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Lina20 [59]
3 years ago
8

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]3 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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Explanation:

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3 years ago
The electric current in a wire is 1.5A. How many electrons flow past a given point in a time of 2s?
kipiarov [429]

Answer:

The amount of electrons that flow in the given time is 3.0 C.

Explanation:

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It is measure in Amperes and can be measured in the laboratory by the use of an ammeter.

In the given question, I = 1.5A, t = 2s, find Q.

From equation 1,

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The amount of electrons that flow in the given time is 3.0 C.

5 0
2 years ago
You have two vectors, which are 2.59 m at 30.0° north of east and 4.18 m at 60.0° north of west. What is the magnitude in meters
andrew11 [14]

Answer:\ec{r}=0.153\hat{i}+4.914\hat{j}v

Explanation:

Given

Vector 1

\vec{a}=2.59\left ( cos30\hat{i}+sin30\hat{j}\right )

Vector 2

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\vec{r}=2.59\left ( cos30\hat{i}+sin30\hat{j}\right )+4.18\left ( -cos60\hat{i}+sin60\hat{j}\right )

\vec{r}=0.153\hat{i}+4.914\hat{j}

|r|=4.916

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3 years ago
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B because it is basically 12 km/ hour
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T=vf-vi/a
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