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attashe74 [19]
2 years ago
7

Imagine holding two identical bricks under water. Brick A is just beneath the surface of the water, while brick B is at a greate

r depth. What is the force needed to hold brick B in place? (Assume the density of water doesn't change with height.) a) the same as b) larger c) smaller than the force required to hold brick A in place.
Physics
1 answer:
FromTheMoon [43]2 years ago
4 0

Answer:

I think it's c

Explanation:

The brick's mass, helps to hold it under the water but brick A is being held up. If you dropped the brick, it would sink to the bottom. You need a stronger force to hold it up.

Hope this helps you! x

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Why must lslamic believers travel to Mecca once during their lives?
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Hajj is very important to islam people so in order to do this they had to go to Mecca. Hajj is rituals and rites that need to be fulfilled. Its not the only pillar for islam but it is one out of five.

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3 years ago
Answer <br> A)<br> B <br> C <br> please
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1 year ago
Blood plasma volume for adults is about 3.1 liters. It has a density of 1.020 gml. How many pounds of blood plasma are in your b
kari74 [83]

Answer:

mass is 6.97 pounds

Explanation:

given data

volume = 3.1 liters

density = 1.020 g/ml = 1.02 kg/l

to find out

How many pounds of blood plasma

solution

we know mass formula from density that is

density = mass / volume

so

mass = density × volume    ...............1

so put all value to get mass

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6 0
3 years ago
A factory worker pushes a crate of mass 31.0 kg a distance of 4.35 m along a level floor at constant velocity by pushing horizon
Debora [2.8K]

Answer:

a. 79.1 N

b. 344 J

c. 344 J

d. 0 J

e. 0 J

Explanation:

a. Since the crate has a constant velocity, its net force must be 0 according to Newton's 1st law. The push force F_p by the worker must be equal to the friction force F_f on the crate, which is the product of friction coefficient μ and normal force N:

Let g = 9.81 m/s2

F_p = F_f = \mu N = \mu mg = 0.26 * 31 * 9.81 = 79.1 N

b. The work is done on the crate by this force is the product of its force F_p and the distance traveled s = 4.35

W_p = F_ps = 79.1*4.35 = 344 J

c. The work is done on the crate by friction force is also the product of friction force and the distance traveled s = 4.35

W_f = F_fs = -79.1*4.35 = -344 J

This work is negative because the friction vector is in the opposite direction with the distance vector

d. As both the normal force and gravity are perpendicular to the distance vector, the work done by those forces is 0. In other words, these forces do not make any work.

e. The total work done on the crate would be sum of the work done by the pushing force and the work done by friction

W_p + W_f = 344 - 344 = 0 J

8 0
3 years ago
Read 2 more answers
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