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xz_007 [3.2K]
4 years ago
15

Which describes a positive or negative net force acting on an object?

Physics
1 answer:
Afina-wow [57]4 years ago
6 0
Positive net force describes acceleration of particle is increasing and negative net force describes that particle is deaccelerating.
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If you have a final velocity of 50 m/s and travelled for 120 seconds. What
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Answer:

a=v-u/t

Explanation:

use this formula and initial velocity is 0

5 0
2 years ago
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A child sits on a merry-go-round, 1.5 meters from the center. The merry-go-round is turning at a constant rate, and the child is
exis [7]

Answer:

5 .07 s .

Explanation:

The child will move on a circle of radius r

r = 1.5 m

Let the velocity of rotation = v

radial acceleration = v² / r

v² / r = 2.3

v² = 2.3 r = 2.3 x 1.5

= 3.45

v = 1.857 m /s

Time of revolution = 2π r / v

= 2 x 3.14 x 1.5 / 1.857

= 5 .07 s .

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3 years ago
Assuming the earth is a uniform sphere of mass M and radius R, show that the acceleration of free fall at the earth's surface is
zimovet [89]

use Newton's gravitational law and 2nd law .....

8 0
2 years ago
023 (part 1 of 2) 10.0 points
Annette [7]

Answer:

Part 1

The angular speed is approximately 1.31947 rad/s

Part 2

The change in kinetic energy due to the movement is approximately 675.65 J

Explanation:

The given parameters are;

The rotation rate of the merry-go-round, n = 0.21 rev/s

The mass of the man on the merry-go-round = 99 kg

The distance of the point the man stands from the axis of rotation = 2.8 m

Part 1

The angular speed, ω = 2·π·n = 2·π × 0.21 rev/s ≈ 1.31947 rad/s

The angular speed is constant through out the axis of rotation

Therefore, when the man walks to a point 0 m from the center, the angular speed ≈ 1.31947 rad/s

Part 2

Given that the kinetic energy of the merry-go-round is constant, the change in kinetic energy, for a change from a radius of of the man from 2.8 m to 0 m, is given as follows;

\Delta KE_{rotational} = \dfrac{1}{2}  \cdot I \cdot \omega ^2 = \dfrac{1}{2}  \cdot m \cdot v ^2

I  = m·r²

Where;

m = The mass of the man alone = 99 kg

r = The distance of the point the man stands from the axis, r = 2.8 m

v = The tangential velocity = ω/r

ω ≈ 1.31947 rad/s

Therefore, we have;

I = 99 × 2.8² = 776.16 kg·m²

\Delta KE_{rotational} = 1/2 × 776.16 kg·m² × (1.31947)² ≈ 675.65 J

3 0
3 years ago
Why will a sheet of paper fall slower than one that is crumbled into a ball
Zarrin [17]
It’s coming in contact with more air molecules than I would if it was in a ball because there is less surface area
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3 years ago
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