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belka [17]
4 years ago
11

Red, green, and blue light rays each enter a drop of water from the same direction. Which light ray's path through the drop will

bend the most, and which will bend the least? Why?
Physics
1 answer:
erik [133]4 years ago
7 0
Blue light will bend the most because it has a shorter wavelength and refracts at a higher angle.
Red light will bend the least because it has a longer wavelength and refracts at a lower angle. 
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A rocket is moving away from the solar system at a speed of 7.5 ✕ 103 m/s. It fires its engine, which ejects exhaust with a spee
Elodia [21]

Answer:

a.2.5x 10^3 m/s

b.mr=48kg/s

Explanation:

A rocket is moving away from the solar system at a speed of 7.5 ✕ 103 m/s. It fires its engine, which ejects exhaust with a speed of 5.0 ✕ 103 m/s relative to the rocket. The mass of the rocket at this time is 6.0 ✕ 104 kg, and its acceleration is 4.0 m/s2. What is the velocity of the exhaust relative to the solar system? (B) At what rate was the exhaust ejected during the firing?

velocity of the exhaust relative to the solar system

velocity of the rocket -velocity of the exhaust relative to the rocket.

7.5 ✕ 103 m/s-5.0 ✕ 103 m/s

2.5x 10^3 m/s

. b  we will look for the thrust of the rocket

T=ma

T=6.0 ✕ 104 kg*4.0 m/s2

T=2.4*10^5N

f=mass rate *velocity of the exhaust

T=2.4*10^5N=mr*5.0 ✕ 10^3 m/s

mr=2.4*10^5N/5.0 ✕ 10^3

mr=48kg/s

5 0
3 years ago
A 25.0 g marble sliding to the right at 20.0 cm/s overtakes and collides elastically with a 10.0 g marble moving in the same dir
ikadub [295]
In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,

   KE1  = KE2

The kinetic energy of the system before the collision is solved below.

  KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
  KE1 = 6125 g cm²/s²

This value should also be equal to KE2, which can be calculated using the conditions after the collision.

KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)

The value of x from the equation is 17.16 cm/s.

Hence, the answer is 17.16 cm/s. 
6 0
4 years ago
Which motor and body should Devon use to build the car with the greatest acceleration?
PilotLPTM [1.2K]

<u>Complete Question:</u>

Devon has several toy car bodies and motors. The motors have the same mass, but they provide different amounts of force, as shown in this table.  

The bodies have the masses shown in this table (refer attached figure).  

Which motor and body should Devon use to build the car with the greatest acceleration?

motor 1, with body 1

motor 1, with body 2

motor 2, with body 1

motor 2, with body 2

<u>Answer:</u>

Devon should build the car with motor 2 and body 1 for having the greatest acceleration.

<u>Explanation:</u>

As per Newton's second law of motion, the acceleration of any object is directly proportional to the force on the object and inversely proportional to the mass of the object.

It can be seen that motor 2 has greater force than the force provided by motor 1. Similarly, the mass of body 1 is found to be lesser compared to mass of body 2. So,

          acceleration =\frac{\text { Force }}{\text { mass }}

It gives, the system with motor 2 and body 1 the maximum acceleration. So the car should be built with motor 2 and body 1.

5 0
3 years ago
Read 2 more answers
A railroad car moving at a speed of 3.46 m/s overtakes, collides and couples with two coupled railroad cars moving in the same d
I am Lyosha [343]

Answer:

2.09\ \text{m/s}

22298.4\ \text{J}

Explanation:

m = Mass of each the cars = 1.6\times 10^4\ \text{kg}

u_1 = Initial velocity of first car = 3.46 m/s

u_2 = Initial velocity of the other two cars = 1.4 m/s

v = Velocity of combined mass

As the momentum is conserved in the system we have

mu_1+2mu_2=3mv\\\Rightarrow v=\dfrac{u_1+2u_2}{3}\\\Rightarrow v=\dfrac{3.46+2\times 1.4}{3}\\\Rightarrow v=2.09\ \text{m/s}

Speed of the three coupled cars after the collision is 2.09\ \text{m/s}.

As energy in the system is conserved we have

K=\dfrac{1}{2}mu_1^2+\dfrac{1}{2}2mu_2^2-\dfrac{1}{2}3mv^2\\\Rightarrow K=\dfrac{1}{2}\times 1.6\times 10^4\times 3.46^2+\dfrac{1}{2}\times 2\times 1.6\times 10^4\times 1.4^2-\dfrac{1}{2}\times 3\times 1.6\times 10^4\times 2.09^2\\\Rightarrow K=22298.4\ \text{J}

The kinetic energy lost during the collision is 22298.4\ \text{J}.

6 0
3 years ago
Look at the picture below. The boy is swinging back and forth on the swing. At which point is the potential energy of a swing th
Ilia_Sergeevich [38]
The potential energy of a swing is greatest at the top of the swing. (Point A).
5 0
3 years ago
Read 2 more answers
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