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3 years ago

If you push a shopping cart with all of your strength will it move faster if it's empty or full of items? Explain

Physics

2 answers:

ss7ja [257]3 years ago

8 0

It will go faster if it was empty because it would weigh less

Strike441 [17]3 years ago

7 0

Empty; there will be a less mass to push.

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When is the universal theme of a story often revealed?

elena-14-01-66 [18.8K]

It is often revealed <span>at the resolution of the story, when the reader can see how the story ends.</span>

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3 years ago

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What happens to the work done when a force is doubled and the distance moved remain the same?

san4es73 [151]

Answer:

Work done gets doubled.

Explanation:

The work done by a force is given by :

W = Fd

Where

F is force and d is distance move

If the force is doubled and the distance moved remain the same, it would mean that the work done becomes double of the initial work done.

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3 years ago

Is lateral shift or lateral displacement same ?

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Answer:

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Explanation:

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3 years ago

An object is propelled along a straight-line path by a force. If the new force were doubled, is acceleration would

Aneli [31]

Still go straight but would obviously go up in speed!!

Hope this helps plz mark as brainlist and 5 star

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3 years ago

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1) Subatomic particles called muons can be created in the upper atmosphere by collisions of cosmic rays (energetic particles com

Vsevolod [243]

Answer and explanation:

Muon travelled straight down towards the earth. Therefore the tree moves up in the rest frame of muon (option a)

In muon rest frame it travels Zero meters

Distance, d = Velocity, v * Time, s

$where, v = 0.9c = 0.9 \times 8 \times 10^8 , s = 2.2 \mu s$

$d = 0.9 \times 3 \times 10^8 \times 2.2 \times 10^{-6}\\\\d = 594m$

Distance from the top of the mountain to the tree is the same as the distance travelled by the tree in the muons rest frame

that is same as in part C which is 594m

Using lorentz contraction

In the rest frame of someone standing on the mountain

the distance is given by

$d' = \frac{d}{\gamma} = d\sqrt{1 - \frac{v^2}{c^2}}, where, \frac{1}{\gamma}= \sqrt{1 - \frac{v^2}{c^2}}$

$d' = 594\sqrt{1 - \frac{(0.9c)^2}{c^2}}$

$d' = 594\sqrt{1 - 0.81}$

$d' = 594 \times 0.4359$

d' = 258.92m

in the rest frame of someone standing on the mountain,

muon moves straight down

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4 years ago