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2 years ago

9

Bill is learning to play tennis. He does pretty well hitting the ball back to his opponent but, many times he misses the ball wh

en he tries to hit it on his serve. How can he improve his success rate of striking the ball every time on his tennis serve?
Hold the ball and toss it with his dominant hand
Place his feet parallel to the baseline prior to tossing the ball
Toss the ball gently into the air prior to striking it with the racquet
Use a wooden racquet instead of an aluminum racquet

1 answer:

KiRa [710]2 years ago

5 0

Answer: Place his feet parallel to the baseline prior to tossing the ball

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Work needed: 720 J

Explanation:

The work needed to stretch a spring is equal to the elastic potential energy stored in the spring when it is stretched, which is given by

$E=\frac{1}{2}kx^2$

where

k is the spring constant

x is the stretching of the spring from the equilibrium position

In this problem, we have

E = 90 J (work done to stretch the spring)

x = 0.2 m (stretching)

Therefore, the spring constant is

$k=\frac{2E}{x^2}=\frac{2(90)}{(0.2)^2}=4500 N/m$

Now we can find what is the work done to stretch the spring by an additional 0.4 m, that means to a total displacement of

x = 0.2 + 0.4 = 0.6 m

Substituting,

$E'=\frac{1}{2}kx^2=\frac{1}{2}(4500)(0.6)^2=810 J$

Therefore, the additional work needed is

$\Delta E=E'-E=810-90=720 J$

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2 years ago

13.<br>14. Why does a person feel weightlessness during freefall?

Sveta_85 [38]

Answer:

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3 years ago

State Newton's three laws of motion.

tresset_1 [31]

Answer:

Newton's first law states that, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.

Newton's second law states that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.

Newton's third law states that if an object A exerts a force on object B, then object B must exert a force of equal magnitude and opposite direction back on object A. This law represents a certain symmetry in nature: forces always occur in pairs, and one body cannot exert a force on another without experiencing a force itself.

Explanation:

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2 years ago

Body 1 has a mass m, and its moving in a circle with a radius r at a speed v. It has a centripetal force. F acting on it. If bod

cluponka [151]

Answer:

The centripetal force on body 2 is 8 times of the centripetal force in body 1.

Explanation:

Body 1 has a mass m, and its moving in a circle with a radius r at a speed v. The centripetal force acting on it is given by :

$F=\dfrac{mv^2}{r}$

Body 2 has a mass 2m and its moving in a circle of radius 4r at a speed 4v. The centripetal force on body 2 is :

$F'=\dfrac{2m\times (4v)^2}{4r}\\\\F'=\dfrac{2m\times 16v^2}{4r}\\\\F'=8\dfrac{mv^2}{r}\\\\F'=8F$

So, the centripetal force on body 2 is 8 times of the centripetal force in body 1.

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2 years ago

The front 1.20 m of a 1,600-kg car is designed as a "crumple zone" that collapses to absorb the shock of a collision. (a) If a c

eimsori [14]

To develop the problem it is necessary to apply the kinematic equations for the description of the position, speed and acceleration.

In turn, we will resort to the application of Newton's second law.

PART A) For the first part we look for the time, in a constant acceleration, knowing the speeds and the displacement therefore we know that,

$X_f = X_i +\frac{1}{2}(V_i+V_f)t$

Where,

X = Desplazamiento

V = Velocity

t = Time

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$X_f = \frac{1}{2} (V_i+V_f)t$

Clearing for time,

$t = \frac{2X_f}{(V_i+V_f)}$

$t = \frac{2*1.2}{24+0}$

$t = 0.1s$

PART B) This is a question about the impulse of bodies, where we turn to Newton's second law, because:

F = ma

Where,

m=mass

a = acceleration

Acceleration can also be written as,

$a= \frac{\Delta V}{t}$

Then

$F = m\frac{\Delta V}{t}$

$F = m\frac{V_f-V_i}{t}$

$F = m\frac{-V_i}{t}$

$F = \frac{(1600kg)(-24m/s)}{(0.1s)}$

$F = -384000N$

Negative symbol is because the force is opposite of the direction of moton.

PART C) Acceleration through kinematics equation is defined as

$V_f^2=V_i^2-2ax$

$0 = (24m/s)^2-2*a(1.2m)$

$a = \frac{(24m/s)^2}{1.2m}$

$a=480m/s^2$

The gravity is equal to 0.8, then the acceleration is

$a = 480*\frac{g}{9.8}$

$a = 53.3g$

3 0

2 years ago