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3 years ago

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An object is moving with the speed of light around the Earth. How much time will it take to complete one round trip along the eq

uator on the surface of earth if radius of earth is 6400km

1 answer:

dimulka [17.4K]3 years ago

8 0

Answer:

0.14 seconds

Explanation:

The speed of light in vacuum is approximately 3.0*10^8. The distance that would be covered by the object would be equivalent to the circumference of the cross-section of the earth on the equator.

Circumference = 2 $\pi$ *6400000 =4.02*10^7

Time = distance/speed = 4.2*10^7 / 3.0*10^8 =0.14s

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A bowling ball of mass 5.8 kg moves in a straight line at 4.34 m/s How fast must a Ping-Pong ball of mass 2.214 g move in a stra

lilavasa [31]

Answer: 11369.46 m/s

Explanation:

We have the following data:

$m_{1}=5.8 kg$ is the mass of the bowling ball

$V_{1}=4.34 m/s$ is the velocity of the bowling ball

$m_{2}=2.214 g \frac{1 kg}{1000 g}=0.002214 kg$ is the mass of the ping-pong ball

$V_{2}$ is the velocity of the ping-pong ball

Now, the momentum $p_{1}$ of the bowling ball is:

$p_{1}=m_{1}V_{1}$ (1)

$p_{1}=(5.8 kg)(4.34 m/s)$

$p_{1}=25.172 kg m/s$ (2)

And the momentum $p_{2}$ of the ping-pong ball is:

$p_{2}=m_{2}V_{2}$ (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

$p_{1}=p_{2}$ (4)

$m_{1}V_{1}=m_{2}V_{2}$ (5)

Isolating $V_{2}$ :

$V_{2}=\frac{m_{1}V_{1}}{m_{2}}$ (6)

$V_{2}=\frac{25.172 kg m/s}{0.002214 kg}$ (7)

Finally:

$V_{2}=11369.46 m/s$

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2 years ago

A 80-kg person sits on a 2.7kg chair. Each leg of the chair makes contact with the floor on a circle that is 1.3 cm in diameter.

Rashid [163]

Answer:

need help with this 2

Explanation:

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2 years ago

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Background information about reflection and refraction of light

KIM [24]

Answer:

reflection :

angle of reflection and angle of incidence is equal
Incident ray and reflect d ray plus normal lies on the same plan on a same point
light reflect uniformly if it is incident on a plan surface

refraction :

light will refrect if light goes from rare medium to a dense medium or vice versa
after the light is refracted, it retunes to its original direction
angle of incidence is not equal to the angle of refraction

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3 years ago

You toss a 0.40-kg ball at 9.0 ms/ to a 14-kg dog standing on an iced-over pond. The dog catches the ball and begins to slide on

just olya [345]

Answer:

a) $v_{f}$ = 0.25 m / s b) u = 0.25 m / s

Explanation:

a) To solve this problem let's start with the conservation of the moment, for this we define a system formed by the ball plus the dog, in this case all the forces are internal and the moment is conserved

We will write the data

m₁ = 0.40 kg

v₁₀ = 9.0 m / s

m₂ = 14 kg

v₂₀ = 0

Initial

po = m₁ v₁₀

Final

$p_{f}$ = (m₁ + m₂) vf

po = pf

m₁ v₁₀ = (m₁ + m₂) $v_{f}$

$v_{f}$ = v₁₀ m₁ / (m₁ + m₂)

$v_{f}$ = 9.0 (0.40 / (0.40 +14)

$v_{f}$ = 0.25 m / s

b) This is the reference frame of the center of mass of the system in this case the speed of this frame is the speed of the center of mass

u = 0.25 m / s

In the direction of movement of the ball

c) Let's calculate the kinetic energy in both moments

Initial

K₀ = ½ m₁ v₁₀² +0

K₀ = ½ 0.40 9 2

K₀ = 16.2 J

Final

$K_{f}$ = ½ (m₁ + m₂) $v_{f}$ 2

$K_{f}$ = ½ (0.4 +14) 0.25 2

$K_{f}$ = 0.45 J

ΔK = K₀ - $K_{f}$

ΔK = 16.2-0.445

ΔK = 1575 J

These will transform internal system energy

d) In order to find the kinetic energy, we must first find the velocities of the individual in this reference system.

v₁₀’= v₁₀ -u

v₁₀’= 9 -.025

v₁₀‘= 8.75 m / s

v₂₀ ‘= v₂₀ -u

v₂₀‘= - 0.25 m / s

$v_{f}$ ‘= $v_{f}$ - u

$v_{f}$ = 0

Initial

K₀ = ½ m₁ v₁₀‘² + ½ m₂ v₂₀‘²

Ko = ½ 0.4 8.75² + ½ 14.0 0.25²

Ko = 15.31 + 0.4375

K o = 15.75 J

Final

$k_{f}$ = ½ (m₁ + m₂) vf’²

$k_{f}$ = 0

All initial kinetic energy is transformed into internal energy in this reference system

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