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Rudiy27
2 years ago
15

If a wood is drowned in water up to 2/3 of its volume and the same wood is dipped in a oil upto 9/10 of its volume.What's the de

nsity of wood and the relative oil?
Density of water -1000kgm-3
Physics
1 answer:
MrMuchimi2 years ago
7 0
S.G. or specific gravity is the ratio of the density of an object to a fluid. It also is the fraction of the object's volume that would be submerged if the object were placed in the fluid. (note: if S.G.>1, the object will sink)

S.G. = density(wood) / density(water)

2/3 = density(wood) / 1000

density(wood) = 2000/3 kg/m^3 = ~666.67 kg/m^3

S.G. = density(wood) / density(oil)

9/10 = 2000/3 / density(oil)

density(oil) = 2000/3 / 9/10 = 20000/27 kg/m^3 = ~740.74 kg/m^3
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8 0
3 years ago
Suppose a grower sprays (2.2x10^1) kg of water at 0 °C onto a fruit tree of mass 180 kg. How much heat is released by the water
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<h3>Heat released by the water when it freezes</h3>

The heat released by the water when it freezes is calculated as follows;

Q = mcΔФ

where;

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Initial temperature of water, Фi = 0 °C

when water freezes, the final temperature, Фf = 0 °C

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Learn more about heat flow here: brainly.com/question/14437874

6 0
2 years ago
Please help! I'm not sure what equation or the process to do this question.
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Answer:

The momentum is 1.94 kg m/s.

Explanation:

To solve this problem we equate the potential energy of the spring with the kinetic energy of the ball.

The potential energy U of the compressed spring is given by

U = \dfrac{1}{2} kx^2,

where x is the length of compression and k is the spring constant.

And the kinetic energy of the ball is

K.E = \dfrac{1}{2}mv^2.

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\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2,

solving for v we get:

v = x \sqrt{\dfrac{k}{m} }.

And since momentum of the ball is p=mv,

p =mx \sqrt{\dfrac{k}{m} }.

Putting in numbers we get:

p =(0.5kg)(0.25m) \sqrt{\dfrac{(120N/m)}{0.5kg} }.

\boxed{p=1.94kg\: m/s}

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