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3 years ago

It is measured that 3/4 of a body's volume is submerged in oil of density 800kg/m³

Physics

1 answer:

Evgesh-ka [11]3 years ago

8 0

Complete question:

It is measured that 3/4 of a body's volume is submerged in oil of density 800kg/m³. What is the specific gravity of oil?

Answer:

The specific gravity of the oil is 0.8.

Explanation:

Given;

density of the oil, $\rho_o$ = 800 kg/m³

density of water, $\rho_w$ = 1000 kg/m³

The specific gravity of any substance is the ratio of the substance density to the density of water.

Specific gravity of the oil = density of the oil / density of water

Specific gravity of the oil = 800/1000

Specific gravity of the oil = 0.8

Therefore, the specific gravity of the oil is 0.8.

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Answer:

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2 years ago

1) You slam on the brakes of your car in a panic, and skid a certain distance on a straight level road. If you had been travelin

aleksandr82 [10.1K]

Answer:

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In that case as the car stops v_f = 0

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2 years ago

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olga55 [171]

Answer:

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Explanation:

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3 years ago

Read 2 more answers

A 2.55 kg block starts from rest on a rough inclined plane that makes an angle of 60 degree with the horizontal. the coefficient

lorasvet [3.4K]

Answer:

Change in mechanical energy = work done by friction

so it is equal to

$W = -8.16 J$

Explanation:

As we know that change in mechanical energy must be equal to the work done by non conservative forces only

So here when block moves down the inclined plane then the work done by friction force is given as

$W = F.d$

here we have

$F = \mu F_n$

here we know that

$F_n = mg cos\theta$

so we have

$F_n = 2.55(9.81)(cos60)$

$F_n = 12.5 N$

Now the friction force on the block is given as

$F_f = \mu F_n$

$F_f = 0.25 \times 12.5$

$F_f = 3.13 N$

now work done by the friction is given as

$W = -(3.13)(2.61)$

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3 years ago

A child is sitting on the seat of a swing with ropes 5 m long. Her father pulls the swing back until the ropes make a 30o angle

morpeh [17]

Answer:

v = 3.7 m/s

Explanation:

As the swing starts from rest, if we choose the lowest point of the trajectory to be the zero reference level for gravitational potential energy, and if we neglect air resistance, we can apply energy conservation as follows:

m. g. h = 1/2 m v²

The only unknown (let alone the speed) in the equation , is the height from which the swing is released.

At this point, the ropes make a 30⁰ angle with the vertical, so we can obtain the vertical length at this point as L cos 30⁰, appying simply cos definition.

As the height we are looking for is the difference respect from the vertical length L, we can simply write as follows:

h = L - Lcos 30⁰ = 5m -5m. 0.866 = 4.3 m

Replacing in the energy conservation equation, and solving for v, we get:

v = √2.g.(L-Lcos30⁰) = √2.9.8 m/s². 4.3 m =3.7 m/s

7 0

3 years ago