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Nimfa-mama [501]
3 years ago
15

Mass of object is 50g moves in a circular path of radius 10cm find work done

Physics
1 answer:
topjm [15]3 years ago
3 0

Answer:

Work done = 0.3142 Nm

Explanation:

Mass of Object is 50 g

Circular path of radius is 10 cm ⇒ 0.1 m

Work done = Force × Distance = ?

*Distance moved (circular path) ⇒ Circumference of the circular path

2πr = 2 × 3.142 × 0.1 ⇒ 0.6284 m

*Force that is enough to move a 50 g must be equal or more than its weight.

therefore convert 50 grams to newton = 0.5 N

Recall that; work done is force times distance

∴ 0.5 N × 0.6284 m

Work done = 0.3142 Nm

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The function​ s(t) represents the position of an object at time t moving along a line. Suppose s (2 )equals 146and s (6 )equals
sineoko [7]

Answer:

v(t) = 27 units

Explanation:

The function s(t) represents the position of an object at time t moving along a line such that,

s(2)=146

and

s(6)=254

We need to find the average velocity of the object over the interval of time [2,6]. The velocity of the object is equal to the total distance divided by time. It is given by :

v(t)=\dfrac{s(6)-s(2)}{6-2}

v(t)=\dfrac{254-146}{6-2}

v(t) = 27 units

So, the  average velocity of the object is 27 units. Hence, this is the required solution.

5 0
3 years ago
An earthquake is a natural rapid shaking of the _______ caused by the release of stored energy in rocks
REY [17]

Answer:

An earthquake is a natural rapid shaking of the tectonic plates caused by the release of stored energy in rocks

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8 0
2 years ago
Firecrackers A and B are 600 m apart. You are standing exactly halfway between them. Your lab partner is 300 m on the other side
pishuonlain [190]

Answer:

See the explanation

Explanation:

Given:

Distance of Firecrackers A and B = 600 m

Event 1 = firecracker 1 explodes

Event 2 = firecracker 2 explodes

Distance of lab partner from cracker A = 300 m

You observe the explosions at the same time

to find:

does event 1 occur before, after, or at the same time as event 2?

Solution:

Since the lab partner is at 300 m distance from the firecracker A and Firecrackers A and B are 600 m apart

So the distance of fire cracker B from the lab partner is:

600 m  + 300 m = 900 m

It takes longer for the light from the more distant firecracker to reach so

Let T1 represents the time taken for light from firecracker A to reach lab partner

T1 = 300/c

It is 300 because lab partner is 300 m on other side of firecracker A

Let T2 represents the time taken for light from firecracker B to reach lab partner

T2 = 900/c

It is 900 because lab partner is 900 m on other side of firecracker B

T2 = T1

900 = 300

900 = 3(300)

T2 = 3(T1)

Hence lab partner observes the explosion of the firecracker A before the explosion of firecracker B.

Since event 1 = firecracker 1 explodes and event 2 = firecracker 2 explodes

So this concludes that lab partner sees event 1 occur first and lab partner is smart enough to correct for the travel time of light and conclude that the events occur at the same time.

8 0
3 years ago
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