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soldi70 [24.7K]
3 years ago
15

If an object floats, the volume of the displaced water is equal to the volume of the whole object. True or false?

Physics
1 answer:
azamat3 years ago
8 0

Answer:

True

Explanation:

This is in fact what the Archimedes principle states, and what he used in the famous anecdote about finding if  the crown of the King was really made of gold.

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If your front lawn is 18.0 feet wide and 20.0 feet long, and each square foot of lawn accumulates 1050 new snowflakes every minu
Sedbober [7]

Answer:

47628 kg/hr

Explanation:

Total area = 18*20 =360 ft^(2)

no .of snowflakes per minute = 1050*360 =378000

mass of snowflakes per minute = 378000*2.1*10^(-3) =793.8 kg/min

mass accumulated per hour = 793.8kg/min * 60min/hr =47628 kg/hr

7 0
3 years ago
Which ia the best example of potential energy
sweet-ann [11.9K]
Without the choices this is difficult to answer but some examples of potential energy are:

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- Stretching a rubber band and holding it. It has potential energy.

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7 0
3 years ago
Read 2 more answers
Is this right? Please tell me why its wrong or right
madam [21]
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8 0
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Determine the mass of the earth from the known period and distance of the moon.
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6 0
3 years ago
Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 25 kg and the larger bottom crate has a m
faust18 [17]

Answer:

Explanation:

Let the bigger crate be in touch with the ground which is friction less. In the first case both m₁ and m₂ will move with common acceleration because m₁ is not sliding over m₂.

1 ) Common acceleration a = force / total mass

= 234 / ( 25 +91 )

= 2.017 m s⁻².

2 ) Force on m₁ accelerating it , which is nothing but friction force on it by m₂

= mass x acceleration

= 25 x 2.017

= 50.425 N

The same force will be applied by m₁ on m₂ as friction force which will act in opposite direction.

3 ) Maximum friction force that is possible between m₁ and m₂

= μ_s m₁g

= .79 x 25 x 9.8

= 193.55 N

Acceleration of m₁

= 193 .55 / 25

= 7.742 m s⁻²

This is the common acceleration in case of maximum tension required

So tension in rope

= ( 25 +91 ) x 7.742

= 898 N

4 ) In case of upper crate sliding on m₂ , maximum friction force on m₁

=  μ_k m₁g

= .62 x 25 x 9.8

= 151.9 N

Acceleration of m₁

= 151.9 / 25

= 6.076 m s⁻².

6 0
2 years ago
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