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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.

You might be interested in
Find your acceleration from 8.3 m/s to 12.5 m/s in 1.24 seconds.
balu736 [363]

The <u>average</u> acceleration for an object undergoing this change in velocity is

(12.5 m/s - 8.3 m/s) / (1.24 s) = (4.2 m/s) / (1.24 s) ≈ 3.4 m/s²

4 0
2 years ago
Light of wavelength 400 nm is incident on a single slit of width 15 microns. If a screen is placed 2.5 m from the slit. How far
olganol [36]

Answer:

0.0667 m

Explanation:

λ = wavelength of light = 400 nm = 400 x 10⁻⁹ m

D = screen distance = 2.5 m

d = slit width = 15 x 10⁻⁶ m

n = order = 1

θ = angle = ?

Using the equation

d Sinθ = n λ

(15 x 10⁻⁶) Sinθ = (1) (400 x 10⁻⁹)

Sinθ = 26.67 x 10⁻³

y = position of first minimum

Using the equation for small angles

tanθ = Sinθ = y/D

26.67 x 10⁻³ = y/2.5

y = 0.0667 m

5 0
3 years ago
Why is it important to be able to trace the pole connection on a meter back to the same type of pole at the electrical source?
lozanna [386]

Answer:

Explanation:

A grounded wire is sometimes strung along the tops of the towers to provide lightning protection.

In areas where the neutral is grounded or earthed, it is essential to endure that the neutral and the live or hot wires are not confused for each other.

When this happens, the fuses on the transformer will not operate unless the fault is very close to the transformer. The fuses in the consumer's intake box, will not operate.

6 0
3 years ago
Please help Need good grade
ivann1987 [24]

Answer:

The other angle is 120°.

Explanation:

Given that,

Angle = 60

Speed = 5.0

We need to calculate the  range

Using formula of range

R=\dfrac{v^2\sin(2\theta)}{g}...(I)

The range for the other angle is

R=\dfrac{v^2\sin(2(\alpha-\theta))}{g}....(II)

Here, distance and speed are same

On comparing both range

\dfrac{v^2\sin(2\theta)}{g}=\dfrac{v^2\sin(2(\alpha-\theta))}{g}

\sin(2\theta)=\sin(2\times(\alpha-\theta))

\sin120=\sin2(\alpha-60)

120=2\alpha-120

\alpha=\dfrac{120+120}{2}

\alpha=120^{\circ}

Hence, The other angle is 120°

6 0
4 years ago
A large airplane typically has three sets of wheels: one at the front and two farther back, one on each side under the wings. Co
Tems11 [23]

(a) The force the ground exerts on each set of rear wheels when the plane is at rest on the runway is 0.743 MN.

(b) The force the ground exerts on the front set of wheels is 0.239 MN.

<h3>Center mass of the airplane</h3>

The concept of center mass of an object can be used to dtermine the mass distribution of the airplane along the line through the center.

<h3>Some assumptions</h3>
  • The wheels under the wind do not pass through the center line.
  • The position of the front wheel is constant and it is zero mark (origin).
  • The rear wheels are at 21.7 m mark

Position of the center mass of the plane is calculated as follows;

Let the position of the center mass, Xcm = y

the center mass is 3 m in front of rear wheels, that is

21.7 - y = 3

y = 21.7 - 3

y = 18.7 m

Xcm = 18.7 m

<h3>Mass of the plane at the position of the rear wheels</h3>

Let the mass of the plane at front wheels = M1

Let the mass of the plane at rear wheels = M2

X_{cm} = \frac{M_1x_1 + M_2x_2}{M_1 + M_2}

18.7 = \frac{M_1(0) + M_2(21.7)}{177000} \\\\3,309,900 = 21.7M_2\\\\M_2 = \frac{3,309,900}{21.7} \\\\M_2 = 152,529.95 \ kg

<h3>Force exerted by the ground on each rear wheel</h3>

There are two rear wheels, and the force exerted on each wheel due to mass of the airplane at this position is calculated as follows;

W = mg\\\\W_1 = W_2 = \frac{1}{2} (mg) = \frac{1}{2} (152,529.95 \times 9.8) = 743,396.76 \ N= 0.743 \ MN

<h3>Mass of the plane at the position of the front wheel</h3>

M1 + M2 = 177,000

M1 = 177,000 - M2

M1 = 177,000 - 152,529.95

M1 = 24,470.05 kg

<h3>Force exerted by the ground on the front wheel</h3>

W = mg

W = 24,470.05 x 9.8

W = 239,806.5 N = 0.239 MN

Learn more about center mass here: brainly.com/question/13499822

7 0
2 years ago
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