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rosijanka [135]
3 years ago
12

What is the momentum of a bowling ball with mass 5 kg and velocity 10 m/s

Physics
2 answers:
kondor19780726 [428]3 years ago
8 0
50 kg% that is your answer to the question you asked
Rashid [163]3 years ago
7 0

Answer:

50kg.m/s

Explanation:

to find the momentum you need this formula

p=mv p=momentum m=mass v=velocity

multiply the mass by the velocity

p=5kg x 10m/s

10 x 5 is 50 and the momentum unit is kg.m/s

p=50kg.m/s

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Earth sits motionless in the universe at the center of a revolving globe of starts , with the moon and planets in orbit around the earth, is the surrounding model of the uninverse
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Maksim231197 [3]

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If the bond enthalpy for a C-H bond is 413 kJ, what will happen when the C-H bond is broken?
pentagon [3]

Explanation:

Bond Enthalpy : It is defined as amount of energy required to break a the particular bond in there gaseous state. It is also known as bond energy. It units are kJ/mol.

  • Breaking of a bond is an Endothermic process (energy absorbed from the surroundings).
  • Formation of bond is an Exothermic process (energy is released to the surroundings).

If the average bond enthalpy for a C-H bond is 413 kJ/mol, When the C-H bond breaks in which energy will be required ,which will be an endothermic reaction.

8 0
3 years ago
A piece of transparent meterial which has one or two spherical surfaces is called​
Iteru [2.4K]

Answer:

Lens

Explanation:

Lenses fit this description.

4 0
3 years ago
At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?
sergey [27]

1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

g=9.8 m/s^2 is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

a'=6.9 m/s^2

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

m' is the new mass of the rocket

Re-arranging the equation and solving for m', we find

m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg

And since the initial mass of the rocket was

m=1.9 \cdot 10^4 kg

This means that the mass of fuel burned is

\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg

3 0
3 years ago
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