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Readme [11.4K]
3 years ago
8

An example of an anaerobic exercise is what?

Physics
1 answer:
san4es73 [151]3 years ago
6 0

Answer:

1st: Theatre History

4th Quarter

Upcoming

Due today

Syllabus

Due Sunday1st: Theatre History

4th Quarter

Upcoming

Due today

Syllabus

Due Sunday1st: Theatre History

4th Quarter

Upcoming

Due today

Syllabus

Due Sunday

Explanation:

1st: Theatre History

4th Quarter

Upcoming

Due today

Syllabus

Due Sunday

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Since the moon has less mass than the earth what happens to objects on the moon
postnew [5]
When you drop an object on the moon, it falls to the ground.
But it only falls about 1/6 as fast as it falls on Earth.
4 0
3 years ago
Some one answer this question !
nirvana33 [79]

Answer:

A

Explanation:

The answer would be A because when feet push down on anything, the force will push down the skateboard

5 0
3 years ago
Can a body have constant speed and still be acclerating ? Give an Example
Allushta [10]






Hi Pupil Here's Your answer :::





➡➡➡➡➡➡➡➡➡➡➡➡➡





An object moving with constant speed can be accelerated if direction of motion changes. For example, an object moving with a constant speed in a circular path has an acceleration because its direction of motion changes continuously.





⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅⬅






Hope this Helps . . . . . . . . .
3 0
3 years ago
An electron is accelerated within a particle accelerator using a 100 MV electric potential. The 100 MeV electron moves along an
Delicious77 [7]

Answer:

The length of the tube is 3.92 m.

Explanation:

Given that,

Electric potential = 100 MV

Length = 4 m

Energy = 100 MeV

We need to calculate the value of \gamma

Using formula of relativistic energy

E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)

Put the value into the formula

1.6\times10^{-15}= 9.1\times`10^{-31}\times9\times10^{16}(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)

(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)=\dfrac{1.6\times10^{-15}}{9.1\times10^{-31}\times9\times10^{16}}

Here, \gamma-1=(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)

\gamma-1=0.01953

\gamma=0.01953+1

\gamma=1.01953

We need to calculate the length

Using formula of length

L'=\dfrac{L}{\gamma}

Put the value into the formula

L'=\dfrac{4}{1.01953}

L'=3.92\ m

Hence, The length of the tube is 3.92 m.

8 0
3 years ago
a golfer hits a 0.05 kg golf ball with an impulse of 3 N-s what is the change in velocity of the ball
Rainbow [258]

so your saying the start is 0 N and when he/she hits the ball its inertia is 3 N. if that is so m*v=

.05*3=<u>.15</u>

7 0
3 years ago
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