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2 years ago

If a 7 kg bowling ball is lifted 2 m into the air and dropped, what speed will it strike the ground? (

Physics

1 answer:

Naily [24]2 years ago

4 0

Answer:

6.32m/s

Explanation:

note:Now these calculations are based in the fact that acc. due to gravity is 10m/s²

okay so I'm thinking you think the speed of a body depends on the mass of the body also,umh... well it doesn't at all!

when two bodies of different masses fall from the same height,they fall at the same time( this is just to say)

now enough of the talking let solve....

so the ball was dropped .ie from rest to the ground through a distance of 2m,

the formula for calculating the distance if a body moving in a straight line is given by:

S=ut + ½at² where u is initial velocity, a is acceleration ( of the body or due to gravity, but since its falling freely under the influence of gravity its " we use the acceleration due to gravity ,which is 10m/s²) and t is the time taken to cover the distance.

from our question the ball was dropped from rest thus its u is 0 therefore we use this equation to find the time it took to touch ground (S=½at²)

solving ....

we get t to be 0.632s

to find the speed we substitute t in the equation below:

V=u+at ,but since u=0

V=at =10•0.632=6.32m/s

therefore the speed the body uses to strike the ground is 6.32m/s

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3 years ago

Each tyre of a car has an area of 100 cm2 in contact with the ground. The car has a mass of 1600 kg. The weight of the car is eq

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3 years ago

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A 7950-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s . A 2950-kg load, initially

Alchen [17]

The new speed of car is 10.9 m/s

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According to the principle of momentum conservation, momentum is only modified by the action of forces as they are outlined by Newton's equations of motion; momentum is never created nor destroyed inside a problem domain.

Mass of the railroad car, m₁ = 7950 kg

Mass of the load, m₂ = 2950 kg

It can be assumed as the speed of the car, u₁ = 15 m/s

Initially, it is at rest, u₂ = 0

Let v is the speed of the car. It can be calculated using the conservation of momentum as :

$m_1u_1 + m_2u_2 = (m_1 + m_2) v$

$v =\frac{m_1u_1}{m_1+m_2}$

$v = \frac{7950*15}{7950+2950}$

$v= 10.9 m/s$

Therefore, the new speed of care is 10.9 m/s

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2 years ago

A man pushed the box with a force of 20 N. He doubled his force to 40 N. The box did not

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Answer:

Explanation:

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2 years ago

The tops of the towers of the Golden Gate Bridge in San franciscoo , are 227 m above the water. Suppose a worker drops a 655 g w

yaroslaw [1]

Answer:

Kinetic energy is 1425.11 J.

Explanation:

Given:

Mass of the wrench is, $m=655\ g=0.655\ kg$

Height of fall is, $h=227\ m$

Force of resistance is, $F=0.141\ N$

Now, the total energy at the top is equal to the potential energy of the wrench at the top since the kinetic energy at the top is 0.

Now, potential energy at the top is given as:

$U=mgh\\U=0.655\times 9.8\times 227\\U=1457.113\ J$

Now, the potential energy at the top is converted to kinetic energy at the bottom and some energy is wasted in overcoming the resistance force by air.

Potential Energy = Kinetic energy + Energy to overcome resistance.

⇒ Kinetic energy = Potential Energy - Energy to overcome resistance.

Energy to overcome resistance force is the work done by the wrench against the resistance force and is given as:

$W_{res}=Fh=0.141\times 227=32.007\ J$

Therefore, Kinetic energy at the bottom is given as:

$K=U-W_{res}\\\\K=1457.113\ J - 32.007\ J\\\\K=1425.106\approx1425.11\ J$

Hence, the kinetic energy of the wrench be when it hits the water is 1425.11 J.

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3 years ago