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Soloha48 [4]
2 years ago
6

PLEASE HELP ASAP!!! NO LINKS! Which type of wave does the illustration depict?

Physics
1 answer:
Xelga [282]2 years ago
3 0

Answer:

Transverse wave

Explanation:

The wave is moving forwards from the hand to the point of attachment

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Kinetic energy is energy of motion while temperature is a measure of that energy in substances.
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Last night Mookie Betts hit a baseball at 32.5 m/s at a 45° angle. Betts
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Answer:

a) Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) The total distance traveled by the baseball was 108.7 m.

Explanation:

a) To know if the hit was a home run we need to calculate the height of the ball at 99 m:

y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}

Where:

y_{f}: is the final height =?

y_{0}: is the initial height = 1 m

v_{0_{y}: is the initial vertical velocity = v₀sin(45)

v₀: is the initial velocity = 32.5 m/s

g: is the gravity = 9.81 m/s²

t: is the time    

First, we need to find the time by using the following equation:

t = \frac{x}{v_{0_{x}}} = \frac{99 m}{32.5 m/s*cos(45)} = 4.31 s

Now, the height is:

y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2} = 1m + 32.5 m/s*sin(45)*4.31 s - \frac{1}{2}9.81 m/s^{2}*(4.31 s)^{2} = 8.93 m      

Since the height of the baseball at 99 m was 8.93 m and the fence at that distance is 3m tall, the hit was a home run.

b) To find the distance traveled by the baseball first we need to find the time of flight:

y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}

0 = 1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2}

1 m + 32.5m/s*sin(45)t - \frac{1}{2}9.81 m/s^{2}t^{2} = 0

By solving the above quadratic equation we have:

t = 4.73 s

Finally, with that time we can find the distance traveled by the baseball:

x = v_{0_{x}}*t = 32.5 m/s*cos(45)*4.73 s = 108.7 m

Hence, the total distance traveled by the baseball was 108.7 m.

I hope it helps you!                                                                                  

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Answer:

weight at height = 100 N .

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Let g₀ and g₁ be acceleration due to gravity , m is mass of the object .

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Applying Newton's law of gravitation

mg₀ = G Mm / R²

At height h from centre

mg₁ = G Mm /h²

Given mg₀ = 400 N

400 = G Mm / R²

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G Mm = 400 x (6400 x 10³ )²

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weight at height = 100 N

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