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2 years ago

How is momentum conserved is a Newton's cradle when one steel ball hits the other

Physics

1 answer:

Vikentia [17]2 years ago

4 0

Answer:

Newton's Cradle experiment perfectly demonstrates the law of conservation of momentum which states that in a closed system, momentum before the collision is equal to momentum after the collision of the system.

As the first ball swings in the air, it gains momentum. When it strikes the second ball, it loses momentum and second ball gains equal amount of momentum. The second ball transfers the momentum to third, then fourth and till the last. The last ball when gains the same momentum swings up in the air. This continues. This experiment is done in drag free condition. This means there is no loss of momentum or opposing forces present.

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The formula v = √2.5r models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radiu

mel-nik [20]

Answer:

The maximum safe speed of the car is 30.82 m/s.

Explanation:

It is given that,

The formula that models the maximum safe speed, v, in miles per hour, at which a car can travel on a curved road with radius of curvature r r, is in feet is given by :

$v=\sqrt{2.5\ r}$ .........(1)

A highway crew measures the radius of curvature at an exit ramp on a highway as 380 feet, r = 380 feet

Put the value of r in equation (1) as :

$v=\sqrt{2.5\ \times 380}$

v = 30.82 m/s

So, the maximum safe speed of the car is 30.82 m/s. Hence, this is the required solution.

4 0

3 years ago

If two particles have equal kinetic energies, are their momenta necessarily equal? explain.

Mandarinka [93]

Answer:

No the given statement is not necessarily true.

Explanation:

We know that the kinetic energy of a particle of mass 'm' moving with velocity 'v' is given by

$K.E=\frac{1}{2}mv^{2}$

Similarly the momentum is given by $m\times v$

For 2 particles with masses $m_{1},m_{2}$ and moving with velocities $v_{1},v_{2}$ respectively the respective kinetic energies is given by

$K.E_{1}=\frac{1}{2}m_{1}v_{1}^{2}$

$K.E_{2}=\frac{1}{2}m_{2}v_{2}^{2}$

Similarly For 2 particles with masses $m_{1},m_{2}$ and moving with velocities $v_{1},v_{2}$ respectively the respective momenta are given by

$p_{1}=m_{1}\times v_{1}$

$p_{2}=m_{2}\times v_{2}$

Now since it is given that the two kinetic energies are equal thus we have

$\frac{1}{2}m_{1}v_{1}^{2}=\frac{1}{2}m_{2}v_{2}^{2}\\\\(m_{1}v_{1})\times v_{1}=(m_{2}v_{2})\times v_{2}\\\\p_{1}\times v_{1}=p_{2}\times v_{2}\\\\\therefore \frac{p_{1}}{p_{2}}=\frac{v_{2}}{v_{1}}............(i)$

Thus we infer that the moumenta are not equal since the ratio on right of 'i' is not 1 , and can be 1 only if the velocities of the 2 particles are equal which becomes a special case and not a general case.

5 0

3 years ago

The Hubble Telescope The Hubble Telescope is a large telescope that orbits Earth. It was named after Edwin Hubble. Hubble was an

Marrrta [24]

Quoting from the article itself:

"Since it is above Earth's atmosphere, it gives us clearer pictures of space than telescopes on Earth can."

3 0

3 years ago

A car sitting at a red light begins to accelerate at 2 m/s2 when the light turns green. It continues with this acceleration unti

jek_recluse [69]

Answer:

x_total = 600 m

Explanation:

This is an exercise and kinematics, let's find the time it takes to reach the velocity 20 m / s

v = v₀ + a t

as part of rest v₀ = 0

t = v / a

t = 20/2

t = 10 s

let's find the distance traveled in this time

x₁ = vo t + ½ a t2

x₁ = 0 + ½ 2 10²

x₁ = 100 m

The remaining time is

t₂ = 35 - t

t₂ = 35 - 10

t₂ = 25 s

as in this range it has a constant speed

v = x₂ / t₂

x₂ = v t₂

x₂ = 20 25

x₂ = 500 m

the total distance traveled is

x_total = x₁ + x₂

x_total = 100 + 500

x_total = 600 m

7 0

2 years ago

A student wants to start a small business in school. Write down six items that

Fiesta28 [93]

Answer:

packets of pen

packets of pencil

copies

books

bottles

mask

3 0

2 years ago