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3 years ago

A tennis ball is hit into the air with a racket. When is the balls kinetic energy the greatest?

Physics

1 answer:

madam [21]3 years ago

3 0

Answer:

Kinetic energy is maximum when the player hits the ball.

Explanation:

Kinetic energy $=\frac{1}{2} mv^2$ , where m is the mass and v is the velocity.

So kinetic energy is proportional to square of velocity.

Velocity is maximum when the player hits the ball.

So kinetic energy is maximum when the player hits the ball.

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Cecily is inflating one of her bicycle tyres with the pump below. When she pushes the plunger down, does the volume of the gas i

Alexxandr [17]

Answer:

See the explanation below

Explanation:

This problem can be better understood graphically, so in the attached image we will use a diagram of a positive displacement air pump.

We can see that when she pushes the plunger, we see that the volume decreases.

8 0

3 years ago

Answer and mark you as brainliest <br> it's that easy

Oduvanchick [21]

Answer:

$t=\frac{v-u}{a}$

b) 35m/s

c) 0

d) -35m/s

e) Velocity is decreasing.

f) 70 seconds

3 0

3 years ago

Read 2 more answers

If the distance between two charged particles is doubled, the force between them changes by a factor of ___. (1 point)

MatroZZZ [7]

Answer:

decreases 1/4 of the original value

4 0

3 years ago

An object of mass 700700 kg is released from rest 20002000 m above the ground and allowed to fall under the influence of gravity

qwelly [4]

Answer:

59.503987 seconds

Explanation:

b = Proportionality constant = 50 Ns/m

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of object = 700 kg

We have the equation of velocity

$v(t)=\dfrac{mg}{b}+\left(V_0-\dfrac{mg}{b}\right)e^{\dfrac{bt}{m}}$

The equation of motion

$x(t)=\dfrac{mg}{b}+\dfrac{m}{b}\left(V_0-\dfrac{mg}{b}\right)(1-e^{\dfrac{bt}{m}})$

$x(t)=\dfrac{700\times 9.81}{50}+\dfrac{9.81}{50}\left(0-\dfrac{700\times 9.81}{50}\right)(1-e^{\dfrac{50t}{700}})$

when x(t)=2000

$2000=\dfrac{700\times 9.81}{50}+\dfrac{9.81}{50}\left(0-\dfrac{700\times 9.81}{50}\right)(1-e^{\dfrac{50t}{700}})\\\Rightarrow 2000\times \:50=\frac{700\times \:9.81}{50}\times \:50+\frac{9.81}{50}\left(0-\frac{700\times \:9.81}{50}\right)\left(1-e^{\frac{50t}{700}}\right)\times \:50\\\Rightarrow 6867-\frac{67365.27}{50}\left(1-e^{\frac{50t}{700}}\right)=100000\\\Rightarrow 50\left(-\frac{67365.27}{50}\left(1-e^{\frac{50t}{700}}\right)\right)=93133\times \:50\\\Rightarrow \frac{-67365.27\left(1-e^{\frac{50t}{700}}\right)}{-67365.27}=\frac{4656650}{-67365.27}\\\Rightarrow 1-e^{\frac{50t}{700}}=-69.12538\dots\\\Rightarrow -e^{\frac{50t}{700}}=-70.12538\dots\\\Rightarrow t=14\ln \left(70.12538\dots \right)\\\Rightarrow t=59.50398\ s$

The time taken is 59.503987 seconds

8 0

3 years ago

A small sphere with mass 5.00.×10−7kg and charge +3.00μC is released from rest a distance of 0.400 m above a large horizontal in

julia-pushkina [17]

I will assume you are asking what the initial acceleration of the sphere is since the information provided seems to indicate that.
First we need to know Newton's Law
F=ma.
We know the mass of the sphere and we want a so we solve to get
a=F/m.
Now we need the force on the charged sphere. This is given by the electric field, E and the charge, Q. The relationship is F=Q×E. (Recall that the electric field units can be expressed in Newtons/Coulomb).
Now the electric field above a large (~infinite) sheet of charge with a known charge density σ, is given by
E = σ/(2ε0)
Plug in your values of σ, to get E, then the sphere charge Q to get F, the the mass into a = F/m to get the acceleration

6 0

3 years ago