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bija089 [108]
3 years ago
5

A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The bal

l moves so fast that the string is always taut and perpendicular to the velocity of the ball. As the ball swings from its lowest point to its highest point Group of answer choices the work done on it by gravity is -118 J and the work done on it by the tension in the string is zero. the work done on it by gravity is -118 J and the work done on it by the tension in the string is 118 J. the work done on it by gravity and the work done on it by the tension in the string are both equal to -118 J. the work done on it by gravity is 118 J and the work done on it by the tension in the string is -118 J. the work done on it by gravity and the work done on it by the tension in the string are both equal to zero.
Physics
1 answer:
Hoochie [10]3 years ago
5 0

Answer:

The ball moves from lowest to highest point:

W = M g h = 3 * 9.8 * 4 = 118 J

This is work done "against" gravity so work done by gravity is -118 J

The tension of the string does no work because the tension does not

move thru any distance   W = T * x = 0 because the length of the string is fixed.

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Answer:

C

Explanation:

4 0
3 years ago
A 0.500-kg object connected to a light spring with a spring constant of 20.0 N/m oscillates on a frictionless horizontal surface
givi [52]

Answer:

a = 0.009 J

b = 0.19 m/s

c = 0.005 J and 0.004 J

Explanation:

Given that

Mass of the object, m = 0.5 kg

Spring constant of the spring, k = 20 N/m

Amplitude of the motion, A = 3 cm = 0.03 m

Displacement of the system, x = 2 cm = 0.02 m

a

Total energy of the system, E =

E = 1/2 * k * A²

E = 1/2 * 20 * 0.03²

E = 10 * 0.0009

E = 0.009 J

b

E = 1/2 * k * A² = 1/2 * m * v(max)²

1/2 * m * v(max)² = 0.009

1/2 * 0.5 * v(max)² = 0.009

v(max)² = 0.009 * 2/0.5

v(max)² = 0.018 / 0.5

v(max)² = 0.036

v(max) = √0.036

v(max) = 0.19 m/s

c

V = ±√[(k/m) * (A² - x²)]

V = ±√[(20/0.5) * (0.03² - 0.02²)]

V = ±√(40 * 0.0005)

V = ±√0.02

V = ±0.141 m/s

Kinetic Energy, K = 1/2 * m * v²

K = 1/2 * 0.5 * 0.141²

K = 1/4 * 0.02

K = 0.005 J

Potential Energy, P = 1/2 * k * x²

P = 1/2 * 20 * 0.02²

P = 10 * 0.0004

P = 0.004 J

4 0
3 years ago
a body dropped from a height reaches a velocity of 13m/s just before touching the ground. What is the initial height of the ball
SashulF [63]

Hi there!

We can use the following (derived) equation to solve for the final velocity given height:

vf = √2gh

We can rearrange to solve for height:

vf² = 2gh

vf²/2g = h

Plug in the given values (g = 9.81 m/s²)

(13)²/2(9.81) = 8.614 m

We can calculate time using the equation:

vf = vi + at, where:

vi = initial velocity (since dropped from rest, = 0 m/s)

a = acceleration (in this instance, due to gravity)

Plug in values:

13 = at

13/a = t

13/9.81 = 1.325 sec

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2 years ago
Help Plsss
sattari [20]
On the Newtonian theory of gravity, gravitation affects anything with mass. Assuming that none of the answer choices is the only thing that exists in the universe, all of the answer choices are subject to the law of universal gravitation (hence “universal”).

Satellites, water, frogs, and stars all have mass as they are all composed of matter. Thus, all four answer choices should be circled.
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3 years ago
Does a sled have inertia while sitting still
madam [21]

Answer:

Yes, a sled has inertia while sitting still.

Explanation:

From Newton's law of inertia, an object at rest will remain at rest unless it is acted upon by an external force. The reason the object will remain at rest unless an external force acts is because of inertia. Inertia means the resistance of an object to motion.

Thus, a sled hammer at rest will remain at rest unless it is acted upon by an external force. So we can conclude that it has Inertia.

7 0
3 years ago
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