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3 years ago

Iron and steel went out c

Physics

1 answer:

Anton [14]3 years ago

7 0

Explanation:

a) 50km

b) 6hrs

c) speed a to b is 10km/h

b to c is rest

c to d is 20km/h, hence fastest.

d) 2 hours

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ser-zykov [4K]

Answer:

How was the result of your physical fitness test in Flexibility and muscular

strength compared to your previous assessments?

Explanation:

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3 years ago

A rubber ball that sits motionless near the edge of a tall bookshelf has no kinetic energy. However, it does have mechanical

klio [65]

It is possible because the rubber ball has mechanical energy which is equal to potential energy.

<h3>What is mechanical energy?</h3>

Mechanical energy of an object is the total energy possessed by the object, including the potential energy and the kinetic energy.

M.A = K.E + P.E

<h3>What is kinetic energy?</h3>

Kinetic energy is the energy possessed by an object due to its motion.

<h3>What is potential energy?</h3>

Potential energy is the energy possessed by an object due to its position.

When kinetic energy (K.E) = 0

M.A = P.E

Thus, it is possible because the rubber ball has mechanical energy which is equal to potential energy.

Learn more about mechanical energy here: brainly.com/question/24443465

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3 years ago

A mango fruit of mass 20g hangs on a tree 8m high. Calculate its potential energy.

steposvetlana [31]

Answer:

1.568 J

Explanation:

PEg= mgΔh

= 0.02×9.8×8

= 1.568 J

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4 years ago

A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±qseparated by a distance s. T

atroni [7]

Answer:

The magnitude of the force on the dipole due to the charge Q = $\rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.$

The magnitude of the torque on the dipole = $\rm \dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi}\dfrac{2qQs^2}{r^3}.$

Explanation:

Given that a point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q, separated by a distance s and the charge Q is located in the plane that bisects the dipole.

The magnitude of the electric field that the dipole exerts at the position where the charge Q is held is given by

$\rm E = \dfrac{k2qs}{(r^2+s^2)^{3/2}}.$

<em>where</em>,

k is the Coulomb's constant, having value = $\dfrac{1}{4\pi \epsilon_o}$

$\epsilon_o$ is the electrical permittivity of free space.

Also, r>>s, therefore, $\rm r^2+s^2\approx r^2.$

$\rm E = \dfrac{k2qs}{(r^2)^{3/2}}=\dfrac{k2qs}{r^3}.$

The magnitude of the electric force F on a charge q placed at a point and the magnitude of the electric field E at that point are related as

$\rm F=qE$

Therefore, the electric force on the charge Q due to the dipole is given by

$\rm F=Q\dfrac{k2qs}{r^3}=\dfrac{1}{4\pi \epsilon_o}\dfrac{2qQs}{r^3}.$

According to Newton's third law of motion, the magnitude of the force exerted by the dipole on the charge Q is same as the magnitude of the force exerted by the charge on the dipole.

Thus, the magnitude of the force on the dipole due to the charge Q = $\dfrac{1}{\epsilon_o}\times \dfrac{1}{4\pi }\dfrac{2qQs}{r^3}.$