.04 m

So the frequency in the speed velocity is

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2 years ago
Say you want to make a sling by swinging a mass M of 1.7 kg in a horizontal circle of radius 0.048 m, using a string of length 0
DENIUS [597]

Answer:

Tension in the string is equal to 58.33 N ( this will be the strength of the string )

Explanation:

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radius of the circle r = 0.48 mF=\frac{mv^2}{r}=\frac{1.7\times 4.05^2}{0.48}=58.33N

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Kinetic energy is equal to KE=\frac{1}{2}mv^2

So \frac{1}{2}\times 1.7\times v^2=14

v^2=16.47

v = 4.05 m/sec

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Answer:

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8 0
2 years ago
Read 2 more answers
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