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3 years ago

SOMEBODY PLEASE HELP!! 20 PTS!!!

Physics

2 answers:

Neporo4naja [7]3 years ago

4 0

1. D
2. C
3. D
4. B
5. D

aliya0001 [1]3 years ago

3 0

Answer:

1. Reflection

2. travel from one medium to another

3. Same waves to travel in opposite direction.

Explanation:

1. When a wave strikes a solid barrier, it bounces back in the same medium. This wave behavior of bouncing back is known as reflection. Its like a basketball hitting a backboard. The ball bounces back at the same angle as it was incident. ∠i = ∠r

2. For refraction to occur in a wave, the wave must travel from one medium to another. When light travels from through mediums of different optical densities, it bends. The wave bends away normal when it enters from denser medium to rarer medium. The wave bends towards the normal when it enters from rarer to denser medium. The angle of refraction and angle of incidence are related by Snell's law.

$\frac{sin(i)}{sin(r)} = \frac{\mu_2}{\mu_1}$

3. The formation of standing wave requires two same waves to travel in the opposite direction and interfere. The incident wave and reflected wave when interfere, form standing waves. There waves are also resonances or harmonics. A standing wave oscillates at one place and does not transfers any energy.

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A railroad track and a road cross at right angles. An observer stands on the road and watches an eastbound train traveling at 60

mamaluj [8]

Answer:

After 4 s of passing through the intersection, the train travels with 57.6 m/s

Solution:

As per the question:

Suppose the distance to the south of the crossing watching the east bound train be x = 70 m

Also, the east bound travels as a function of time and can be given as:

y(t) = 60t

Now,

To calculate the speed, z(t) of the train as it passes through the intersection:

Since, the road cross at right angles, thus by Pythagoras theorem:

$z(t) = \sqrt{x^{2} + y(t)^{2}}$

$z(t) = \sqrt{70^{2} + 60t^{2}}$

Now, differentiate the above eqn w.r.t 't':

$\frac{dz(t)}{dt} = \frac{1}{2}.\frac{1}{sqrt{3600t^{2} + 4900}}\times 2t\times 3600$

$\frac{dz(t)}{dt} = \frac{1}{sqrt{3600t^{2} + 4900}}\times 3600t$

For t = 4 s:

$\frac{dz(4)}{dt} = \frac{1}{sqrt{3600\times 4^{2} + 4900}}\times 3600\times 4 = 57.6\ m/s$

4 0

3 years ago

These the flow of electrons (the current) and where some of the electrons' energy gets converted into heat. (Lessons 5.01-5.03)

liq [111]

Answer:

Conductors.

Explanation:

Conductors allow the flow of electrons i. e. the current and where some of the electrons energy gets converted into heat while some electrons energy converted into light. Conductors are the materials which has the ability to allow heat and electricity to flow through them. Metals such as gold, copper, Silver, Aluminum, Mercury, Steel, Iron and salty water etc are good conductors of heat and electricity.

3 0

3 years ago

Which quantities are scalars?

Fofino [41]

Answer:

A and B

Explanation:

Because both of them have only magnitude not direction.

7 0

3 years ago

Read 2 more answers

A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)

kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

$W=Fd cos \theta$

where

F is the magnitude of the force applied

d is the displacement of the crate

$\theta$ is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied horizontally. Therefore, the work done is

$W=(230)(4.0)(cos 0^{\circ})=920 J$

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

$W=(1300)(4.0)(cos 0^{\circ})=5200 J$