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pashok25 [27]
3 years ago
6

What is the momentum of a 5 kg object that has a velocity of 1. 2 m/s? 3. 8 kg • m/s 4. 2 kg • m/s 6. 0 kg • m/s 6. 2 kg â

€˘ m/s.
Physics
2 answers:
nevsk [136]3 years ago
7 0

The momentum of a 5kg object that has a velocity of 1.2m/s is 6.0kgm/s.

<h3> MOMENTUM:</h3>

Momentum of a substance is the product of its mass and velocity. That is;

Momentum (p) = mass (m) × velocity (v)

According to this question, an object has a mass of 5kg and velocity of 1.2m/s. The momentum is calculated thus:

Momentum = 5kg × 1.2m/s

Momentum = 6kgm/s.

Therefore, the momentum of a 5kg object that has a velocity of 1.2m/s is 6.0kgm/s.

Learn more about momentum at: brainly.com/question/250648?referrer=searchResults

Paul [167]3 years ago
5 0

Answer:

6 kg.m/s

Explanation:

Momentum is the product of mass and velocity. So,

  • p = mv

Here,

  • <em>p</em> denotes momentum [?]
  • <em>m</em> denotes mass [5kg]
  • <em>v</em> denotes velocity [1.2 m/s]

→ p = 5 kg × 1.2 m/s

→ p = (5 × 1.2) kg.m/s

→ p = (5 × 12/10) kg.m/s

→ p = 12/2 kg.m/s

→ <u>p</u><u> </u><u>=</u><u> </u><u>6</u><u> </u><u>kg</u><u>.</u><u>m</u><u>/</u><u>s</u>

Therefore, momentum of the object is 6 kg.m/s.

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Scalar quantities have a size or magnitude only and need no other information to specify them. Thus, 10 cm, 50 sec, 7 litres and 3 kg are all examples of scalar quantities.

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4. Which of the following would be a good reference point to describe the motion of a dog?
saul85 [17]

ANOTHER RUNNING DOG

Explanation:

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4 0
3 years ago
A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this
wlad13 [49]

Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

r=r_o\times A^{\frac{1}{3}}

r_o=1.25 \times 10^{-15} m = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}

r=4.6656\times 10^{-15} m=4.6656 fm

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) F=k\times \frac{q_1q_2}{a^2}

k=9\times 10^9 N m^2/C^2 = Coulombs constant

q_1,q_2 = charges kept at distance 'a' from each other

F = electrostatic force between charges

q_1=+1.602\times 10^{-19} C

q_2=+1.602\times 10^{-19} C

Force of repulsion between two protons on opposite sides of the diameter

a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m

F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}

F=2.6527 N

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

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