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garik1379 [7]
3 years ago
12

An upward force of 32.6 N is applied via a string to lift a ball with a mass of 2.8 kg. (a) What is the gravitational force acti

ng on the ball? (b) What is the net force acting upon the ball? (c) What is the acceleration of the ball?
Physics
1 answer:
Igoryamba3 years ago
3 0

Answer:

a) Fg = -27.4 N

b) Fnet = 5.2 N

c) a = 1.9 m/s2

Explanation:

a)

  • There are two forces acting on the ball, one directed upward (assuming this direction as positive, along the y-axis) which is the tension on the string (lifting force), and another aimed downward, which is the attractive force due to gravity.
  • Applying the Newton's Universal Law of Gravitation to a mass close to the surface of the Earth (in this case the ball), we can take the acceleration due to gravity like a constant, that we call by convention g, equal to -9.8 m/s2.
  • So, we can write the following expression for Fg:

       F_{g} = m*g = 2.8 kg*(-9.8m/s2) = -27.4 N (1)

b)

  • The net force on the ball, will be just the difference between the lifting force (32.6 N) and the force due to gravity, Fg:

       F_{net} = T -F_{g} = 32.6 N - 27.4 N = 5.2 N  (2)

c)

  • According Newton's 2nd Law, the acceleration caused by a net force on a point mass (we can take the ball as one) is given by the following expression:

       a = \frac{F_{net} }{m} = \frac{5.2N}{2.8kg} = 1.9 m/s2  (3)

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Explanation:

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3 0
3 years ago
What is the net displacement of the particle between 0 seconds and 80 seconds?
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After a thorough research, there exists the same question that has choices and the link of the graph (http://i37.servimg.com/u/f37/16/73/53/52/graph410.png)

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3 years ago
If a student flicks a stationary 0.1 kg ball with 5N of force for 0.1 seconds. What is its final
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5m/s

Explanation:

Given parameters:

Mass of ball = 0.1kg

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Unknown:

final speed of the ball = ?

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According to newton's second law "the net force on a body is the product of its mass and acceleration".

  Force = mass x acceleration      equation 1

Acceleration =

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3 0
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timofeeve [1]

Answer:

Destructive interference

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Destructive interference occurs when waves come together in such a way that they completely cancel each other out. When two waves interfere destructively, they must have the same amplitude in opposite directions.

6 0
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distance = 95 / 3 m

displacement = 95/3 m or 32 m (2 s.f.)
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