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Makovka662 [10]
3 years ago
8

A force pair is produced when a tennis racket strikes a tennis ball. Which of the following best explains why the tennis ball do

es not have zero net force acting on it?
-The force exerted on the ball is greater than that exerted on the racket.
-Each half of the force pair acts on a different object.
-The two forces act in the same direction.
-The forces act perpendicular to each other.
Physics
1 answer:
Serga [27]3 years ago
4 0

Answer:

Each half of the force pair acts on a different object.

Explanation:

When a tennis racket strikes a tennis ball a pair force is produced. when the racket strikes the ball the racket exerts an action force on the tennis ball, according to Newton's third law for every action there is an equal and opposite reaction force, as a reaction the ball exert an equal and opposite force on the racket. These forces are often called pair forces.

As the forces acts on different bodies (Action force act on ball and reaction force act on racket) so the net force tennis ball is never zero.

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3 years ago
Two charged particles separated by a distance of = 3 and experienced electrostatic forces of = 60 . What would be this force if
klemol [59]

Answer: 539.4 N

Explanation:

Let's begin by explaining that Coulomb's Law establishes the following:  

"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them"

What is written above is expressed mathematically as follows:

F_{E}= K\frac{q_{1}.q_{2}}{d^{2}} (1)

Where:

F_{E}=60 N  is the electrostatic force

K=8.99(10)^{9} Nm^{2}/C^{2} is the Coulomb's constant  

q_{1} and q_{2} are the electric charges

d=3 m is the separation distance between the charges  

Then:

60 N= 8.99(10)^{9} Nm^{2}/C^{2}\frac{q_{1}.q_{2}}{(3 m)^{2}} (2)

Isolating q_{1} and q_{2}:

q_{1}q_{2}=6(10)^{-8} C^{2} (3)

Now, if we keep the same charges but we decrease the distance to d_{1}=1 m, (1) is rewritten as:

F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{6(10)^{-8} C^{2}}{(1 m)^{2}} (4)

Then, the new electrostatic force will be:

F_{E}= 539.4 N (5) As we can see, the electrostatic force is increased when we decrease the distance between the charges.

4 0
3 years ago
A flat sheet of ice has a thickness of 1.4 cm. It is on top of a flat sheet of crown glass that has a thickness of 3.0 cm. Light
MAXImum [283]

Answer:

t = 2.13 10-10 s , d = 6.39 cm

Explanation:

For this exercise we use the definition of refractive index

        n = c / v

Where n is the refraction index, c the speed of light and v the speed in the material medium.

The refractive indices of ice and crown glass are 1.13 and 1.52, respectively, therefore the speed of the beam in the material medium is

        v = c / n

As the beam strikes perpendicularly, the beam path is equal to the distance of the leaves, there is no refraction, so we can use the uniform motion relationships

        v = d / t

        t = d / v

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Let's look for the times on each sheet

Ice

        t₁ = 1.4 10⁻² 1.31 / 3 10⁸

        t₁ = 0.6113 10⁻¹⁰ s

Crown glass (BK7)

        t₂ = 3.0 10⁻² 1.52 / 3.0 10⁸

        t₂ = 1.52 10⁻¹⁰ s

Time is a scalar therefore it is additive

         t = t₁ + t₂

         t = (0.6113 + 1.52) 10⁻¹⁰

         t = 2.13 10-10 s

The distance traveled by this time in a vacuum would be

        d = c t

       d = 3 10⁸ 2.13 10⁻¹⁰

       d = 6.39 10⁻² m

       d = 6.39 cm

3 0
3 years ago
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