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konstantin123 [22]
3 years ago
13

Albert is piloting his spaceship, heading east with a speed of 0.92 cc relative to Earth. Albert's ship sends a light beam in th

e forward (eastward) direction, which travels away from his ship at a speed cc. Meanwhile, Isaac is piloting his ship in the westward direction, also at 0.92 cc, toward Albert's ship
With what speed does Isaac see Albert's light beam pass his ship?
Physics
1 answer:
ohaa [14]3 years ago
4 0

Answer:

He sees the light as 1c

Explanation:

According to relativity, the speed of light is the same in all inertial frame of reference.

If we were to add the velocities as applicable to a normal moving bodies, the relative speed of the light beam will exceed c which will break relativistic law since nothing can go past the speed of light.

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I have 2 bulbs one coverts 60% of energy to light 45% which is the most efficient and what happens to the rest of the energy
Leya [2.2K]

Answer:

The first one (60%)

Explanation:

The first one converts 15% more energy than the other one. Therefore, it is more efficient.

Hope this helps!

5 0
2 years ago
Read 2 more answers
Determine the inductance l of a 0.55-m-long air-filled solenoid 2.7 cm in diameter containing 9000 loops.
faltersainse [42]

Self inductance of a solenoid does not depend on magnetic field or current. When deriving the equation you can see that its self inductance is only dependent on geometric factors

4 0
3 years ago
A 30-g bullet is fired with a horizontal velocity of 450 m/s and becomes embedded in block B, which has a mass of 3 kg. After th
Zolol [24]

Answer:

(a) The velocity of the bullet and B after the first impact is 4.4554 m/s.

(b) The velocity of the carrier is 0.40872 m/s.

Explanation:

(a) To solve the question, we  apply the principle of conservation of linear momentum as follows.

we note that the distance between B and C is 0.5 m

Then we  have

Sum of initial momentum = Sum of final momentum

0.03 kg × 450 m/s = (0.03 kg + 3 kg) × v₂

Therefore v₂ = (13.5 kg·m/s)÷(3.03 kg) = 4.4554 m/s

The velocity of the bullet and B after the first impact = 4.4554 m/s

(b) The velocity of the carrier is given as follows

Therefore from the conservation of linear momentum we also have

(m₁ + m₂)×v₂  = (m₁ + m₂ + m₃)×v₃

Where:

m₃ = Mass of the carrier = 30 kg

Therefore

(3.03 kg)×(4.4554 m/s) = (3.03 kg+30 kg) × v₃

v₃ = (13.5 kg·m/s)÷ (33.03 kg) = 0.40872 m/s

The velocity of the carrier = 0.40872 m/s.

3 0
3 years ago
Vector B~ has x, y, and z components of 8.7,
blagie [28]

Answer:

The magnitude of B is 10.95\ units

Explanation:

we know that

The magnitude of Vector B is

B=\sqrt{x^{2}+y^{2}+z^{2}}

where

x,y and z are the components of vector B

we have

x=8.7\ units, y=1.4\ units,z=6,5\ units

substitute

B=\sqrt{8.7^{2}+1.4^{2}+6.5^{2}}

B=\sqrt{119.9}

B=10.95\ units

3 0
3 years ago
A curve of radius 166 m is banked at an angle of 11°. An 736-kg car negotiates the curve at 81 km/h without skidding. Neglect th
lianna [129]

Answer:

F_n = 7509.33\ N

Explanation:

given,

radius of curve = 166 m

angle of the banked road = 11°

mass of car = 736 Kg

speed of the curve = 81 km/h

                                = 81 x 0.278 = 22.52 m/s

normal force acting on the tires

on tire there will be two force acting on it

first one will be force acting due to weight and the other force acting on the tire is due to centripetal force.

F_n = m g cos \theta+ \dfrac{mv^2}{r} sin \theta

F_n = 736 \times 9.8 \times cos 11^0+\dfrac{736 \times 22.52^2}{166} sin 11^0

F_n = 7080.28+429.047

F_n = 7509.33\ N

6 0
2 years ago
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