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2 years ago

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

Physics

1 answer:

valentina_108 [34]2 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.

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An object with mass M is attached to the

zysi [14]

Answer:

W = 5/4 Mgℓ

Explanation:

Sum of forces:

∑F = ma

T − Mg = M (g/4)

T = 5/4 Mg

Work = force × distance

W = Tℓ

W = 5/4 Mgℓ

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2 years ago

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A basketball is filled with air and has a small hole. Explain why the basketball would leak.

Assoli18 [71]

Because it has a small hole and air is like water, an area of higher pressure wants to move to a lower pressure area.

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What traction of the radioisotope<br>remains in the body after one day?

r-ruslan [8.4K]

The fraction of radioisotope left after 1 day is $(\frac{1}{2})^{\frac{1}{\tau}}$ , with the half-life expressed in days

Explanation:

The question is incomplete: however, we can still answer as follows.

The mass of a radioactive sample after a time t is given by the equation:

$m(t)=m_0 (\frac{1}{2})^{\frac{t}{\tau}}$

where:

$m_0$ is the mass of the radioactive sample at t = 0

$\tau$ is the half-life of the sample

This means that the mass of the sample halves after one half-life.

We can rewrite the equation as

$\frac{m(t)}{m_0}=(\frac{1}{2})^{\frac{t}{\tau}}$

And the term on the left represents the fraction of the radioisotope left after a certain time t.

Therefore, after t = 1 days, the fraction of radioisotope left in the body is

$\frac{m(1)}{m_0}=(\frac{1}{2})^{\frac{1}{\tau}}$

where the half-life $\tau$ must be expressed in days in order to match the units.

Learn more about radioactive decay:

brainly.com/question/4207569

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3 years ago

~~~~NEED HELP ASAP~~~~

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the above three pictures may help you

go through the attachments

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2 years ago

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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:

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

So,

Time in which train will come to rest = 20 seconds

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2 years ago