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3 years ago

How to do this question plz

Physics

1 answer:

Aleks04 [339]3 years ago

6 0

Answer:

Explanation:

Most discussions of refraction will have a diagram similar to that of C.

The angle of incidence is measured from the normal to the boundary, as is the angle of refraction. The product of the sine of the angle and the index of refraction is the same for the media on either side of the boundary.

n₁·sin(θ₁) = n₂·sin(θ₂)

For media, such as optical fiber, that has an index of refraction greater than 1, the angle of refraction will be smaller in that media than the angle of incidence coming from air.

Figure C applies.

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How is an appliance's power related to the amount of energy it uses?

zysi [14]

The more powerful an appliance is, the more amounts of energy it uses. For instance a jack hammer that does not create a lot of destruction or change does not use a lot of energy. If a jack hammer that creates a lot of destruction and change probably uses a lot more energy and makes more sound.

6 0

3 years ago

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

5 0

3 years ago

A ship is towed through a narrow channel by applying forces to three ropes attached to its bow. Determine the magnitude and orie

Y_Kistochka [10]

This question is solved using an available similar problem as data provided for the forces was not given.

Repeat the same steps outlined for your problem.

Regards.

Answer:

F = 1.598 KN , Q = 90 degree (+ y-axis)

Explanation:

Sum of Forces in x-direction to the left (+)

2 cos (30) + 3cos (60) + F*cos (Q) = F_a ..... 1

Sum of Forces in y-direction to the up (+)

2 sin (30) + F*sin (Q) - 3 sin (60) ...... 2

Using Eq 2 and solve:

F*sin (Q) = 1.598 KN

F_min when sin (Q) is max, max possible value of sin(Q) = 1 @ Q = 90 degrees.

Hence,

F_min = 1.598 KN

Using Eq 1 @ Q = 90 degrees and F = 1.598 KN:

F_a = 2 cos (30) + 3cos (60) = 3.2 KN

6 0

3 years ago

A car travelled a distance of 200 m with initial velocity of 216 km/hr. Calculate the acceleration

pogonyaev

Answer:

$a = 16\ m/s^2$

Explanation:

When the velocity changes uniformly, the object has a constant acceleration. The acceleration, the velocities, and the distance are related by the equation:

$v_f^2=v_o^2+2ax$

Where:

vf = final velocity

vo = initial velocity

a = acceleration

x = distance

Solving for a:

$\displaystyle a=\frac{v_f^2-v_o^2}{2x}$

The car travels a distance of x=200 m and the velocities are:

vo = 216 Km/h

vf = 360 Km/h

Both velocities must be converted to meters by seconds.

vo = 216 Km/h *1000/3600 = 60 m/s

vf = 360 Km/h *1000/3600 = 100 m/s

The acceleration is:

$\displaystyle a=\frac{100^2-60^2}{2*200}$

$\displaystyle a=\frac{10000-3600}{400}$

$\displaystyle a=\frac{6400}{400}$

$\mathbf{a = 16\ m/s^2}$

7 0

3 years ago

Date the law of conversation of energy

Margaret [11]

Answer:

In its most compact form, it states: matter is neither created nor destroyed. In 1842, Julius Robert Mayer discovered the Law of Conservation of Energy.

Explanation:

3 0

3 years ago

How to do this question plz ​

How to do this question plz