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sukhopar [10]
2 years ago
6

In which case is the speed of laser light the slowest relative to the observer? a person walking in the street and who sees lase

r light emitted from a low‑speed car that moves in the same direction as the person a student sitting in the classroom and who see laser light from the pointer used by the instructor laser light emitted from a high‑speed aircraft and detected by another high‑speed aircraft traveling in the opposite direction a person who sees laser light passing through a bucket full of water it is the same for all of the above
Physics
1 answer:
AlekseyPX2 years ago
8 0

Answer:

<em>a person who sees laser light passing through a bucket full of water.</em>

Explanation:

According to relativity, the speed of light is the same irrespective of the relative speed between the source, and the observer. The only exception is when light travel from a less dense medium to a denser medium as in air into the bucket full of water. So the speed of light is slowest relative to the observer in the laser that passes through the bucket of water.

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a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres
shusha [124]

Answer:

Part a) When collision is perfectly inelastic

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b) When collision is perfectly elastic

v_m = \frac{m + M}{2m}\sqrt{5Rg}

Explanation:

Part a)

As we know that collision is perfectly inelastic

so here we will have

mv_m = (m + M)v

so we have

v = \frac{mv_m}{m + M}

now we know that in order to complete the circle we will have

v = \sqrt{5Rg}

\frac{mv_m}{m + M} = \sqrt{5Rg}

now we have

v_m = \frac{m + M}{m} \sqrt{5Rg}

Part b)

Now we know that collision is perfectly elastic

so we will have

v = \frac{2mv_m}{m + M}

now we have

\sqrt{5Rg} = \frac{2mv_m}{m + M}

v_m = \frac{m + M}{2m}\sqrt{5Rg}

6 0
2 years ago
on the surface of planet x a body with a mass of 10 kilograms weighs 40 newtons. The magnitude of the acceleration due to gravit
tamaranim1 [39]
Based on the Newton's second law of motion, the value of the net force acting on the object is equal to the product of the mass and the acceleration due to gravity. If we let a be the acceleration due to gravity, the equation that would allow us to calculate it's value is,
      W = m x a
where W is weight, m is mass, and a is acceleration. Substituting the known values,
    40 kg m/s² = (10 kg) x a
Calculating for the value of a from the equation will give us an answer equal to 4. 
ANSWER: 4 m/s². 

6 0
2 years ago
Read 2 more answers
A rock with density 1900 kg/m3 is suspended from the lower end of a light string. When the rock is in air, the tension in the st
wel

Answer:

the tension T2 when the rock is completely immersed is T2 =  29.05 N

Explanation:

from Newton's second law

F= m*a

where F= force , m= mass , a= acceleration

when the rock is suspended ,a=0 since it is at rest. Then

T1 - m*g = 0 , T1= tension when suspended in air , g= gravity

assuming constant density of the rock

m= ρ rock *V , where  ρ rock = density of the rock , V= volume

thus

T1= m*g = ρ rock *g*V

V=  T1/(ρ rock *g)

when the rock is submerged in oil , it receives an upward force that equals the weight of the volume of displaced oil (V displaced). Since it is completely submerged the volume displaced is the volume of the rock V=Vdisplaced  

When the rock is at rest , then

F= m*a=0

T2 + ρ oil *g*V displaced - ρ rock *g*V  =0

T2 = ρ rock *g*V - ρ oil *g*V = g*V (ρ rock - ρ oil)

T2 = g*V (ρ rock - ρ oil) = T1/(ρ rock *g) *g * (ρ rock - ρ oil)

T2 = T1 * (ρ rock - ρ oil)/ρ rock

replacing values

T2 = 48 N * (1900 kg/m3- 750 kg/m3)/ 1900 kg/m3 = 29.05 N

T2 =  29.05 N

3 0
2 years ago
The square loop shown in the figure moves into a 0.80T magnetic field at a constant speed of 10m/s. The loop has a resistance of
PolarNik [594]
The question ask to find and calculate the induced current in the loop as a function time and the best answer would be that the induced current in the loop is 0.08 amperes. I hope you are satisfied with my answer and feel free to ask for more if you have clarifications and further questions
7 0
2 years ago
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A 1000-kg car is driving toward the north along a straight horizontal road at a speed of 20.0 m/s. The driver applies the brakes
OlgaM077 [116]

Answer:

Explanation:

mass of car, m = 1000 kg

initial velocity, u = 20 m/s

final velocity, v = 0 m/s

distance, s = 120 m

Let a be the acceleration of motion

use third equation of motion

v² = u² + 2 as

0 = 20 x 20 + 2 x a x 120

a = - 1.67 m/s²

Let F be the force

Force, F  mass x acceleration

F = - 1000 x 1.67

F = - 1666.67 N

The direction of force is towards south and the magnitude of force is 1666.67 N.

8 0
2 years ago
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