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NemiM [27]
3 years ago
11

You are in Paris, 60 m up in the Eiffel Tower. If you throw a euro downward at a velocity of 2.0 m/s, how long would it take the

euro to hit the ground? Neglect air resistance.
Physics
1 answer:
kondor19780726 [428]3 years ago
4 0

Answer:

t = 3.29 seconds

Explanation:

It is given that,

Height of the Eiffel tower is 60 m

Initial speed of a euro, u = 2 m/s

It will move under the action of gravity in the downward direction. Firstly, we can find the final velocity as follows :

v^2-u^2=2ad\\\\v=\sqrt{u^2+2ad} \\\\v=\sqrt{(2)^2+2\times 9.81\times 60} \\\\v=34.36\ m/s

Let t is the time taken by the euro to hit the ground. It can be calculated as :

v=u+at\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{34.36-2}{9.81}\\\\t=3.29\ s

Hence, it will take 3.29 seconds to hit the ground.

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The Whirlpool galaxy is about 30 million light-years away. If you were in a spaceship that could travel at half of the speed of
Nataly_w [17]

Answer:

51.96 years

2) 30 million of years

Explanation:

First we must know the travel time of the ship seen from the earth. The spaceship travels at half the speed of light, this means that the amount of time the spacecraft must spend to travel the same distance is double compared to the light, that is 60 years.

Now due to the speed of the ship, we must take into account relativistic effects, such as time dilation, this is given by:

t'=\frac{t}{\sqrt{1-\frac{v^2}{c^2}}}

Where t is the time measured in the ship, t' is the time measured in the earth, inertially moving with velocity v.

Rewriting for t:

t=t'\sqrt{1-\frac{v^2}{c^2}}\\t=60\sqrt{1-\frac{(0.5c)^2}{c^2}}\\t=60\sqrt{1-0.5^2}\\t=51.96 years

This is the amount of time it would take you reach the Whirlpool galaxy in the spaceship.

2) a light year is a measure of distance, which indicates the kilometers that light travels in a year. Thus, the light emitted by Whirlpool galaxy takes 30 million of years reaches our planet.

6 0
3 years ago
If I push a box at a constant rate is there friction force acting on it?
yanalaym [24]
Yes, the friction is acting in the opposite direction you are pushing.
3 0
3 years ago
A car slid off an icy 10m bridge and landed 12m away from the bridge. How much time was the car in the air? (Hunt: Projectile)
Contact [7]

You will use the height of the bridge from the ground.

Solution:

Formula to be used is y=Viy(t)+g(t^2)/2

Where:

Vi=initial velocity which is 0 m/s

 y=10 m

Gravitational acceleration or g =9.8m/s^2

T= time you need

Substitute all the given to the formula

10m=(0m/s)(t)+(9.8m/s^2)(t^2)/2

10mx2=9.8m/s^2(t^2)

Now isolate the variable you want to find which is T or time

10mx2/9.8m/s^2=t^2

20m/9.8m/s^2=t^2

Square root of 2.04= square root of t^2

T=1.43 secs

The answer is 1.43 seconds

8 0
3 years ago
Read 2 more answers
Can someone explain how they got their answer or how I get the change in number? :(
enyata [817]

Answer:

See Below

Explanation:

Okay, I thinkkk what it is asking by what you summarzied for me issss:

They split the total time into four quarters. They then took (for the first quarter) the start time. Then when the first quarter ends and the second quarter starts is the "end" time.

They then subtract the start time of the second quarter from the end time of the first quarter.

I hope this helps, good luck! :D

5 0
2 years ago
PLEASE HELP! I'LL GIVE BRAINLEST​
erma4kov [3.2K]

Answer:

8.162

Explanation:

W=mg

  = 2.2 x 3.71

  = 8.162 N

5 0
3 years ago
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