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katrin2010 [14]
3 years ago
12

I have a table that I recorded of my data and my partner's data

Physics
1 answer:
Iteru [2.4K]3 years ago
4 0
<span>My table                                  My partner's table

position (m)     Time (s)           position (m)     time (s)

0m                    0                      0m                   0
2m                    2.30                 2m                   2.74
4m                    4.28                 4m                   2.39
6m                    5.00                 6m                   2.20
8m                    5.89                 8m                   1.80
10m                  6.59                 10m                  3.21

a. Explain how you can use your graph to figure out who had the fast average speed.

Compare the slopes of the lines from the origin to the last point.

You will verify that this slope for the second graph is steeper, so the average speed of the second partner is faster.

b. Explain how you can use your data table to figure out who had the faster average speed

Compare the final times for the two people. Given that the final distances are equal, that who has the lesser time is the person with faster average speed.

c. Look at each person's line on the graph. How can you use the graph to tell whether you moved at a constant speed? Did you move at a constant speed? Did your partner?

Constant speed implies constant slope, which is a straight line. So, none ot the persons moved at constant speed.

d. Calculate each person's average speed in meters per second and in centimeters per second.

1) You:

Average speed = displacement / time = 10m / 6.59 s = 1.52 m/s = 152 cm/s

2) Partner:

Average speed = displacement / time = 10m / 3.21s = 3.12 m/s = 312 cm/s
</span>
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A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of
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Answer:

(a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

Explanation:

Given that,

Speed = 33 m/s

Angle = 60°

Time = 3.0 sec

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Using equation of motion

h=ut-\dfrac{1}{2}gt^2

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h=33\times\sin60\times3.0-\dfrac{1}{2}\times9.8\times(3.0)^2

h=41.6\ m

(b). We need to calculate the maximum height of the ball

Using formula of height

h_{max}=\dfrac{(u\sin\theta)^2}{2g}

Put the value into the formula

h=\dfrac{(33\sin60)^2}{2\times 9.8}

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(c). We need to calculate the vertical component of velocity of ball

Using equation of motion

v=u-gt

v=u\sin\theta-gt

Put the value into the formula

v_{y}=33\times\sin 60-9.8\times3.0

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We need to calculate the horizontal component of velocity of ball

Using formula of velocity

v_{x}=u\cos\theta

Put the value into the formula

v_{x}=33\times\cos60

v_{x}=16.5\ m/s

We need to calculate the ball's impact speed

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v=\sqrt{v_{x}^2+v_{y}^2}

Put the value into the formula

v=\sqrt{(16.5)^2+(-0.82)^2}

v=16.52\ m/s

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(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

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Answer:

A)6.15 cm to the left of the lens

Explanation:

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q is the distance of the image from the lens

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In this problem, we have

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