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Makovka662 [10]
2 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]2 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.

You might be interested in
An object initially at rest experiences a constant horizontal acceleration due to the action of a resultant force applied for 10
Marianna [84]

Answer:

a = 18.28 ft/s²

Explanation:

given,

time of force application, t= 10 s

Work = 10 Btu

mass of the object = 15 lb

acceleration, a =  ? ft/s²

1 btu = 778.15 ft.lbf

10 btu = 7781.5 ft.lbf

m = \dfrac{15}{32.174}\ slug

m = 0.466 slug

now,

work done  is equal to change in kinetic energy

W = \dfrac{1}{2} m (v_f^2-v_i^2)

7781.5 = \dfrac{1}{2}\times 0.466\times v_f^2

 v_f = 182.75\ ft/s

now, acceleration of object

  a = \dfrac{v_f-v_o}{t}

  a = \dfrac{182.75-0}{10}

         a = 18.28 ft/s²

constant acceleration of the object is equal to 18.28 ft/s²

3 0
2 years ago
Derive an expression for the total mechanical energy of the system as the monkey reaches the top of the motion, Etop, in terms o
ipn [44]

Answer:

U =  0.5 * k *(x + d - h_max)^2 + m*g*h_max

Explanation:

Given:

- The extension in spring @ equilibrium = x m

- The spring constant = k

- The amount of distance pulled down = d

- mass of the toy = m

Find:

- The total mechanical energy E_top at the top position h_max in terms of the available variables.

Solution:

- First we need to determine the types of Energy that are in play:

- The Elastic potential Energy E_p in a spring is given:

                              E_p: 0.5 * k * (ext)

- In our case when the toy at the top most position h_max will have a net extension ext, by summing displacement of spring:

             ext = Equilibrium + distance pulled - h_max = (x + d - h_max)

Hence, the elastic potential energy will be:

                              E_p = 0.5 * k *(x + d - h_max)^2

- The gravitational potential energy E_g is given by:

                              E_g = m*g*h_max

Where, bottom most position is taken as reference (datum).

- The kinetic Energy E_k is given by:

                              E_k = 0.5*m*v_top^2

- Since we know that the maximum height is reached when velocity is zero

Hence,                   E_k = 0.5*m*0^2 = 0.

The total Energy of the system U is sum of all energies and play:

                               U = E_p + E_k + E_g

                               U =  0.5 * k *(x + d - h_max)^2 + m*g*h_max

8 0
2 years ago
Which trailer has more downward pressure where it attaches to the car?
VARVARA [1.3K]

The one that is loaded worst. The overall weight is not important; tongue weight is a matter of loading. Our 12,000 lb snow cat trailer, which has stops to position the cat properly, has under 100 lbs tongue weight. Excessive tongue weight is a Bad Thing because it reduces weight on the towing vehicle's front wheels, leading to instability.

4 0
2 years ago
A cart, which has a mass of 2.30 kg is sitting at the top of an inclined plane, which is 4.50 meters long and meets the horizont
expeople1 [14]

Answer:

a) The gravitational potential energy before the cart rolls down the incline is 24.6 J.

b) The magnitude of the force that causes the cart to roll down is 5.47 N.

c) The acceleration of the cart is 2.38 m/s²

d) It takes the cart 1.94 s to reach the bottom of the incline.

e) The velocity of the cart at the bottom of the inclined plane is 4.62 m/s.

f) The kinetic energy of the cart as it reaches the bottom of the incline is 24.6 J.

g) The work done by the gravitational force is 24.6 J.

Explanation:

Hi there!

a) The gravitational potential energy is calculated using the following equation:

EP = m · g · h

Where:

EP = gravitational potential energy.

m = mass of the object.

g = acceleration due to gravity.

h = height at which the object is located.

The height of the inclined plane can be calculated using trigonomoetry:

sin 14.0° = height / lenght

sin 14.0° = height / 4.50 m

4.50 m · sin 14.0° = height

height = 1.09 m

Then, the gravitational potential energy will be:

EP = m · g · h

EP = 2.30 kg · 9.81 m/s² · 1.09 m = 24.6 J

The gravitational potential energy before the cart rolls down the incline is 24.6 J.

b) Please, see the attached figure for a graphical description of the problem and the forces acting on the cart. The force that causes the cart to accelerate down the incline is the horizontal component of the weight (Fwx in the figure). The magnitude of this force can be obtained using trigonometry:

sin 14° = Fwx / Fw

The weight of the cart (Fw) is calculated as follows:

Fw = m · g

Fw = 2.30 kg · 9.81 m/s²

Fw = 22.6 N

Then, the x-component of the weight will be:

FW · sin 14° = Fwx

22.6 N · sin 14° = Fwx

Fwx = 5.47 N

The magnitude of the force that causes the cart to roll down is 5.47 N.

c)Using the equation of Fwx we can calculate the acceleration of the cart:

Fwx = m · a

Where "m" is the mass of the cart and "a" is the acceleration.

Fwx / m = a

5.47 N / 2.30 kg = a

a = 2.38 m/s²

The acceleration of the cart is 2.38 m/s²

d) To calculate the time it takes the cart to reach the bottom of the incline, let´s use the equation of position of the cart:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the cart at time t.

x0 = initial position.

v0 = initial velocity.

a = acceleration.

t = time.

Considering the initial position as the point at which the cart starts rolling (x0 = 0) and knowing that the cart starts from rest (v0 = 0), let´s find the time it takes the cart to travel the 4.50 m of the inclined plane:

x = 1/2 · a · t²

4.50 m = 1/2 · 2.38 m/s² · t²

2 · 4.50 m / 2.38 m/s² = t²

t = 1.94 s

It takes the cart 1.94 s to reach the bottom of the incline.

e) The velocity of the cart at the bottom of the inclined plane can be obtained using the following equation:

v = v0 + a · t

v = 0 m/s + 2.38 m/s² · 1.94 s

v = 4.62 m/s

The velocity of the cart at the bottom of the inclined plane is 4.62 m/s.

f) The kinetic energy can be calculated using the following equation:

KE = 1/2 · m · v²

Where:

KE =  kinetic energy.

m = mass of the cart.

v = velocity of the cart.

KE = 1/2 · 2.30 kg · (4.62 m/s)²

KE = 24.6 J

The kinetic energy of the cart as it reaches the bottom of the incline is 24.6 J.

The gain of kinetic energy is equal to the loss of gravitational potential energy.

g) The work done by the gravitational force can be calculated using the work-energy theorem: the work done by the gravitational force is equal to the negative change in the gravitational potential energy:

W = -ΔPE

W = -(final potential energy - initial potential energy)

W = -(0 - 24.6 J)

W = 24.6 J

This can also be calculated using the definition of work:

W = Fw · d

Where "d" is the distance traveled in the direction of the force, that is the height of the inclined plane:

W = 22.6 N · 1.09 m = 24.6 J.

The work done by the gravitational force is 24.6 J.

4 0
2 years ago
A person weighing 0.70 kn rides in an elevator that has an upward acceleration of 1.5 m/s2. what is the magnitude of the normal
creativ13 [48]
First of all, we can find the mass of the person, since we know his weight W:
W=mg=0.70 kN=700 N
And so
m= \frac{W}{g}= \frac{700 N}{9.81 m/s^2}=71.4 kg

We know for Newton's second law that the resultant of the forces acting on the person must be equal to the product between the mass and the acceleration a of the person itself:
\sum F =  ma
There are only two forces acting on the person: his weight W (downward) and the vincular reaction Rv of the floor against the body (upward). So we can rewrite the previous equation as
R_v -W = ma
We know the acceleration of the system, a=1.5 m/s^2 (upward, so with same sign of Rv), so we can solve to find the value of Rv, the normal force exerted by the elevator's floor on the person:
R_v = ma+W=(71.4 kg)(1.5 m/s^2)+700 N =807N
8 0
2 years ago
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