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kolbaska11 [484]
3 years ago
8

A spherical ball is dropped through a liquid, explain why it reaches terminal velocity.

Physics
1 answer:
Alekssandra [29.7K]3 years ago
8 0

Probably because of the drag coefficient and the density of the liquid.

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A brick is dropped from a high scaffold. a. What is its velocity after 4.0s ?
Ilia_Sergeevich [38]

Answer:

A: 1.962

B: 3.924

Explanation:

g = G *M /R^2

g = 9.807*M/R^2 the gravitational constant of ground level on earth is about 9.807

g = 9.807*5lbs/R^2 the average brick is about 5 pounds.

g = 9.807*5*10^2.   I'm assuming the height is around ten feet to help you out.

with these numbers plugged in you get an acceleration of 0.4905 a final velocity after 4 seconds 1.962. It's height fallen after 4 seconds is 3.924.

( M = whatever the brick weighs it's not specified in the question)

(R = the distance from the ground or how high the scaffold is)

(hopefully you can just plug your numbers in there hope this helps)

6 0
3 years ago
A box is sliding down an incline tilted at a 11.1° angle above horizontal. The box is initially sliding down the incline at a sp
raketka [301]

Answer:s=0.68 m

Explanation:

Given

Inclination \theta =11.1^{\circ}

Speed of block(u)=1.6 m/s

Coefficient of kinetic Friction \mu _k=0.39

deceleration provided by friction=g\sin \theta -\mu _kg\cos \theta [/tex]

Using v^2-u^2=2as

Final velocity v=0

0-1.6^2=2(g\sin \theta -\mu _kg\cos \theta )s

s=\frac{-1.6^2}{2\cdot (9.8\sin 11.1-0.39\times 9.8\times \cos 11.1)}

s=0.68 m

5 0
3 years ago
Plants absorb water from the soil through their roots. _______ of this water is used for photosynthesis; _______ of this water e
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THE CORRECT CHOICE IS B
3 0
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An astronaut on Pluto attaches a small brass ball to a 1.00-m length of string and makes a simple pendulum. She times 10 complet
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Answer:

The acceleration due to gravity at Pluto is 0.0597 m/s^2.

Explanation:

Length, L = 1 m

10 oscillations in 257 seconds

Time period, T = 257/10 = 25.7 s

Let the acceleration due to gravity is g.

Use the formula of time period of simple pendulum

T = 2\pi\sqrt{\frac{L}{g}}\\\\25.7 = 2 \times 31.4\sqrt{\frac{1}{g}}\\\\g = 0.0597 m/s^2

7 0
3 years ago
How many items are present in the compound (NH4)2Cr2O7
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