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Snezhnost [94]
2 years ago
5

A cylinder with a radius of 1.8 cm is floating in water as shown below. The mass of the cylinder is 1.80 kg. Calculate the depth

of the bottom end of the cylinder
Physics
1 answer:
Westkost [7]2 years ago
6 0

Answer:

Explanation:

I will ASSUME that the cylinder floats with the axis through the center of its flat ends is perpendicular to the water surface.

The cylinder will displace it's mass in water.

water density has a value of 1000 kg/m³

if the depth of the bottom is "h" in meters

1.80 kg = π(0.018² m²)(h m)(1000) kg/m³

h = 1.768388256...

or about 1.77 m

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Answer:

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Explanation:

Because it is impossible for it to show the real depth of the ocean and how deep it is

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Which of the following are true for acceleration?
Fittoniya [83]

The SI unit for acceleration is m/s2 ( D)

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Suppose that you can throw a projectile at a large enough v0 so that it can hit a target a distance R downrange. Given that you
NikAS [45]

Answer:

Theta1 = 12° and theta2 = 168°

The solution procedure can be found in the attachment below.

Explanation:

The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).

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3 years ago
Pete Zaria works on weekends at Barnaby's Pizza Parlor. His primary responsibility is to fill drink orders for customers. He fil
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Answer:

W_n_e_t=7.648512 \approx 7.6J

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v=0.7844645406 \approx 0.78m/s

Explanation:

From the question we are told that

Mass of pitcher   M= 2.6kg

Force on pitcher f=8.8N

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Coefficient of friction \mu=0.28

a)Generally frictional force is mathematically given by

F=\mu N

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F=7.1344N

Generally work done on the pitcher is mathematically given as

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W_f=8.8*0.48=>4.224N\\W_F=7.1344*0.48=>3.424512N

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K.E= W_n_e_t

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Velocity as subject

v=\sqrt{\frac{K.E*2}{m} }

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