Is the magnitude of buoyant force same in all liquids????

Physics

1 answer:

valentina_108 [34]3 years ago

8 0

Answer:

No. Buoyant force is equal to the weight of the displaced fluid.

B = mg

If ρ is the density of the fluid and V is the displaced volume:

B = ρVg

For fully submerged objects, where the displaced volume equals the volume of the object, buoyant force will be different in different fluids.

For floating objects, buoyant force equals the weight of the object, so it will be the same for any liquid that the object floats in.

You might be interested in

A waitperson carrying a tray with a platter on it tips the tray at an angle of 12° below the horizontal. If the gravitational fo

Nadusha1986 [10]

Answer:

The answer is 1.0 N

Explanation:

inclination of tray=12^{\circ}

gravitational Force=5 N

Now this gravitational force has two component i.e.

5\sin \theta is parallel to the tray =1.039 N

5\cos \theta is perpendicular to the tray =4.890 N

7 0

3 years ago

Julie throws a ball to her friend Sarah. The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial spe

sattari [20]

I can't make sense of this question. Julie's throwing the ball, so it's leaving her rather than arriving at her ???

3 0

3 years ago

Which statements describe the results of the effect of angle of insolation on absorption experiment? Check all that apply.

lilavasa [31]

Answer:

At 6 minutes, the soil temperature at a 45° angle of insolation is higher than at 0°.

At 15 minutes, the soil temperature at a 90° angle of insolation is higher than at 45°

Explanation:

5 0

3 years ago

Read 2 more answers

a train is moving with an initial velocity of 30 m/s, the brakes are applied so as to produce a uniform acceleration of -1.5 m/s

Pepsi [2]

Answer:

$\boxed{\sf Time \ in \ which \ train \ will \ come \ to \ rest = 20 \ sec}$

Given:

Initial velocity (u) = 30 m/s

Final speed (v) = 0 m/s

Acceleration (a) = - 1.5 m/,s²

To Find:

Time in which train will come to rest (t).

Explanation:

$\sf From \ equation \ of \ motion: \\ \sf \implies \bold{v = u + at} \\ \\ \sf Substituting \ value \ of \ v, \ u \ and \ a: \\ \sf \implies 0 = 30 + ( - 1.5)(t) \\ \sf \implies 0 = 30 - 1.5(t) \\ \sf \implies 30 - 1.5(t) = 0 \\ \\ \sf Subtract \: 30 \: from \: both \: sides: \\ \sf \implies (30 - \boxed{ \sf 30}) - 1.5(t) = \boxed{ \sf - 30} \\ \\ \sf 30 - 30 = 0 : \\ \sf \implies - 1.5(t) = - 30 \\ \\ \sf Divide \: both \: sides \: of \: - 1.5(t) = - 30 \: by \: - 1.5 : \\ \sf \implies \frac{ - 1.5(t)}{ \boxed{ \sf - 1.5}} = \frac{ - 30}{ \boxed{ \sf -1.5 }} \\ \\ \sf \frac{ \cancel{ \sf 1.5}}{\cancel{ \sf 1.5}} = 1 : \\ \sf \implies t = \frac{ - 30}{ - 1.5} \\ \\ \sf \frac{ - 30}{ - 1.5} = \frac{\cancel{ \sf 1.5} \times 20}{\cancel{ \sf 1.5}} = 20 : \\ \sf \implies t = 20 \: sec$