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2 years ago

An object moves in a circle with a period of 0.025 hours. What is its frequency in Hz?

Physics

1 answer:

Sedaia [141]2 years ago

7 0

Answer:

40 Hz

Explanation:

f = 1/T = 1 / 0.025 = 40 Hz

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What keeps both the cars pressed down on the road?

mihalych1998 [28]

Answer:

Gravity

Explanation:

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3 years ago

What is the atomic weight of a penny?

Gnom [1K]

I actually know the answer to this one, you use pennies to find the atomic weight of a penny, it really doesn't have a weight. LOL

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3 years ago

Read 2 more answers

A dolphin in an aquatic show jumps straight up out of the water at a velocity of 13.0 m/s.

SVEN [57.7K]

Answer:

a) Knowns, initial speed $v_{i}=13.0 m/s$ , final speed $v_{f}=0 m/s$ and gravity due it is a constant $g=9.8m/s^{2}$

b) The maximum high reached by the dolphin is $y_{max}=8.62 m$

c) Total time is $t=2.65s$

Explanation:

a) First of all the initial speed is given at the start of the problem, gravity is constant and final speed is known any object thrown straight up reaches its max high at $0m/s$ speed.

b) Second, now that we know final speed we use $v_{f} =v_{i}-gt$ , as we clear for $t=\frac{v_{i}-v_{f} }{g}=\frac{20.0m/s}{9.8m/s^{2} }=1.32 s$ .

Then we use $y=v_{i}t-\frac{1}{2} gt^{2}=(20.0m/s)(1.32s)-\frac{1}{2} (9.8m/s^{2} ) (1.32s)^{2} =8.62m$

c)Third, finally we can use $y=v_{i}t-\frac{1}{2} gt^{2}$ , as we know $y=0m$ when the dolphin fall into the water again and $v_{i} =13.0m/s$ , then we have $0=(13m/s)t-\frac{1}{2} (9.8m/s^{2} )t^{2}$ is a quadratic form $0=t(13.0-4.9t)$ so we have $t_{i}=0s$ and $t_{f}=\frac{13}{4.90} =2.65s$

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3 years ago

What is a successful result of science

Vlada [557]

I think the answer is discovery.

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2 years ago

On the moon, the acceleration of gravity is 1.6 m/s2. If an object has a

Amanda [17]

Answer:

5.1 kg

Explanation:

Its mass on the moon is 5.1 kg because mass is an intrinsic property of a material and does not change with location. Although, its weight might vary because its acceleration of gravity g is dependent on the mass M and radius r of the planet(in this case, moon) involved g = GM/r². Since weight W = mg is dependent o g, weight varies but mass remains constant.

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3 years ago