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2 years ago

5

How do you give an example for finding net force?

1 answer:

Crazy boy [7]2 years ago

7 0

-- 6 people all trying to push a car out of snow

-- a Tug-o-War with 30 people of different sizes pulling on each end of the rope

-- you and your sister both pulling on the same doll (or Transformer)

-- lifting a book up from the table to a high shelf
taking a book down from a high shelf to the table
(one force is you; another force is gravity)

-- grabbing your big dog by his collar and trying to bring him inside

-- three people at the table all grab the ketchup bottle at the same time

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Can you find a vector quantity that has a magnitude of zero but components that are not zero? Explain. Can the magnitude of a ve

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It is not possible to find a vector quantity of magnitude zero but components different from zero

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2 years ago

A frog falls from its rainforest tree. If we ignore wind resistance, (a) how much time does it take the frog to fall a distance

gtnhenbr [62]

Answer:

Explanation:

a) Using the equation of motion

S = ut + 1/2gt²

S is the distance of fall

g is the acceleration due to gravity

t is the time taken

Given S = 12.0m, g = 9.81m/s^2, un= 0m/s

12 = 0+1/2(9.81)t²

12 = 4.905t²²²

t² = 12/4.905

t² = 2.446

t = √2.446

t = 1.56secs

b) To determine how fast is the frog falling at this point, we need to calculate the speed of the frog. Using the equaton v = u+gt

v = 0+9.81(1.56)

v = 15.34m/s

Hence the frog is falling at the rate of 15.34m/s

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2 years ago

A truck, initially at rest, rolls down a frictionless hill and attains a speed of 20 m/s at the bottom. To achieve a speed of 40

Crank

Answer:

To achieve the velocity of 40 m/sec height will become 4 times

Explanation:

We have given initially truck is at rest and attains a speed of 20 m/sec

Let the mass of the truck is m

At the top of the hill potential energy is mgh and kinetic energy is $\frac{1}{2}mv^2$

So total energy at the top of the hill $=mgh+0=mgh$

At the bottom of the hill kinetic energy is equal to $\frac{1}{2}mv^2$ and potential energy will be 0

So total energy at the bottom of the hill is equal to $0+\frac{1}{2}mv^2$

Form energy conservation $mgh=\frac{1}{2}mv^2$

$v=\sqrt{2gh}$ , for v = 20 m/sec

$20=\sqrt{2\times 9.8\times h}$

Squaring both side

$19.6h=400$

h = 20.408 m

Now if velocity is 0 m/sec

$40=\sqrt{2gh}$

$19.6h=1600$

h = 81.63 m

So we can see that to achieve the velocity of 40 m/sec height will become 4 times

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