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Stels [109]
3 years ago
6

Using -10 m/s2 for acceleration due to gravity, what would be the total displacement of the object if it took 8 seconds before h

itting the water?
Physics
1 answer:
ASHA 777 [7]3 years ago
6 0
If it starts from 0m/s...
s=?
u=0
a=-10
t=8
s=ut +1/2at^2
so s=(0×8)+ (0.5×-10×64)
s=0+(32×-10)
s=32×-10
s=-320metres
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Bc the equator is constantly facing the sun while the earth rotates
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215 = 25x – 35, solve for x.
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Add 35 to 215. then divide by 25. you should get x=10
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How does the radius of a string affect centripetal force.
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because a raduis is half of 25% of a cicrle.

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5 0
2 years ago
What is the final position of the object if its initial position is x = 0.40 m and the work done on it is equal to 0.21 J? What
r-ruslan [8.4K]

Answer:

a) Final position is x = 0.90 m

b) Final position is x = 0.133 m

Explanation:

The workdone between two points is usually approximated as the area under the force-distance curve between those two points.

From the graph,

As at the initial position, x = 0.40 m and the corresponding F = 0.8 N,

The area from that point onwards up to the end of that particular bar = 0.8 (0.5 - 0.4) = 0.08 J

The next bar has force = 0.4 N and the width of the bar = (0.75 - 0.50) = 0.25 m

Work done under this bar = 0.4 × 0.25 = 0.1 J

Total work done from the starting position up to this point now = 0.08 + 0.1 = 0.18 J, still less than 0.21 J

So, the final position has to be on the last bar. Let the position be x. The force on the last bar = 0.2 N

0.21 = 0.18 + 0.2 (x - 0.75)

0.03 = 0.2x - 0.15

0.2x = 0.18

x = 0.9 m

Therefore, the final position of the object, to do 0.21 J worth of work, starting from x = 0.4 m is 0.90 m.

b) For this part, negative work is done, this means, we will move in the negative direction to try and trace this total work done.

From the starting point where the initial position is 0.40 m, the force here is 0.80 N

The workdone under this bar to the left is

The workdone = 0.8 (0.25 - 0.4) = - 0.12 J

Since we're tracing -0.19 J, the final position has to be on the last bar (on the left), Let the position be x. The force on the last bar on the left (could also be referred to as the first bar) = 0.60 N

- 0.19 = -0.12 + 0.6 (x - 0.25)

-0.07 = 0.6x - 0.15

0.6x = 0.08

x = (0.08/0.6) = 0.133 m

Therefore, the final position of the object, after doing -0.19 J worth of work, starting from x = 0.4 m is 0.133 m.

Hope this Helps!!!

4 0
3 years ago
A race car's velocity increases from +28 m/s to +36 m/s over a 2.0-s time interval. What is the car's
BigorU [14]

Answer:

4 m/s^{2}

Explanation:

Acceleration, a=\frac {v-u}{t}

Where v and u are the final and initial velocities of the race car respectively, t is the time taken for the race car to attain velocity of 36 m/s.

Substituting 36 m/s for v, 28 m/s for u and 2 s for t then

a=\frac {36 m/s-28 m/s}{2}=4 m/s^{2}

3 0
3 years ago
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