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Ksivusya [100]
4 years ago
9

I need the answer to this question what has the student plotted on the vertical axis?

Physics
1 answer:
expeople1 [14]4 years ago
7 0

Answer:

The correct option is D

Explanation:

In trying to achieve what the student wanted to see, which is to see the relationship between the weight the cord can hold and how long the cord will stretch. Since the origin of the graph is from zero, the value plotted on the vertical axis would be just the length caused by each weights. Thus, <u>the original length would have to be subtracted from the measured length to determine the actual length caused by the weight added to the cord</u>.

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What is the value of a conversion factor ratio?<br>A.1<br>B.3<br>C.10<br>D.12
Alecsey [184]
A is the correct answer :D

4 0
4 years ago
A 3-ton Toyota land cruiser travelled at a speed of 45m/s, collides with a 2- ton comfort taxi which was parked in front of the
Olin [163]

Answer:

v = 27 m/s

Explanation:

To find the speed of cars after the collision  you take into account the momentum conservation law. Total momentum of both cars before the collision must be equal to the total momentum of both cars after the collision.

After the collision both cars traveled together, then you have:

m_1v_1+m_2v_2=(m_1+m_2)v   (1)

m1: mass of the Toyota = 3-ton = 3000 kg

m2: mass of the taxi = 2-ton = 2000kg

v1: speed of the Toyota before the collision = 45m/s

v2: speed of the car before the collision = 0 m/s (it is  at rest)

v: speed of both cars after the collision = ?

You solve the equation (1) for v:

v=\frac{m_1v_1+m_2v_2}{m_1+m_2}

Next, you replace the values of the rest of the variables:

v=\frac{(3000kg)(45m/s)+0kgm/s}{3000kg+2000kg}=27\frac{m}{s}

hence, just after the collision both cars have a speed of 27m/s

7 0
3 years ago
Starting from rest, a racecar moves 112 m in the first 7 s of uniform acceleration. What is the car's acceleration?​
Leviafan [203]

Answer:

215mph

Explanation:

multiply 112milws by 2 and multiple that by seven, then divide it by two

6 0
2 years ago
Q15.17 Both wave intensity and gravitation obey inverse-square laws. Do they do so for the same reason?Discuss the reason for ea
s2008m [1.1K]

(a) Intensity obeys inverse square law from basis of light passing through a given surface.

(b) Gravitation obeys inverse square law from the basis of force between two masses.

(c) The maximum magnitude of the acceleration of the block is 126.75 m/s².

<h3>What is intensity?</h3>

Intensity is the ratio is ratio of power to area of a given surface.

I = P/A (W/m²)

where;

  • P is power
  • A is area
<h3>Universal gravitation law</h3>

F = \frac{Gm_1m_2}{r^2}

Intensity and gravitation do not obey inverse square law for same reason;

  • Intensity obeys inverse square law from basis of light passing through a given surface.
  • Gravitation obeys inverse square law from the basis of force between two masses.
<h3>Acceleration of the block</h3>

a = v²/A

a = (3.9²)/0.12

a = 126.75 m/s²

Learn more about acceleration here: brainly.com/question/605631

#SPJ1

7 0
2 years ago
PLS HELP ME<br> How does one stage in a stars life lead to another
Oxana [17]
Just as during formation, when the material contracts, the temperature and pressure increase. This newly generated heat temporarily counteracts the force of gravity, and the outer layers of the star are now pushed outward ( in not sure tho ) I hope this helps
3 0
3 years ago
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