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Assoli18 [71]
3 years ago
12

An object accelerates downward at a constant rate of 3.75 m/s2 while completely submerged in water. Neglect the viscosity (fluid

friction) of the water. The average density of the object is ______ kg/m3. The density of water is 1000 kg/m3 and g
Physics
1 answer:
guajiro [1.7K]3 years ago
7 0

Answer:

Do = density of obect      Dw = density of water

Weight of object = V g Do

Weight of water = V g Dw

F = m a = V Do a = V g Do - V g Dw

Acceleration of object = mass * force on object

Do (g - a) = Dw g

Do = Dw ( 1 / (1 - a / g) = 1000 * 1 / (1 - 3.75 / 9.80)  kg/m^3

Do = 1441 kg / m^3

If no acceleration then density of object = density of water

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Answer:

A fundamental theory that provides a description of the physical properties of nature at the scale of atoms and subatomic particles.

Explanation:

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6 0
3 years ago
In which one of the following situations is zero net work done? A) A ball rolls down an inclined plane. B) A physics student str
lidiya [134]

Answer:

Option D

Explanation:

The work done can be given by the mechanical energy used to do work, i.e., Kinetic energy and potential energy provided to do the work.

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In option D, the box is being pulled with constant velocity, making the acceleration zero and thus Kinetic energy of the system is zero. Hence work done in this case is zero.

6 0
3 years ago
Read 2 more answers
Two particles are located on the x axis. particle 1 has a mass m and is at the origin. particle 2 has a mass 2m and is at x = +l
wlad13 [49]

The solution would be like this for this specific problem:

<span>
The force on m is:</span>

<span>
GMm / x^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] -> 1

The force on 2m is:</span>

<span>
GM(2m) / (L - x)^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] -> 2

From (1), you’ll get M = 2mx^2 / L^2 and from (2) you get M = m(L - x)^2 / L^2 

Since the Ms are the same, then 

2mx^2 / L^2 = m(L - x)^2 / L^2 

2x^2 = (L - x)^2 

xsqrt2 = L - x 

x(1 + sqrt2) = L 

x = L / (sqrt2 + 1) From here, we rationalize. 

x = L(sqrt2 - 1) / (sqrt2 + 1)(sqrt2 - 1) 

x = L(sqrt2 - 1) / (2 - 1) 


x = L(sqrt2 - 1) </span>

 

= 0.414L

 

<span>Therefore, the third particle should be located the 0.414L x axis so that the magnitude of the gravitational force on both particle 1 and particle 2 doubles.</span>

8 0
3 years ago
A sailboat moves north for a distance of 10.00 km when blown by a wind from the exact south with a force of 5.00 x 10^4 n. how m
Nataliya [291]
Work is defined as the force times the distance which is mathematically expressed W = Fxd. The given force is 5x10^4 and the distance is 10000 m (the distance is converted as meter because Nm = J) the work done by the wind is W = 5x10^4 N (10000) = 500 x 10^6 Joules. I hope it answered your question
8 0
3 years ago
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