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Alinara [238K]
3 years ago
8

Which number is right

Physics
1 answer:
KatRina [158]3 years ago
7 0
'C' is the correct choice. Please, do me a favor ... take a few minutes to read the other choices, A, B and D, and understand what's wrong with each of those. That's MUCH MUCH more important than knowing that C is correct.
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According to the text, there is no energy shortage now, nor will there ever be. what reason (s) is given to support this stateme
Bas_tet [7]
The reason why there is no energy shortage nor will there ever be is because energy is being preserved and conserved and only changes form. It never gets lost or increased.
8 0
3 years ago
What is the resistance of a light bulb if a potential difference of 120 V will produce a current of 0.5 A in the bulb?
Yakvenalex [24]

Explanation:

Ohm's law:

V = IR

120 V = (0.5 A) R

R = 240 Ω

6 0
3 years ago
Read 2 more answers
A 100 N force is applied to a 50 kg crate resting on a level floor. The coefficient of kinetic friction is 0.15. What is the acc
Nookie1986 [14]

The acceleration of the crate after it begins to move is 0.5 m/s²

We'll begin by calculating the the frictional force

Mass (m) = 50 Kg

Coefficient of kinetic friction (μ) = 0.15

Acceleration due to gravity (g) = 10 m/s²

Normal reaction (N) = mg = 50 × 10 = 500 N

<h3>Frictional force (Fբ) =?</h3>

Fբ = μN

Fբ = 0.15 × 500

<h3>Fբ = 75 N</h3>

  • Next, we shall determine the net force acting on the crate

Frictional force (Fբ) = 75 N

Force (F) = 100 N

<h3>Net force (Fₙ) =?</h3>

Fₙ = F – Fբ

Fₙ = 100 – 75

<h3>Fₙ = 25 N</h3>

  • Finally, we shall determine the acceleration of the crate

Mass (m) = 50 Kg

Net force (Fₙ) = 25 N

<h3>Acceleration (a) =?</h3>

a = Fₙ / m

a = 25 / 50

<h3>a = 0.5 m/s²</h3>

Therefore, the acceleration of the crate is 0.5 m/s²

Learn more on friction: brainly.com/question/364384

8 0
2 years ago
Two asteroids are 75000 m apart one has a mass of 8 x 10^7 kg if the force of gravity between them is 1.14 what is the mass of t
vaieri [72.5K]
I’m going to say B

I hope this helped
4 0
3 years ago
Read 2 more answers
What is the ideal banking angle for a gentle turn of 1.20-km radius on a highway with a 105 km/h speed limit (about 65 mi/h), as
Mnenie [13.5K]

Answer:

4.14°

Explanation:

given:

r = 1.2 km

v = 105 km/h

1) <em>convert your given </em>

a) r = 1.2 km to m = 1200m

b) v = 105 km/h  to m/s = 29.2 m/s

2) <em>plug into your ideal banking angle equation</em>

tan^-1(\frac{v^2}{rg}) = \frac{29.2^2}{(1200)(9.8)} = 4.14°

8 0
2 years ago
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