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antoniya [11.8K]

3 years ago

10

An apple is initially sitting on the bottom shelf of a pantry. A hungry physics student picks up the apple to eat it, but change

s her mind and puts the apple down on a shelf somewhere above its original location. During this process, the total work done on the apple by all forces is:

1 answer:

andre [41]3 years ago

8 0

Answer:

Zero

Explanation:

Work done is given by multiplying force and distance moved. The distance is moved both positive and negative and it's equal distance. Since force used is the same hence work

W=F*d+ (F*-d)=0

Therefore, total work done is zero

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Electromagnetic waves are ____________ (one thousand/one million) times faster than sound.

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one million

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What happens if you throw a car in a tornado. What happens to the car. What movements does it undertake. Does it spin in the tor

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The car a heavy tornado would spin and spin.Would a car really stay on the ground to a tornado,think of the possibilities that would happen yourself,which opinion would happen

a.the car would stay on the ground

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2 years ago

Orlov [11]

Answer:

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Explanation:

Pа: 5. Mass of the earth is 6x10-kg and its radius is 6400km. What is the mass of a man weighting 977N? (99.99kg) ma. Orth be.

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2 years ago

While on the moon, the Apollo astronauts enjoyed the effects of a gravity much smaller than that on

Ainat [17]

Answer:

1.628 $\frac{m}{sec^{2} }$

Explanation:

Anywhere in the universe, In a closed system, <u>Conservation of energy</u> is applicable.

In this case

Neil is initially on the surface of moon and has a velocity of 1.51 $\frac{m}{sec}$ in upward direction.

⇒He has Kinetic energy= $K_{i}$ = $\frac{1}{2} m{v^{2} }$ J

But with respect to the surface of the moon,

where m=mass of moon

v=velocity of Neil

He has Potential energy= $P_{i}$ =0 J

At the highest point of his jump, his velocity =0

⇒ Kinetic energy= $K_{f}$ =0 J

His Potential energy with respect to the surface of moon= $P_{f}$ = $m \times g\times h$

where m=mass of moon

g= gravitational acceleration on moon

h=height from moon's surface

By Conservation Energy Principle

$K_{i}$ + $P_{i}$ = $K_{f}$ + $P_{f}$

$K_{i}$ +0=0+ $P_{f}$

$\frac{1}{2} m{v^{2} }$ = $m \times g\times h$

$\frac{v^{2} }{2}$ = $g\times h$

$\frac{1.5^{2} }{2}$ J= $g\times 0.7 m$

⇒ g = $\frac{1.14}{0.7}$ = 1.628 $\frac{m}{sec^{2} }$

8 0

2 years ago

A crate, heavier than you are, rests on a rough floor. The coefficient of static friction between the crate and the floor is the

KIM [24]

Answer:

No the crate can not be pushed across the floor.

Explanation:

The static frictional force ( $F_{s}$ ) is the product of the coefficient of static friction ( $\mu_{s}$ ) and the normal force ( $N$ ) produced by the floor on the object. Mathematically,

$F_{s} = \mu_{s}N$

For static friction the normal force ( $N$ ) is balanced by the weight of the object.

So, for the crate the static frictional force ( $F_{sC}$ ) is written as

$F_{sC} = \mu_{s} N_{C} = \mu_{C} Mg$

where ' $M$ ' is the mass of the crate.

and for the person having mass ' $m$ ' the static frictional force ( $F_{sP}$ ) is written as

$F_{sP} = \mu_{s}N_{P} = \mu_{s}mg$

Comparing bot the equations,

$\dfrac{F_{sC}}{F_{sP}} = \dfrac{M}{m}$

As, $M > m$ so, $F_{sC}$ > $F_{sP}$ .

Therefore it is not possible for the person to push the crate across the floor.

6 0

2 years ago