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V125BC [204]
3 years ago
14

At the instant a 2.0-kg particle has a velocity of 4.0 m/s in the positive x direction, a 3.0-kg particle has a velocity of 5.0

m/s in the positive y direction. What is the speed of the center of mass of the two-particle system
Physics
1 answer:
satela [25.4K]3 years ago
5 0

Answer:

the speed of the center of mass of the two-particle system is 3.4 m/s.

Explanation:

Given that,

Mass of particle 1, m = 2 kg

Velocity of the particle 1, v = 4 m/s in +x direction

Mass of particle 2, m' = 3 kg

Velocity of the particle 2, v' = 5 m/s in +y direction.

The x -coordinate of velocity of the centre of mass is given by :

v_x=\dfrac{mv+m'v'}{m+m'}\\\\v_x=\dfrac{2\times 4+3\times 0}{2+3}\\\\v_x=1.6\ m/s

The y -coordinate of velocity of the centre of mass is given by :

v_y=\dfrac{mv+m'v'}{m+m'}\\\\v_y=\dfrac{2\times 0+3\times 5}{2+3}\\\\v_y=3\ m/s

So, the the speed of the center of mass of the two-particle system given by :

v=\sqrt{v_x^2+v_y^2} \\\\v=\sqrt{1.6^2+3^2} \\\\v=3.4\ m/s

So, the speed of the center of mass of the two-particle system is 3.4 m/s.

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How long does it take for an airplane to change its velocity from 140 m/s to 180 m/s if its
kobusy [5.1K]

Answer:

Explanation:

a = 4ms⁻²,  Vf = 180 m/s  &  Vi = 140m/s

a = \frac{Vf-Vi}{t}

4 = \frac{180-140}{t}

t = 40/4

t = 10sec

To Measure Distance Use third Equation of Motion:

2aS = Vf²-Vi²

S = \frac{180*180 - 140*140}{2(4)}

S = 12800/8 = 1600m

4 0
2 years ago
What us the difference between accurate data and reproducible data
Nikolay [14]

It's just in the name! Accurate data is helpful, and correct, but reproducible data is all of that, and is able to be given to other people through different sources! At least, that's what my understanding of them are. Hope it helps!

4 0
3 years ago
(b) How much energy must be supplied to boil 2kg of water? providing that the specific latent heat of vaporization of water is 3
Lelu [443]

Complete question:

(b) How much energy must be supplied to boil 2kg of water? providing that the specific latent heat of vaporization of water is 330 kJ/kg. The initial temperature of the water is 20 ⁰C

Answer:

The energy that must be supplied to boil the given mass of the water is 672,000 J

Explanation:

Given;

mass of water, m = 2 kg

heat of vaporization of water, L =  330 kJ/kg

initial temperature of water, t = 20 ⁰C

specific heat capacity of water, c = 4200 J/kg⁰C

Assuming no mass of the water is lost through vaporization, the energy needed to boil the given water is calculated as;

Q = mc(100 - 20)

Q = 2 x 4200 x (80)  

Q = 672,000 J

Q = 672,000 J

Q = 672,000 J

Therefore, the energy that must be supplied to boil the given mass of the water is 672,000 J

8 0
2 years ago
What engine thrust is required for a rocket of mass 35 kg to leave the launching pad? 3.5 N. 35 N. 351 N. 3,500 N. 35,000 N.
soldi70 [24.7K]

In order for the object to move upward, it needs an upward force
that's at least equal to its own weight.

Weight = (mass) x (gravity) = (35 kg) x (9.8 m/s²) = 343 N.

The engine thrust has to be more than 343 N.

7 0
2 years ago
Challenge: Some but not all of the gasoline used by car's enegine is transformed into kinetic energy. Where might some of the en
Levart [38]
Most of the excess energy is released as waste heat into the air surrounding the engine. Small amounts of excess energy are also released as sound energy, and as electrical energy generated by the alternator in a car's engine.
8 0
3 years ago
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