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alexandr1967 [171]

3 years ago

13

When is a train on Earth in motion relative to the sun? A It is relative to the sun only when the train is not moving. B It is n

ever in motion relative to the sun. C It is relative to the sun only when the train is moving. D It is always in motion relative to the sun.

1 answer:

grandymaker [24]3 years ago

7 0

Answer:

A. It is relative to the sun only when the train is not moving.

Explanation:

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harkovskaia [24]

Answer:

the weight is 49.1 N

Explanation:

The computation of the weight is shown below:

As we know that

= 5kg of potatoes × gravitational acceleration

= 5kg of potatoes × 9.82 m/s

= 49.1 N

Hence, the weight is 49.1 N

We simply applied the above formula in order to determine the weight

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Rank the objects below in order from smallest to largest, where 1 is the smallest and 5 is the largest. local group of galaxies

sasho [114]

1. earth
2. solar system
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4. local group of galaxies
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4 0

3 years ago

Four springs with the following spring constants, 113.0 N/m, 65.0 N/m, 102.0 N/m, and 101.0 N/m are connected in series. What is

Llana [10]

Answer:

$K_e_q=22.75878093\frac{N}{m}$

$f=1.363684118Hz$

Explanation:

In order to calculate the equivalent spring constant we need to use the next formula:

$\frac{1}{K_e_q} =\frac{1}{K_1} +\frac{1}{K_2} +\frac{1}{K_3} +\frac{1}{K_4}$

Replacing the data provided:

$\frac{1}{K_e_q} =\frac{1}{113} +\frac{1}{65} +\frac{1}{102} +\frac{1}{101}$

$K_e_q=22.75878093\frac{N}{m}$

Finally, to calculate the frequency of oscillation we use this:

$f=\frac{1}{2(pi)} \sqrt{\frac{k}{m} }$

Replacing m and k:

$f=\frac{1}{2(pi)} \sqrt{\frac{22.75878093}{0.31} } =1.363684118Hz$

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3 years ago

A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s. It then crosses a rough patch of snow that exerts a fricti

uysha [10]

Answer:The sled slides 16.875m before rest.

Explanation:

$a=\frac{F}{m} =\frac{12N}{20kg}$

a=0.6 m/s²

$Vf=0=Vi-a.t$

$t=\frac{Vi}{a} =t=\frac{4.5m/s}{0.6 m/s2} =t=7.5seg$

$d= Vi.t - \frac{a.t^{2}}{2}$

$d= 4.5 * 7.5 - \frac{0.6*7.5^{2} }{2} \\\\d=16.875m$

3 0

3 years ago

HURRY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Levart [38]

It is an example of "Elastic Potential Energy"

In short, Your Answer would be Option C

Hope this helps!

5 0

3 years ago