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LiRa [457]
3 years ago
7

Which best describes the speed of the ball as it is thrown straight up into the air and comes back down

Physics
1 answer:
Andrews [41]3 years ago
4 0
Assuming that up is the positive direction, the velocity of the ball goes from it's initial and approaches zero, which it reaches at the apex of it's height. After peaking, the velocity is now reaching the same numbers it had on the way up, only with a negative value this time. If it had a velocity of 5m/s upon leaving the hand, it will have a velocity of -5m/s when it came back to the hand.
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An object is placed at O ona number line. It moves 3 units to the right, then 4 units to the left, and then 6 units to
kvv77 [185]

Answer:

You have a displacement of 5 units to the right.

Explanation:

First you go three to the right which lands on the 3 mark. Then you move it 4 to the left which substracts 4, landing the object at -1. Finally you move 6 to the right, and you finish at marker 5. Since displacement is not total distance but just final distance from the start point directly to end point, it is only a displacement of 5.

6 0
3 years ago
Read 2 more answers
1. You discover that your iPod batteries keep dying after a few days. You think that the new 48 hour iBattery will work better.
stiks02 [169]

Answer:

Explanation:

It probably would

7 0
3 years ago
Decide which of the following day-to-day activities might be considered aerobic exercise?
erastova [34]

Answer:

3. Mowing the lawn

Explanation:

Aerobic exercise is any type of cardiovascular conditioning.

Mowing the lawn is the only thing that is actively moving.

3 0
3 years ago
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t is the estimated distance traveled through the air by a marshmallow that is launched from a 15- cm long launcher when it is pl
DerKrebs [107]

Answer:

10

Explanation:

15 cm minus 5 cm

5 0
3 years ago
A) 1.2-kg ball is hanging from the end of a rope. The rope hangs at an angle 20° from the vertical when a 19 m/s horizontal wind
Marat540 [252]

Answer:

Part a)

F_v = 4.28 N

Part B)

L = 1.02 m

Part C)

v = 1.25 m/s

Explanation:

Part A)

As we know that ball is hanging from the top and its angle with the vertical is 20 degree

so we will have

Tcos\theta = mg

T sin\theta = F_v

\frac{F_v}{mg} = tan\theta

F_v = mg tan\theta

F_v = 1.2\times 9.81 (tan20)

F_v = 4.28 N

Part B)

Here we can use energy theorem to find the distance that it will move

-\mu mg cos\theta L + mg sin\theta L = -\frac{1}{2}mv^2

(-(0.37)m(9.81) cos15 + m(9.81) sin15)L = - \frac{1}{2}m(1.4)^2

(-3.5 + 2.54)L = - 0.98

L = 1.02 m

Part C)

At terminal speed condition we know that

F_v = mg

bv^2 = mg

2.5 v^2 = 3.9

v = 1.25 m/s

7 0
3 years ago
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