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3 years ago

14

Which of these correctly describes whether a girl holding a ball in the same position is doing work on the ball?

1 answer:

aksik [14]3 years ago

4 0

Answer:

The girl is doing work on the ball because the energy in her muscles changed, even though the ball is not displaced.

Explanation:

The complete question is...

Which of these correctly describes whether a girl holding a ball in the same position is doing work on the ball?

-The girl is doing work on the ball because the energy of the ball changed, even though it is not displaced.

-The girl is doing work on the ball because the energy in her muscles changed, even though the ball is not displaced.

-The girl is doing no work on the ball because the ball is not displaced.

-The girl is doing no work on the ball because she is exerting a net force on the ball.

Holding up a ball costs energy, which is used to counter the work that would have otherwise be done on the ball by gravity. Although no physical distance is moved, we should consider the fact that by holding the ball, the girls hand exerts physical force to hold the ball in place. Also, there is a potential gravitational work on the ball due to gravity, but the force exerted by the girls hand does an equivalent of this gravito-potential work in order to counter it and hold the ball in place. All these activities eventually lead to a change in energy in her hand muscle to show that energy is expended.

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As a block of mass 42 kilograms drops from the edge of a 40-meter-high cliff it experiences a loss of energy due to air resistan

JulsSmile [24]

Answer:

<em>The block hits the ground at 27.9 m/s</em>

Explanation:

<u>Gravitational Potential Energy (GPE)</u>

It's the energy stored in an object because of its height in a gravitational field.

It can be calculated with the equation:

U=m.g.h

Where m is the mass of the object, h is the height with respect to a fixed reference, and g is the acceleration of gravity or $9.8 m/s^2$ .

When the block is at the edge of the cliff it has potential energy that can be transformed into any other type of energy as it starts falling to the ground.

The GPE of the block of mass m=42 Kg at h=40 m is:

U = 42*9.8*40

U = 16,464 J

The block loses 81 J due to air resistance, thus the energy stored when it hits the ground is 16,464 J - 81 J = 16,383 J.

This energy is stored as kinetic energy, whose formula is:

$\displaystyle K=\frac{1}{2}mv^2$

Solving for v:

$\displaystyle v=\sqrt{\frac{2K}{m}}$

$\displaystyle v=\sqrt{\frac{2*16,383 }{42}}$

$v=\sqrt{780.143}$

v = 27.9 m/s

The block hits the ground at 27.9 m/s

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3 years ago

A 1.2 kg block of wood hangs motionless from strings. A 50 g bullet, traveling horizontally, strikes the block and becomes embed

kicyunya [14]

Answer:

A. <u>200m/s</u>

Explanation:

Using the law of conservation of momentum expressed as;

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses of the object

u1 and u2 are the respective velocities

v is the common velocity

Given

m1 = 1.2kg

u1 = 0m/s (block is a stationary object)

m2 = 50g= 0.05kg

u2 = ?

v = 8.0m/s

Substitute the values into the formula and get u2 (speed of the bullet before hitting the block)

1.2(0)+0.05u2 = (1.2 + 0.05)(8)

0.05u2 = 1.25(8)

0.05u2 = 10

u2 = 10/0.05

u2 = 200m/s

Hence the speed of the bullet before it hit the block is <u>200m/s</u>

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What is a electromagnet?

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An electromagnet is a soft metal core which is made into a magnet by passing the electric current through a coil which is surrounding it.

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It is used to classify the climates around the world

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A screen is placed 1.60 m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40 c

Lunna [17]

Answer:

distance between the two second-order minima is 2.8 cm

Explanation:

Given data

distance = 1.60 m

central maximum = 1.40 cm

first-order diffraction minima = 1.40 cm

to find out

distance between the two second-order minima

solution

we know that fringe width = first-order diffraction minima /2

fringe width = 1.40 /2 = 0.7 cm

and

we know fringe width of first order we calculate slit d

β1 = m1λD/d

d = m1λD/β1

and

fringe width of second order

β2 = m2λD/d

β2 = m2λD / ( m1λD/β1 )

β2 = ( m2 / m1 ) β1

we know the two first-order diffraction minima are separated by 1.40 cm

so

y = 2β2 = 2 ( m2 / m1 ) β1

put here value

y = 2 ( 2 / 1 ) 0.7

y = 2.8 cm

so distance between the two second-order minima is 2.8 cm

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