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Tanzania [10]
2 years ago
7

A transverse wave on a string has an amplitude A. A tiny spot on the string is colored red. As one cycle of the wave passes by,

what is the total distance traveled by the red spot?a.) 4Ab.) 1/2 Ac.) 1/4 Ad.) Ae.) 2A
Physics
1 answer:
DiKsa [7]2 years ago
6 0

Since it's a transverse wave, a particle on the string moves left and right as the wave passes by, but the particle doesn't travel forward or backward at all.

So the little red dot moves 'A' to the left, then 'A' back to the center, then 'A' to the right, then 'A' back to the center again.

All together, the red dot moves a total distance of <em>4A . (choice 'a')</em>

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A car traveling at a velocity v can stop in a minimum distance d. What would be the car's minimum stopping distance if it were t
alexira [117]

Answer:

a. 4d.

If the car travels at a velocity of 2v, the minimum stopping distance will be 4d.

Explanation:

Hi there!

The equations of distance and velocity of the car are the following:

x = x0 + v0 · t + 1/2 · a · t²

v = v0 + a · t

Where:

x =  position of the car at time t.

x0 = initial position.

v0 = initial velocity.

t = time.

a = acceleration.

v = velocity of the car at time t.

Let´s find the time it takes the car to stop using the equation of velocity. When the car stops, its velocity is zero. Then:

velocity = v0 + a · t      v0 = v

0 = v + a · t

Solving for t:

-v/a = t

Since the acceleration is negative because the car is stopping:

v/a = t

Now replacing t = v/a in the equation of position:

x = x0 + v0 · t + 1/2 · a · t²     (let´s consider x0 = 0)

x = v · (v/a) + 1/2 · (-a) (v/a)²    

x = v²/a - 1/2 · v²/a

x = 1/2 v²/a

At a velocity of v, the stopping distance is 1/2 v²/a = d

Now, let´s do the same calculations with an initial velocity v0 = 2v:

Using the equation of velocity:

velocity = v0 + a · t

0 = 2v - a · t

-2v/-a = t

t = 2v/a

Replacing in the equation of position:

x1 = x0 + v0 · t + 1/2 · a · t²  

x1 = 2v · (2v/a) + 1/2 · (-a) · (2v/a)²

x1 = 4v²/a - 2v²/a

x1 = 2v²/a

x1 = 4(1/2 v²/a)

x1 = 4x

x1 = 4d

If the car travels at a velocity 2v, the minimum stopping distance will be 4d.

5 0
1 year ago
Carla sees an equation that models a nuclear change.
victus00 [196]

Answer:

C. nuclear fusion, because the equation shows two hydrogen nuclei combining to form a helium nucleus

Explanation:

Nuclear reaction can either be; fission or fusion. Nuclear fission is the process by which a massive nucleus breaks in to two smaller nuclei of almost the same size with the release of high amount of energy. Nuclear fusion is the process by which two nuclei reacts, joins, to produce a massive nucleus (compared to the masses of the reacting elements) with the release of high amount of energy.

From the given equation, two hydrogen isotopes; deuterium and tritium reacts with each other to produce helium nucleus and a neutron.

This reaction is a nuclear fusion which produces a massive nuclei.

5 0
2 years ago
Motion maps for two objects, Y and Z, are shown.
Vesna [10]

Answer:

The answer is B) 3 seconds

Explanation:

I just took the test on 2020 edge and got it right

5 0
2 years ago
The block in the diagram below is AT REST. However, the tension in the cable is not the only thing holding the block back. Stati
Vedmedyk [2.9K]

Answer:

The  tension in the rope is 229.37 N.

Explanation:

Given:

Mass of the block is, m=33.2\ kg

Coefficient of static friction is, \mu = 0.214

Angle of inclination is, \theta = 31.5°

Draw a free body diagram of the block.

From the free body diagram, consider the forces in the vertical direction perpendicular to inclined plane.

Forces acting are mg\cos \theta and normal N. Now, there is no motion in the direction perpendicular to the inclined plane. So,

N=mg\cos \theta\\N=(33.2)(9.8)\cos (31.5)\\N=277.415\ N

Consider the direction along the inclined plane.

The forces acting along the plane are mg\sin \theta and frictional force, f, down the plane and tension, T, up the plane.

Now, as the block is at rest, so net force along the plane is also zero.

T=mg\sin \theta+f\\T=mg\sin \theta +\mu N\\T= (33.2)(9.8)(\sin (31.5)+(0.214\times 277.415)\\T= 170+59.37\\T=229.37\ N

Therefore, the  tension in the rope is 229.37 N.

3 0
1 year ago
You are riding on a roller coaster that starts from rest at a height of 25.0 m and moves along a frictionless track. however, af
djyliett [7]
I attached the missing picture.
We can figure this one out using the law of conservation of energy.
At point A the car would have potential energy and kinetic energy.
A: mgh_1+\frac{mv_1^2}{2}
Then, while the car is traveling down the track it loses some of its initial energy due to friction:
W_f=F_f\cdot L
So, we know that the car is approaching the point B with the following amount of energy:
mgh_1+\frac{mv_1^2}{2}- F_fL
The law of conservation of energy tells us that this energy must the same as the energy at point B. 
The energy at point B is the sum of car's kinetic and potential energy:
B: mgh_2+\frac{mv_2}{2}
As said before this energy must be the same as the energy of a car approaching the loop:
mgh_2+\frac{mv_2}{2}=mgh_1+\frac{mv_1^2}{2}- F_fL
Now we solve the equation for v_1:
v_1^2=2g(h_2-h_1)+v_2^2+\frac{2F_fL}{m}\\&#10;v_1^2=39.23\\&#10;v_1=\sqrt{39.23}=6.26\frac{m}{s}

4 0
1 year ago
Read 2 more answers
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