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GalinKa [24]
2 years ago
15

1. What are the units of area under the line/curve? 2. Does the area have any meaning?

Physics
1 answer:
TiliK225 [7]2 years ago
6 0

The area under the velocity time graph is 125 m and the meaning of the area is displacement.

<h3>What is area under velocity - time graph?</h3>

The area under a velocity time graph represents the displacement of the object.

total area of the graph = A1 + A2

total area of the graph = ¹/₂ (base₁)(height₁) + ¹/₂ (base₂)(height₂)

total area of the graph = ¹/₂(4)(40) + ¹/₂(3)(30)

total area of the graph = 125 m

Thus, the area under the velocity time graph is 125 m and the meaning of the area is displacement.

Learn more about velocity time graph here: brainly.com/question/4710544

#SPJ1

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Svetradugi [14.3K]

Answer:

W = 8.01 × 10^(-17) [J]

Explanation:

To solve this problem we need to know the electron is a subatomic particle with a negative elementary electrical charge (-1,602 × 10-19 C), The expression to calculate the work is given by:

W = q*V

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q = charge = 1,602 × 10^(-19) [C]

V = voltage = 500 [V]

W = work [J]

W = 1,602 × 10^(-19) * 500

W = 8.01 × 10^(-17) [J]

8 0
3 years ago
_______ do not have definite size or always take the shape of their container?
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Liquids do not have a definite size and always take the shape of the container they're in.

6 0
3 years ago
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The A note has a frequency of 440 Hz. What is the wavelength if the speed of sound is 343 m/s?
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3 years ago
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If the velocity of a pitched ball has a magnitude of 44.5 m/sm/s and the batted ball's velocity is 55.5 m/sm/s in the opposite d
Yuliya22 [10]

Incomplete question as the mass of baseball is missing.I have assume 0.2kg mass of baseball.So complete question is:

A baseball has mass 0.2 kg.If the velocity of a pitched ball has a magnitude of 44.5 m/sm/s and the batted ball's velocity is 55.5 m/sm/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Answer:

ΔP=20 kg.m/s

Explanation:

Given data

Mass m=0.2 kg

Initial speed Vi=-44.5m/s

Final speed Vf=55.5 m/s

Required

Change in momentum ΔP

Solution

First we take the batted balls velocity as the final velocity and its direction is the positive direction and we take the pitched balls velocity as the initial velocity and so its direction will be negative direction.So we have:

v_{i}=-44.5m/s\\v_{f}=55.5m/s

Now we need to find the initial momentum

So

P_{1}=m*v_{i}

Substitute the given values

P_{1}=(0.2kg)(-44.5m/s)\\P_{1}=-8.9kg.m/s

Now for final momentum

P_{2}=mv_{f}\\P_{2}=(0.2kg)(55.5m/s)\\P_{2}=11.1kg.m/s

So the change in momentum is given as:

ΔP=P₂-P₁

=[(11.1kg.m/s)-(-8.9kg.m/s)]\\=20kg.m/s

ΔP=20 kg.m/s

3 0
3 years ago
1.) A stone falls from rest from the top of a cliff.
KengaRu [80]

Answer:

Explanation:

ignore air resistance

Let t be the time of fall for the dropped stone.

½(9.8)t² = 43.12(t - 2.2) + ½(9.8)(t - 2.2)²

4.9t² = 43.12t - 94.864 + 4.9(t² - 4.4t + 4.84)

4.9t² = 43.12t - 94.864 + 4.9t² - 21.56t + 23.716

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t = 71.148/21.56 = 3.3 s

h = ½(9.8)3.3² = 53.361 = 53 m

or

h = 43.12(3.3 - 2.2) + ½(9.8)(3.3 - 2.2)² = 53.361 = 53 m

4 0
3 years ago
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