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xenn [34]
3 years ago
12

You are asked to calculate an object's velocity, in order to do so you must know the object's ?

Physics
1 answer:
elixir [45]3 years ago
5 0
There are so many answers. It depends on which field of physics you are working on.
You could calculate velocity from momentum or kinetic energy knowing also the object's mass for example.
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Which of the following is a physical property?
coldgirl [10]
Flammability, I think!
6 0
3 years ago
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If ram is 40kg and travels 200m in 30 sec find his power​
WARRIOR [948]

Work done

  • N×200
  • mg200
  • 40(10)(200)
  • 400(200)
  • 80000J

Power

  • Work done/Time
  • 80000/30
  • 8000/3
  • 2668W
8 0
2 years ago
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How does a large star create the elements? at what element does the star stop making new elements?
Dahasolnce [82]
They are fused in the core of the star due to great pressures and temperatures.  They are made all the way through iron.  At that point the star dies.  If it is a really large star it will become a supernova when it dies, creating all of the elements beyond iron as well, but only in its death.  No star can create anything beyond iron in its life cycle
4 0
3 years ago
The amount of energy that must be absorbed or lost to raise or lower the temperature of 1 g of liquid water by 1°c _____.
Fiesta28 [93]

Answer:

4.2 J

Explanation:

Specific heat capacity: This is defined as the amount of a heat required to rise a unit mass of a substance through a temperature of 1 K

From specific heat capacity,

Q = cmΔt.............................. Equation 1

Where Q = amount of energy absorbed or lost, c = specific heat capacity of water, m = mass of water, Δt = Temperature rise.

Given: m = 1 g = 0.001 kg, Δt = 1 °C

Constant : c = 4200 J/kg.°C

Substitute into equation 1

Q = 0.001×4200(1)

Q = 4.2 J.

Hence the energy absorbed or lost = 4.2 J

6 0
3 years ago
An air bubble has a volume of 2.0 cm3 when it is released by a submarine 100 m below the surface of a freshwater lake. What is t
UkoKoshka [18]

Answer:

21.35 cm^3

Explanation:

let the volume at the surface of fresh water is V.

The volume at a depth of 100 m is V' = 2 cm^3

temperature remains constant.

density of water, d = 1000 kg/m^3

Pressure at the surface of fresh water is atmospheric pressure,

P = Po = 1.013 x 10^5 N/m^2

The pressure at depth 100 m is P' = Po + hdg

P' = 1.013 \times 10^{5}+ 100 \times 1000 \times 9.8

P' = 10.813 x 10^5 N/m^2

Use the Boyle's law

P V = P' V'

1.013 \times 10^{5}\times V = 10.813 \times 10^{5}\times 2

V = 21.35 cm^3

Thus, the volume of air bubble at the surface of fresh water is 21.35 cm^3.

5 0
3 years ago
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