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TEA [102]
3 years ago
7

After striking both mirrors, at what angle relative to the incoming ray does the outgoing ray emerge?

Physics
1 answer:
PIT_PIT [208]3 years ago
4 0
The appropriate response is Zero degrees. The beam will leave the two mirrors along a way parallel to the one it came in on. This is the guideline of the corner reflector, which is frequently utilized as a radar target. Take note of that the corner reflector utilizes three reflecting surfaces (that are set up at 90o from each other) rather than the two like are being utilized here. Wikipedia has a truly awesome drawing that shows this two-dimentional issue pleasantly. A moment connection is given to the article on the corner reflector and the 3-D angles.
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1. A student lifts a box of books that weighs 185 N. The box is
aksik [14]

1)  148 J

When lifting an object, the work done on the object is equal to its change in gravitational potential energy. Mathematically:

W = \Delta U = (mg) \Delta h

where

mg is the weight of the object

\Delta h is the change in height

For the box in this problem,

mg = 185 N

\Delta h = 0.800 m

Substituting into the equation, we find:

W=(185)(0.800)=148 J

2) (a) 28875 J

The work done by a force applied parallel to the direction of motion of the object is given by

W=Fd

where

F is the magnitude of the force

d is the displacement

In this problem,

F = 825 N is the force applied by the two students together

d = 35 m is the displacement of the car

Substituting,

W=(825)(35)=28875 J

2) (b) 57750 J

As seen previously, the equation that gives the work done by the force is

W=Fd

We see that the work done is proportional to the magnitude of the force: therefore, if the force is doubled, then the work done is also doubled.

The work done previously was

W = 28875 J

Now the force is doubled, so the new work done will be

W' = 2(28875)=57750 J

3) 4.4 J

In this case, the force acting on the ball is the force of gravity, whose magnitude is:

F = mg

where

m = 0.180 kg is the mass of the ball

g=9.8 m/s^2 is the acceleration of gravity

Solving the equation,

F=(0.180)(9.8)=1.76 N

Now we find the work done by gravity using the same formula applied before:

W=Fd

where d = 2.5 m is the displacement of the ball. We can apply this version of the formula since the force is parallel to the displacement. Substituting,

W=(1.76)(2.5)=4.4 J

4) 595.2 kg

In this case, we have the work done on the box:

W = 7.0 kJ = 7000 J

And we also know the change in height of the box:

\Delta h = 1.2 m

As we stated in part a), the work done on the box is equal to its change in gravitational potential energy:

W=mg \Delta h

Solving for m, we find

m=\frac{W}{g \Delta h}

And substituting the numerical values, we find the mass of the box:

m=\frac{7000}{(9.8)(1.2)}=595.2 kg

5) They do the same work

In fact, the net work done by each person on the box is equal to the change in gravitational potential energy of the box:

W=mg \Delta h

Where \Delta h is the difference in height between the final position and the initial position of the box.

This means that the work done on the box depends only on its initial and final position, not on the path taken. The two men carry the box along different paths, however the reach at the end the same position, and they started from the same position: this means that the value of \Delta h is the same for both of them, so the work they have done is exactly the same.

5 0
3 years ago
Father, I come to you worn and weary from the hard times I have walked through recently. I come to you seeking your shelter wher
Romashka [77]
This is not a HES tip
6 0
3 years ago
Read 2 more answers
Consider the sections of two circuits illustrated above. Select True or False for all statements. After connecting c and d to a
Basile [38]

Answer:

Follows are the solution to the given question:

Explanation:

In this question the missing file of the circuit is not be which is defined in attached file please find it.

In Option 1, this statement is true because the current is on R_3 \  and \ R_4, that is the same.

In option 2, this statement is false because Rcd=R_3+R_4 therefore it implies that Rcd is always larger then R_3.

In option 3,  this statement is true because the voltage of R_1 \ and \ R_2 is always equal.

In option 4, this statement is true because Rab= \frac{1}{(\frac{1}{R1}+\frac{1}{R2})}=\frac{R1R2}{(R1+R2)}. \frac{R2}{(R1+R2)}is always smaller then 1 therefore,R_1 \times (\frac{R2}{(R1+R2)}) is always equal to R1.

5 0
2 years ago
At 6: 00 am, a motorbike set off from town A to town B at a speed of 40km/h. At the same time, a car set off from town B to town
Keith_Richards [23]

Answer:

One would need to know how far apart the towns are:

T = SA / 40      time it takes for first cyclist to travel S1

T = SB / 60       time it takes for cyclist B to travel distance S2

SA + SB = S     the distance between the towns

SB = 60 / 40 SA = 1.5 SA

SA + 1.5 SA = S

S = 2.5 SA where cyclist travels distance SA

The time will depend on the separation of the towns.

4 0
1 year ago
g A 475-gram ball is traveling horizontally at 12.0 m/s to the left when it is suddenly struck horizontally by a bat, causing it
AlexFokin [52]

Answer:

7.64 ms

Explanation:

3 0
3 years ago
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