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Black_prince [1.1K]
1 year ago
13

The operator of a space station observes a space vehicle approaching at a constant speed v. The operator sends a light signal at

speed c toward the space vehicle. What is the speed of the light signal as viewed from the space vehicle
Physics
1 answer:
GenaCL600 [577]1 year ago
5 0

Answer:

The speed of the light signal as viewed from the observer is c.

Explanation:

Recall the basic postulate of the theory of relativity that the speed of light is the same in ALL inertial frames. Based on this, the speed of light is independent of the motion of the observer.

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If electromagnetic radiation a has a lower frequency than electromagnetic radiation b the wavelength of a is
Jobisdone [24]
Inversely proportional to its frequency. If electromagnetic radiation A has a lower frequency than electromagnetic B, then compared to B, the wavelength of A is...? - equal - shorter - longer - exactly half the length of
5 0
1 year ago
A section of highway has the following flowdensity relationship q = 50k − 0.156k2 [with q in veh/h and k in veh/mi]. What is the
lions [1.4K]

Answer:

a) capacity of the highway section = 4006.4 veh/h

b) The speed at capacity = 25 mph

c) The density when the highway is at one-quarter of its capacity = k = 21.5 veh/mi or 299 veh/mi

Explanation:

q = 50k - 0.156k²

with q in veh/h and k in veh/mi

a) capacity of the highway section

To obtain the capacity of the highway section, we first find the k thay corresponds to the maximum q.

q = 50k - 0.156k²

At maximum flow density, (dq/dk) = 0

(dq/dt) = 50 - 0.312k = 0

k = (50/0.312) = 160.3 ≈ 160 veh/mi

q = 50k - 0.156k²

q = 50(160.3) - 0.156(160.3)²

q = 4006.4 veh/h

b) The speed at the capacity

U = (q/k) = (4006.4/160.3) = 25 mph

c) the density when the highway is at one-quarter of its capacity?

Capacity = 4006.4

One-quarter of the capacity = 1001.6 veh/h

1001.6 = 50k - 0.156k²

0.156k² - 50k + 1001.6 = 0

Solving the quadratic equation

k = 21.5 veh/mi or 299 veh/mi

Hope this Helps!!!

3 0
2 years ago
An object is projected with initial speed v0 from the edge of the roof of a building that has height H. The initial velocity of
Vsevolod [243]

Answer:

Explanation:

Initial velocity u = V₀ in upward direction so it will be negative

u = - V₀

Displacement s = H . It is downwards so it will be positive

Acceleration = g ( positive as it is also downwards )

Using the formula

v² = u² + 2 g s

v² = (- V₀ )² + 2 g H

= V₀² + 2 g H .

v = √ ( V₀² + 2 g H )

6 0
2 years ago
A test car and its driver, with a combined mass of 600 kg, are moving along a straight,horizontal track when a malfunction cause
ANEK [815]

Answer:

The two of the following measurements, when taken together, would allow engineers to find the total mechanical energy dissipated during the skid

B. The contact area of each tire with the track.

C. The co-efficent of static friction between the tires and the track.

D. The co-efficent of static friction between the tires and the track.

Explanation:

4 0
1 year ago
A solid sphere of brass (bulk modulus of 14.0 ✕ 1010 N/m2) with a diameter of 2.20 m is thrown into the ocean. By how much does
astra-53 [7]

Answer:

Diameter decreases by the diameter of 0.0312 m.

Explanation:

Given that,

Bulk modulus =  14.0 × 10¹⁰ N/m²

Diameter d = 2.20 m

Depth = 2.40 km

Pressure = ρ g h = 1030 × 9.81 × 2.4 × 1000

               = 24.25 × 10⁶  N/m²

Volume = \dfrac{4}{3} \pi r^3

         \dfrac{\Delta V}{V}=\dfrac{(\Delta r)^3}{r^3}

Bulk modulus is equal to

B = -\dfrac{\Delta P}{\dfrac{\Delta V}{V} }

now

B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{r^3} }

B = -\dfrac{24.25 \times 10^6}{\dfrac{(\Delta r)^3}{1.1^3} }

(\Delta r)^3 = \dfrac{24.25 \times 10^6 \times 1.1^3}{14.0 \times 10^{10}}

Δ r = -0.0156 m

change in diameter

Δ d = -2 × 0.0156

Δ d = -0.0312 m

Diameter decreases by the diameter of 0.0312 m.

7 0
2 years ago
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