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kotegsom [21]
2 years ago
12

Name one element that does not need to gain or lose electrons to become stable.

Physics
1 answer:
creativ13 [48]2 years ago
3 0

Answer:

Helium

Explanation:

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For a body falling freely from rest​ (disregarding air​ resistance), the distance the body falls varies directly as the square o
jasenka [17]

Answer:

The answer to the question is

The object would fall 57.625 m in the first 5 seconds

Explanation:

To solve the question, we note that

the height of fall = 490 ft = ‪149.352‬ m

Time to touch the ground = 7 seconds

We are required to find out how far the object falls in the first 5 seconds

We apply the relation

S = u·t + 0.5×g·t ² = We then have

‪149.352‬ = U×7+0.5*9.81*49 From where u = -13 m/s

Therefore to find how far it falls in the first 5 seconds, we have

-13*5 + 0.5*9.81*25 = 57.625 m

5 0
2 years ago
Read 2 more answers
What is the magnification of an astronomical telescope whose objective lens has a focal length of 74 cm and whose eyepiece has a
Novay_Z [31]

Answer:

The magnification of an astronomical telescope is -30.83.

Explanation:

The expression for the magnification of an astronomical telescope is as follows;

M=-\frac{f_o}{f_e}

Here, M is the magnification of an astronomical telescope, f_e is the focal length of the eyepiece lens and f_o is the focal length of the objective lens.

It is given in the problem that an astronomical telescope having a focal length of objective lens 74 cm and whose eyepiece has a focal length of 2.4 cm.

Put f_o=74 cm and f_e=2.4 cm in the above expression.

M=-\frac{74}{2.4}

M=-30.83

Therefore, the magnification of an astronomical telescope is -30.83.

5 0
3 years ago
What is v^2=0.05-4.9 please i need this asap​
Margaret [11]

Answer:

v =2.02

Explanation:

v^2=0.05-4.9

v^2=-4.85

square root both side

v=2.02

^^^^this is a not a perfect square  

7 0
3 years ago
Which of the following is a factor that determines the composition of magma?
Lina20 [59]
The answer is C. I hope I helped!
5 0
3 years ago
A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying p
wel

Answer:

Normal force = 8.75 N

Explanation:

given,

frictional force between the steel spatula and the Teflon frying pan=0.350 N

coefficient of friction between material =0.04

normal force = ?

using formula,

Frictional force = coefficient of friction × normal force

normal\ force = \dfrac{Frictional\ force}{coefficient\ of\ friction}

normal\ force = \dfrac{0.350}{0.04}

Normal force = 8.75 N

8 0
2 years ago
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