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Rashid [163]
4 years ago
11

Despite a very strong wind, a tennis player manages to hit a tennis ball with her racquet so that the ball passes over the net a

nd lands in her opponents court. Consider the following forces: 1 - A downward force of gravity. 2 - A force by the "hit". 3 - A force exerted by the air. Which of the above forces is (are) acting on the tennis ball after it has left contact with the racquet and before it touches the ground?
Physics
1 answer:
kobusy [5.1K]4 years ago
5 0

Answer:

The downward force of gravity and the force exerted by the air.

Explanation:

Gravity, in a Newtonian Framework, acts at distance, so, the ball doesn't has to be connected to Earth to feel its gravitational pull (as we know by experience).

The racket only can exert a force in the ball when is touching it, so, once the ball has left the contact with the racket, there is no force by the "hit".

The air is always touching the ball on the atmosphere (even when physicist pretend that is not). So, there is drag from the wind, and a buoyant force  exerted by the air over the ball at every time.

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A nautical mile is 6076 feet, and 1 knot is a unit of speed equal to 1 nautical mile/hour. How fast is a boat going 8 knots goin
mezya [45]

Answer:

The speed of the boat is equal to 13.50 ft/s.

Explanation:

given,

1  nautical mile = 6076 ft

1 knot = 1 nautical mile /hour

1 knot = 6076 ft/hr

speed of boat = 8 knots

 8 knots = 8 nautical mile /hour

               =8 \times \dfrac{6076\ ft}{1\nautical\ mile}\times \dfrac{1\ hour}{60\times 60\ s}

               = 13.50 ft/s

The speed of the boat is equal to 13.50 ft/s.

5 0
3 years ago
A child bounces a 51 g superball on the sidewalk. The velocity change of the super bowl is from 22 m/s downward to 14 m/s upward
noname [10]

Answer:

F=1.02x10^{-3} N

Explanation:

From the exercise we know:

m=51g*\frac{1kg}{1000g}=0.051kg

v_{1}=-22m/s

v_{2}=14m/s

t_{2}-t_{1}=1800s

So, the average acceleration is:

a=\frac{v_{2}-v_{1}}{t_{2}-t_{1}}=\frac{(14-(-22))m/s}{1800s}=0.02m/s^2

The average force is:

F=m*a=(0.051kg)(0.02m/s^2)=1.02x10^{-3} N

8 0
3 years ago
An apple in a tree has a gravitational store of 8J. As it falls, it accelerates constantly until it hits the ground. What is the
Masja [62]

Answer:

Explanation:

Given that on the tree the gravitational energy stored is 8J

Then, mgh = 8J.

The apple begins to fall and hit the ground, what is the maximum kinetic energy?

Using conservation of energy, as the above is about to hit the ground, the apple is at is maximum speed, and the height then is 0m, so the potential energy at the ground is zero, so all the potential of the apple at the too of the tree is converted to kinetic energy as it is about to hits the ground. Along the way to the ground, both the Kinetic energy and potential energy is conserved, it is notice that at the top of the tree, the apple has only potential energy since velocity is zero at top, and at the bottom of the tree the apple has only kinetic energy since potential energy is zero(height=0)

So,

K.E(max) = 8J

7 0
3 years ago
What are the similarities between strong nuclear force and weak nuclear force
Artemon [7]
Alike because they both act on the quarks making up the nucleons and they have very short ranges. The Strong Nuclear Force is an attractive force between protons and neutrons that keep the nucleus together and the Weak Nuclear Force is responsible for the radioactive decay of certain nuclei. Which also makes them very different
7 0
3 years ago
two masses are kept 2 metre apart there is gravitational force of 2 Newton what is the gravitational force when they are kept at
sukhopar [10]

Answer: 0.5N

Explanation:

Gravitational force is calculated using the formula :

F = Gm1m2/r^2

Where G is the gravitational constant (6.67 × 10^-11)

At a distance 'r' of 2metres apart:

Mass of objects are m1 and m2

Gravitational force 'F1' = 2N

Inputting values into the formula :

2 = Gm1m2 / 2^2 - - - - - (1)

At a distance 'r' of 4meters apart:

Mass of objects are m1 and m2

Gravitational force 'F2' = y

Inputting values

F2 = Gm1m2 / 4^2 - - - - - (2)

Dividing equations 1 and 2

2 = Gm1m2 / 2^2 ÷ F2 = Gm1m2 / 4^2

2 / F2 = (Gm1m2 / 4) / (Gm1m2 / 16)

2 / F2 = (Gm1m2 / 4) × (16 / Gm1m2)

2/F2 = 16 / 4

Cross multiply

2 × 4 = 16 × F2

8 = 16F2

F2 = 8/16

F2 = 0.5N

7 0
3 years ago
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