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alisha [4.7K]
3 years ago
7

As a car heads down a highway traveling at a speed v away from a ground observer, which of the following statements are true abo

ut the measured speed of the light beam from the car's headlights? (Select all that apply.)
a. The ground observer measures the light speed to be c.
b. The driver measures the light speed to be c.
c. The driver measures the light speed to be c − v.
d. The ground observer measures the light speed to be c + v.
e. The ground observer measures the light speed to be c − v.
Physics
1 answer:
aliina [53]3 years ago
5 0

Answer:

a and b

Explanation:

According to theory of relativity by great Scientist Albert Eintein the speed of light is independent of frame of reference that the speed of light shall remain constnat whether the frame is inertial or non-inertial.

Therefore, both the ground observer and the driver shall measure same speed of light that is C.

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Ten seconds after starting from rest, a freely-falling object will have a speed of about ________________________
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Answer:

100 m/s

Explanation:

Speed increases at a rate of 10 m/s (actually 9.8 m/s) every second. Thus after 10 seconds, the speed is 10 x 10 = 100 m/s.

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The net flow of energy into and out of the earths system is referred to as energy budget. which type of energy is lost in space
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3 years ago
Nicolo` works on weekends at a Slow Food Parlor. He fills a pitcher full of Cola, places it on the counter top and gives the 2.6
Slav-nsk [51]

Answer:4.22 J

Explanation:

Given

mass of pitcher m=2.6\ kg

Force applied  F=8.8\ N

distance moved s=48\ cm

Applying work-Energy theorem which states that work done by all the forces is equal to the change in kinetic energy of the object

Work done by force W=F\cdot s

W=8.8\times 0.48 J

change in kinetic Energy =\frac{1}{2}mv^2-0

8.8\times 0.48=\Delta K.E.

K.E.-0=4.22

K.E.=4.22\ J

8 0
4 years ago
A +8.75 μC point charge is glued down on a horizontal frictionless table. It is tied to a -6.50 μC point charge by a light, nonc
lisov135 [29]

(a) The tension on the wire when the two charges have opposite signs is 383.5 N.

(b) The tension on the wire if both charges were negative is 3.640.25 N.

The given parameters;

  • <em>first charge, q₁ = 8.75 μC </em>
  • <em>second charge, q₂ = -6.5 μC  </em>
  • <em>electric field, E = 1.85 x 10⁸ N/C</em>
  • <em>distance between the two charges, r = 2.5 cm</em>

<em />

(a)

The attractive force between the charges is calculated as follows;

F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = -819 \ N

The force on the negative charge due to the electric field is calculated as follows;

F_2 = Eq_2\\\\F_2 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_2 = 1202.5 \ N

The tension on the wire is the resultant of the two forces and it is calculated as follows;

T = F_2 + F_1\\\\T = 1202.5 - 819\\\\T = 383.5 \ N

(b) when the two charges are negative

The repulsive force between the two charges is calculated as follows;

F_1 = \frac{kq_1q_2}{r^2} \\\\F_1 = \frac{(9\times 10^9) \times (-8.75\times 10^{-6})\times (-6.5\times 10^{-6})}{(0.025)^2} \\\\F_1 = 819 \ N

The force on the first negative charge due to the electric field is calculated as follows;

F_2 = Eq_1\\\\F_2 = (1.85 \times 10^8)\times (8.75 \times 10^{-6})\\\\F_2 = 1618.75 \ N

The force on the second negative charge due to the electric field is calculated as follows;

F_3 = Eq_2\\\\F_3 = (1.85 \times 10^8) \times (6.5 \times 10^{-6})\\\\F_3 = 1202.5 \ N

The tension on the wire is the resultant of the three forces and it is calculated as follows;

T= F_1 + F_2 + F_3\\\\T= 819 + 1618.75 + 1202.5\\\\T = 3,640.25 \ N

Learn more here:brainly.com/question/19565286

5 0
3 years ago
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