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Tatiana [17]
3 years ago
12

The process of changing the energy of a system by means of force. Force * Diatance

Physics
1 answer:
Dmitry_Shevchenko [17]3 years ago
7 0
This is known as work
You might be interested in
Compared to its weight on Earth, a 5kg object on the moon will weigh
shutvik [7]

Answer:

8.1 N/49 N=0.1653  which means 16.53% of the weight of the object on Earth.

Explanation:

On the Moon, where the gravitational constant is 1.62 \frac{m}{s^2}, the weight of the 5 kg object will be: weight_M=m*g_M = 5 kg * 1.62 \frac{m}{s^2} =8.1 N

Where the answer is in Newtons (N) since all quantities are given in the SI system.

On Earth, on the other hand, the weight of the object is:

weight_E=m*g_E= 5 kg* 9.8 \frac{m}{s^2} = 49N

Therefore the object's weight on the Moon compared to that on Earth will be:

8.1N/49N=0.1653

That is, 16.53% of the weight the object has on Earth.

5 0
3 years ago
What happens to the temperature of a gas when it is compressed? a the temperature does not change. b the temperature increases.
nasty-shy [4]
I think the temperature increases
5 0
3 years ago
Read 2 more answers
What forces act upon the moon as it orbits the earth
ValentinkaMS [17]

the axis acts against and it would be a contact force



7 0
3 years ago
Read 2 more answers
If we decrease the amount of force applied to an object, and all other factors remain the same, the amount of work completed wil
Nat2105 [25]
A ) decrease.
B ) increase.
C ) increase, then decrease.
D ) not change.

The answer is A) decrease

Take pushing a box, for example-- You  push your hardest then give out, still trying to push the box. You are doing less work than what you have started with!

( Mind marking me for branliest? ; ) )
4 0
3 years ago
A 594 Ω resistor, an uncharged 1.3 μF capacitor, and a 6.53 V emf are connected in series. What is the current in milliamps afte
ivanzaharov [21]

Answer:

6.88 mA

Explanation:

Given:

Resistance, R = 594 Ω

Capacitance = 1.3 μF

emf, V = 6.53 V

Time, t = 1 time constant

Now,

The initial current, I₀ = \frac{\textup{V}}{\textup{R}}

or

I₀ = \frac{\textup{6.53}}{\textup{594}}

or

I₀ = 0.0109 A

also,

I = I_0[1-e^{-\frac{t}{\tau}}]

here,

τ = time constant

e = 2.717

on substituting the respective values, we get

I = 0.0109[1-e^{-\frac{\tau}{\tau}}]

or

I = 0.0109[1-2.717^{-1}]

or

I = 0.00688 A

or

I = 6.88 mA

5 0
3 years ago
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