1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
vovikov84 [41]
3 years ago
14

In which of the following materials is the velocity of light greatest?

Physics
2 answers:
just olya [345]3 years ago
8 0
<span>The velocity of light is by far the greatest in air. There is nothing in air to stop or impede the movement of light in air. When light travels in air it will not stop unless running into clouds, smoke, smoke, droplets, or impurities. Any other substance is harder for light to travel through.</span>
borishaifa [10]3 years ago
7 0
A is the right answer
You might be interested in
How much heat is needed to change 1.25 kg of steak at 100°C to water at 100°C?
cricket20 [7]

The heat required to change 1.25 kg of steak is 2825 kJ /kg.

<u>Explanation</u>:

Given, mass m = 1.25 kg,     Temperature t = 100 degree celsius

To calculate the heat required,

                                 Q = m \times L

where m represents the mass in kg,

          L represents the heat of vaporization.

When a material in the liquid state is given energy, it changes its phase from liquid to vapor and the energy absorbed in this process is called heat of the vaporization. The heat of vaporization of the water is about 2260 kJ/kg.

                                  Q = 1.25 \times 2260

                                  Q = 2825 kJ /kg.

7 0
4 years ago
Which three statements describe electromagnetic waves?
o-na [289]

Answer:

It is most likely option A B and C

3 0
3 years ago
Read 2 more answers
Which of the following periodic properties increases as you move from left to right across the periodic table?
andreyandreev [35.5K]

Answer:

the answer choice is B

Explanation:

as we move from left to right , the atomic size decreases due to higher number of protons in the nucleus, which are able to attract the electrons more strongly. and so the electronegativity and  electron affinity increases for the same reason. the nuclear charge increase due to more protons , and without an increase in inner electrons , there is less shielding effect. so effective nuclear charge increases.

7 0
3 years ago
Read 2 more answers
2. Two toy cars are involved in a race. Car A has mass m while car B has mass 2m. a. The two cars have the same force applied to
ArbitrLikvidat [17]

Answer:

a) The kinetic energy of the two cars is the same

the moment of car 2 is greater than the moment of car 1

b)  the kinetic energy of car 1 is greater than that of car 2

the moment of the two cars is the same

Explanation:

a) to know the kinetic energy of each car, we must find the speed, use Newton's second law to find the acceleration

Car 1

     F = m a

    a = F / m

Let's use kinematics to find the velocity after x = 1 m

       v² = v₀² + 2 a x

The initial speed is zero

       v = √ (2 F/m  x)

For the distance of x = 1 m

        v₁ = √ (2 F / m)

Car 2

      F = 2m a

      a = F / 2m

      v² = 2 a x

      v = √ (F/m  x)

 For x = 1 m

       v₂ = √(F / m)

Let's calculate the kinetic energy of each car

Car 1

      K₁ = ½ m v₁²

      K₁ = ½ m 2F / m

      K₁ = F

Car 2

      K₂ = ½ 2m v₂²

      K₂ = ½ 2m F / m

      K₂ = F

The kinetic energy of the two cars is the same

Let's calculate the moment

Car 1

   P₁ = m v₁

   P₁ = m √ (2F / m)

Car 2

    P₂ = 2m v²

    P₂ = 2m √(F / m)

We see that the moment of car 2 is greater than the moment of car 1

b) in this part the force is applied by t = 10 s

Acceleration is the same, let's find the speed

Car1

          v = v₀ + a t

          v = F / m t

          v₁ = F / m 10

Car 2

           v₂ = F / 2m 10

           v₂ = F / m 5

Let's calculate the kinetic energy of each car

Car 1

           K₁ = ½ m v₁²

           K₁ = ½ m (F / m 10)²

           K₁ = 50 F² / m

Car2

         K₂ = ½ 2m v₂²

         K₂ = m (F / m 5)²

         K₂ = 25 F² / m

In this case we see that the kinetic energy of car 1 is greater than that of car 2

Let's calculate the moment

Car 1

         P₁ = m v₁

         P₁ = m F / m 10

         P₁ = 10 F

 

Car 2

        P₂ = 2m v₂

        P₂ = 2m F / m 5

        P₂ = 10 F

In this case the moment of the two cars is the same

7 0
3 years ago
The engineer of a passenger train traveling at 25.0 m/s sights a freight train whose caboose is 200 m ahead on the same track. T
zaharov [31]

a) The train collide after 22.5 seconds

b) The trains collide at the location x = 537.5 m

c) See graph in attachment

d) The freight train must have a head start of 500 m

e) The deceleration must be smaller (towards negative value) than -0.25 m/s^2

f) The two trains avoid collision if the acceleration of the freight train is at least 0.35 m/s^2

Explanation:

a)

We can describe the position of the passenger train at time t with the equation

x_p(t)=u_p t + \frac{1}{2}at^2

where

u_p = 25.0 m/s is the initial velocity of the passenger train

a=-0.100 m/s^2 is the deceleration of the train

On the other hand, the position of the freight train is given by

x_f(t)=x_0 + v_f t

where

x_0=200 m is the initial position of the freight train

v_f = 15.0 m/s is the constant velocity of the train

The collision occurs if the two trains meet, so

x_p(t)=x_f(t)\\u_pt+\frac{1}{2}at^2=x_0+v_ft\\25t+\frac{1}{2}(-0.100)t^2=200+15t\\0.050t^2-10t+200=0

This is a second-order equation that has two solutions:

t = 22.5 s

t = 177.5 s

We are interested in the 1st solution, which is the first time at which the passenger train collides with the freight train, so t = 22.5 seconds.

b)

In order to find the location of the collision, we just need to substitute the time of the collision into one of the expression of the position of the trains.

The position of the freight train is

x_f(t)=x_0 +v_ft

And substituting t = 22.5 s, we find:

x_f(22.5)=200+(15)(22.5)=537.5 m

We can verify that the passenger train is at the same position at the time of the collision:

x_p(22.5)=(25.0)(22.5)+\frac{1}{2}(-0.100)(22.5)^2=537.5 m

So, the two trains collide at x = 537.5 m.

c)

In the graph in attachment, the position-time graph of each train is represented. We have:

  • The freight train is moving at constant speed, therefore it is represented with a straight line with constant slope (the slope corresponds to its velocity, so 15.0 m/s)
  • The passenger train has a uniformly accelerated motion, so it is a parabola: at the beginning, the slope (the velocity) is higher than that of the freight train, however later it decreases due to the fact that the train is decelerating

The two trains meet at t = 22.5 s, where the position is 537.5 m.

d)

In order to avoid the collision, the freight train must have a initial position of

x_0'

such that the two trains never meet.

We said that the two trains meet if:

x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0' + v_f t

Re-arranging,

\frac{1}{2}at^2+(u_p-v_f)t-x_0'=0\\-\frac{1}{2}at^2+(v_f-u_p)t+x_0'=0

Substituting the values for the acceleration and the velocity,

0.05t^2-10t+x_0'=0

The solution of this equation is given by the formula

t=\frac{+10\pm \sqrt{10^2-4\cdot 0.05 \cdot x_0'}}{2(0.05)}

The two trains never meet if the discrimant is negative (so that there are no solutions to the equation), therefore

10^2-4\cdot 0.05 \cdot x_0'100\\x_0'>500 m

Therefore, the freight train must have a head start of 500 m.

e)

In this case, we want to find the acceleration a' of the passenger train such that the two trains do not collide.

We solve the problem similarly to part d):

x_p(t)=x_f(t)\\u_p t + \frac{1}{2}a't^2=x_0 + v_f t

Re-arranging

\frac{1}{2}a't^2+(u_p-v_f)t-x_0=0\\-\frac{1}{2}a't^2+(v_f-u_p)t+x_0=0

Substituting,

-0.5at^2-10t+200=0

The solution to this equation is

t=\frac{+10\pm \sqrt{10^2-4\cdot (-0.5a') \cdot (200)}}{2(0.05)}

Again, the two trains never meet if the discriminant is negative, so

10^2-4\cdot (-0.5a') \cdot (200)

So, the deceleration must be smaller (towards negative value) than -0.25 m/s^2

f)

In this case, the motion of the freight train is also accelerated, so its position at time t is given by

x_f(t)=x_0 + v_f t + \frac{1}{2}a_ft^2

where a_f is the acceleration of the freight train.

Then we solve the problem similarly to the previous part: the two trains collide if their position is the same,

x_p(t)=x_f(t)\\u_p t + \frac{1}{2}at^2=x_0 + v_f t+\frac{1}{2}a_ft^2

Re-arranging,

\frac{1}{2}(a_f-a)t^2+(v_f-u_p)t+x_0=0\\\\\frac{1}{2}(a_f-0.100)t^2-10t+200=0

And the solution is

t=\frac{+10\pm \sqrt{10^2-4\cdot (0.5a_f-0.05) \cdot (200)}}{2(0.5a_f-0.05)}

Again, the two trains avoid collision if the discriminant is negative, so

10^2-4\cdot (0.5a_f-0.05) \cdot (200)0.35 m/s^2

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0
3 years ago
Other questions:
  • Ted Clubber Lang. A hook in boxing primarily involves horizontal flexion of the shoulder while maintaining a constant angle at t
    9·1 answer
  • A stone was dropped off a cliff and hit the ground with a speed of 136 ft/s. what is the height of the cliff? (use 32 ft/s2 for
    6·1 answer
  • A ball player catches a ball 3.13 s after throwing it vertically upward. a)With what speed did he throw it?
    6·1 answer
  • 2 characteristics of constant speed
    5·1 answer
  • Which graph shows a negative acceleration
    10·1 answer
  • Melanie completes a long distance run at an average speed of 6 mph. If it takes her 3 hours, how far did she run?
    11·1 answer
  • Help plz anyone ?????
    6·1 answer
  • The section of a longitudinal wave in which the particles are densely packed together
    5·1 answer
  • Question is in the pic
    5·2 answers
  • Explain numericals of electricity chapter of class 10 and also diagrams.What is ammeter?What is voltmeter?potential difference?O
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!