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3 years ago

A person swims to the other end of a 20m long pool and back. What is their displacement?

Physics

1 answer:

mafiozo [28]3 years ago

3 0

Answer:

Zero

Explanation:

It is given that,

A person swims to the other end of a 20m long pool and back.

We need to find his displacement.

Displacement = shortest path covered

He reaches at the same position as from where he has started. It means the shortest path covered is equal to 0 i.e. his displacement is zero.

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Explain the uses of Electrophorus <br>

Maslowich

Answer:

Explanation:

Comment

Interesting creature.

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2 years ago

Astronomers think most galaxy interactions took place at redshifts of greater than 1 because

Evgesh-ka [11]

Because they are moving away from us.

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3 years ago

A 0.2 kg cannon ball is fired at an upward angle of 45° from the top of a 165 m cliff with a speed of 175 m/s. (A) Using conserv

sergejj [24]

To solve this problem, it is necessary to apply the concepts related to the work done by a body when a certain distance is displaced and the conservation of energy when it is consumed in kinetic and potential energy mode in the final and initial state. The energy conservation equation is given by:

$\Delta KE_i + \Delta PE_i = \Delta KE_f + \Delta PE_f$

Where,

KE = Kinetic Energy (Initial and Final)

PE = Potential Energy (Initial and Final)

And the other hand we have the Work energy theorem given by

$\Delta KE = W = F*d$

Where

W= Work

F = Force

D = displacement,

PART A) Using the conservation of momentum we can find the speed, so

$\Delta KE_i + \Delta PE_i = \Delta KE_f + \Delta PE_f$

$\frac{1}{2}mv_1^2 + mgh_i = \frac{1}{2}mv_f^2+mg_h2$

The height at the end is 0m. Then replacing our values

$\frac{1}{2}mv_1^2 + mgh_i = \frac{1}{2}mv_f^2+0$

Deleting the mass in both sides,

$\frac{1}{2}v_1^2 + gh_i = \frac{1}{2}v_f^2$

Re-arrange for find $v_f^2,$

$v_f^2= 2(gh_i)+v_1^2$

$v_f^2 = 2(9.8*165)+(175)^2$

$v_f=\sqrt{33859}$

$v_f = 184.008m/s$

PART B) Applying the previous Energy Theorem,

$\Delta KE = W = F*d$

$\frac{1}{2}mv^2 = F*d$

$\frac{1}{2}(0.2)(184.008)^2 = (75)*d$

Solving for d

$d = 45.15 m$

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3 years ago

Calculate the electric field strength required to just support an oil drop of mass

Ray Of Light [21]

Answer:

$E= 3.06 \times 10^{-17} N/C$

Explanation:

As we know that the oil drop experiment tells us the balancing the gravitational force by electrostatic force.

So ,

Electrostatic force = Gravitational force

qE=mg

Here , q is charge , E is electric field , m is mass and g is gravity.

so,

$E=\frac{mg}{q}$

Insert values from question,

$E=\frac{10^{-2}kg \times 9.8 m/s^{2}}{3.2 \times 10^{-19}C}$

$E= 3.06 \times 10^{-17} N/C$

This is the required value of electric field.

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3 years ago

The force of gravity depends mainly on objects _______?

geniusboy [140]

The strength of the gravitational forces between two objects depends on
the masses of both objects and on the distance between their centers.

I don't think you can say that one or the other is the "<em>main</em>" influence.

-- If each mass is multiplied by ' k ', then the forces between them are
multiplied by ' k² '.

-- And if the distance between them is multiplied by ' k ', then the forces
between them are divided by ' k² '.

Seems pretty equal to me.

5 0

3 years ago

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