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2 years ago

Can someone help me with these problems?

Physics

1 answer:

Paladinen [302]2 years ago

3 0

120/60= 2m per s
220/55= 4m per s
720/80= 9m per s
Fastest run: 3rd run
Speed: 9m per s
Slowest: 1st run
Speed: 2m per s

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A client experiences difficulty in performing the prone iso-abs exercise. Which of the following should be suggested as an regre

butalik [34]

Answer:

a. Quadruped arm and opposite leg raise

Explanation:

Quadruped arm and opposite leg lift

- Kneel on the floor, lean forward and place your hands down.

- Keep your knees in line with your hips and hands directly under your shoulders.

- Simultaneously raise one arm and extend the opposite leg, so that they are in line with the spine.

- Go back to the starting position.

This method is usually used as an alternative to iso-abs exercise or also known as a bridge, which allows you to exercise the abdominal and spinal area at the same time.

It is also used together with other exercises for the treatment of hyperlordosis.

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3 years ago

Which formula is used to find fluctuation of the shape of body

Sladkaya [172]

Answer:

varn=n1+1ehvkT–1

Explanation:

This is Einstein's equation.

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3 years ago

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3 years ago

find the speed of a rolling ball that travels a distance of 10 m over the top of a smooth table in 2.0 seconds

umka21 [38]

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3 years ago

Read 2 more answers

You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.

Alja [10]

Answer:

Part a)

$M = 7.25 \times 10^{25} kg$

Part b)

$M = 2\times 10^{30} kg$

Explanation:

Part a)

As we know that the diameter of the planet is given as

$d = 1.8 \times 10^7 m$

now the radius of the planet is given as

$r = 9 \times 10^6 m$

now we know that the acceleration due to gravity of the planet is given by the equation

$g = \frac{GM}{r^2}$

here we know that

$g = 59.7 m/s^2$

now from above equation we have

$59.7 = \frac{(6.67 \times 10^{-11})M}{(9\times 10^6)^2}$

now we have

$M = 7.25 \times 10^{25} kg$

Part b)

Now by kepler's law we know that

time period of revolution of planet about the star is given as

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

so we have

$\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}$

now we have

$\frac{1^2}{402^2} = \frac{(1.5 \times 10^11)^3}{r^3}}$

$rr = 8.17 \times 10^{12} m$

mula of time period

$402\times (365\times 24 \times 3600) = 2\pi \sqrt{\frac{(8.17\times 10^12)^3}{(6.67\times 10^{-11})M}}$

Now we have

$M = 2\times 10^{30} kg$

7 0

3 years ago

Can someone help me with these problems? ​

Can someone help me with these problems?