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torisob [31]
3 years ago
6

Suzette had prepared the graph below to add to her lab

Physics
2 answers:
andre [41]3 years ago
6 0

Answer:

<h2>A title.</h2>

Explanation:

The thing missing here is not regarding the calculations, because Suzette even draw the line that explain the relation between the two variables, which also are included there. However, Suzette didn't include the title of the graph, so basically, we don't know to which problem belongs that graph. This is an important issue, because in lab report, we tend to have a lot of information, and if it's not identified, it's completely useful.

Charra [1.4K]3 years ago
4 0

Answer:

A title

Explanation:

Because this is middle school.

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WILL MARK BRAINLIEST PLEASE HELP!! Which class of lever is a seesaw? What are the parts of a seesaw's lever class? Describe each
kondor19780726 [428]

A seesaw is a class 1 lever

Parts:

Load- The force applied to the lever

Fulcrum- in the middle

Effort- what lifts the load


so in a seesaw with 2 people, whoever is using their feet to push the seesaw up is the effort while the other person is the load. This changes depending on who is pushing the seesaw. the joint/thing between the two people is the fulcrum.



3 0
4 years ago
A pulse of light takes 3.00 ns to travel through air from an emitter to a detector. When a piece of transparent material with a
cupoosta [38]

Answer:

(a) 1.75\times10^8\text{ m/s}

(b) 1.71

Explanation:

(a) The difference in the times of travel in the two case = 3.20 - 3.00 = 0.20\text{ ns} = 2.0\times10^{-9}\text{ s}

This difference is the time in the transparent material. With a thickness of 35.0 cm, the speed in the material is

v = \dfrac{35 cm}{2.0\times10^{-9}\text{ s}}=\dfrac{0.35 m}{2.0\times10^{-9}\text{ s}} = 1.75\times10^8\text{ m/s}

(b) The refractive index of the material is the ratio of the velocity of light in  vacuum to its velocity in the material. Using speed of light in vacuum as c = 3.00\times10^8\text{ m/s}, the refractive index, n, is

n=\dfrac{c}{v} = \dfrac{ 3.00\times10^8\text{ m/s}}{1.75\times10^8\text{ m/s}}= 1.71

4 0
3 years ago
Find the final velocity of a 40 kg skateboarder traveling at an initial velocity of 10 m/s that moves up a hill from a height of
slavikrds [6]

Answer:

vf = 0

Explanation:

Since the initial height hi = 0, we can rewrite the energy equation as

vf^2 = vi^2 - 2ghf = (10 m/s)^2 - 2(10 m/s^2)(5 m) = 0

Therefore, his final velocity vf is

vf = 0

3 0
3 years ago
initially, a particle is moving at 5.33 m/s at an angle of 37.9° above the horizontal. Two seconds later, its velocity is 6.11 m
Ainat [17]

Explanation:

Average acceleration is the change in velocity over the change in time:

a = (v − v₀) / t

In the x direction:

aₓ = (6.11 cos (-54.2°) − 5.33 cos (37.9°)) / 2.00

aₓ = -0.316 m/s²

In the y direction:

aᵧ = (6.11 sin (-54.2°) − 5.33 sin (37.9°)) / 2.00

aᵧ = -4.11 m/s²

7 0
3 years ago
Q:
Shalnov [3]

Answer:

Explanation:

We will use the KE equation you wrote here and fill in what we are given:

36=\frac{1}{2}m(12)^2 and isolating the m:

m=\frac{2(36)}{12^2} which gives us

m = .50 kg

3 0
3 years ago
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