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3 years ago

6. A 50 N block is raised 2 m. If the net work done on the block is 50 J, what is the applied force on the block?

Physics

1 answer:

geniusboy [140]3 years ago

5 0

Answer:

F = 75[J]

Explanation:

We know that work is defined as the product of force by distance.

In this way we have two forces, the weight of the block down, and the force that bring about the block to rise.

$W = -(F_{weight*d})+(F_{upward}*d)$

where:

W = work = 50 [J]

d = distance = 2 [m]

Fweight = 50 [N]

Fupward [N]

Now replacing:

$50=-(50*2)+(F_{upward}*2)\\50+100=F_{upward}*2\\F_{upward}=150/2\\F_{upward}=75[J]$

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2 years ago

A ball initially at rest rolls down a hill with an acceleration of 3.3 m/s2. If it accelerates for 7.5 s, how far will it move?

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<u>Answer:</u>

Ball will move 92.8125 meter along the cliff in 7.5 seconds.

<u>Explanation:</u>

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In this case initial velocity = 0 m/s, acceleration = 3.3 $m/s^2$ , we need to calculate displacement when time = 7.5 seconds.

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3 years ago

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3 years ago

f an arrow is shot upward on the moon with velocity of 35 m/s, its height (in meters) after t seconds is given by h(t)=35t−0.83t

inysia [295]

Answer:

The velocity of the arrow after 3 seconds is 30.02 m/s.

Explanation:

It is given that,

An arrow is shot upward on the moon with velocity of 35 m/s, its height after t seconds is given by the equation:

$h(t)=35t-0.83t^2$

We know that the rate of change of displacement is equal to the velocity of an object.

$v(t)=\dfrac{dh(t)}{dt}\\\\v(t)=\dfrac{d(35t-0.83t^2)}{dt}\\\\v(t)=35-1.66t$

Velocity of the arrow after 3 seconds will be :

$v(t)=35-1.66t\\\\v(t)=35-1.66(3)\\\\v(t)=30.02\ m/s$

So, the velocity of the arrow after 3 seconds is 30.02 m/s. Hence, this is the required solution.

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3 years ago

A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path wit

makvit [3.9K]

Answer:

Final velocity of the block = 2.40 m/s east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Mass of bullet = 0.0140 kg

Consider east as positive.

Initial velocity of bullet = 205 m/s

Mass of Block = 1.8 kg

Initial velocity of block = 0 m/s

Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

Final velocity of bullet = -103 m/s

We need to find final velocity of the block( u )

Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u

We have

2.87 = -1.442 + 1.8 u

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Final velocity of the block = 2.40 m/s east.

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3 years ago