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steposvetlana [31]

3 years ago

6

A juggler throws a ball straight up into the air. The ball remains in the air for a time (t) before it lands back in the juggler

s hand. What is the acceleration of the ball during the entire time the ball is in the air?

1 answer:

natka813 [3]3 years ago

5 0

Answer:

$9.8 m/s^2$ , downward

Explanation:

There is only one force acting on the ball during its motion: the force of gravity, which is given by

$F=mg$

where

m is the mass of the ball

$g=9.8 m/s^2$ is the acceleration of gravity (downward)

According to Newton's second law,

$F=ma$

where F is the net force on the object and a is its acceleration. Rearranging for a,

$a=\frac{F}{m}$

As we said, the only force acting on the ball is gravity, so F = mg and the acceleration of the ball is:

$a=\frac{mg}{m}=g$

Therefore, the ball has a constant acceleration of $9.8 m/s^2$ downward for the entire motion.

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A charge of 32.0 nC is placed in a uniform electric field that is directed vertically upward and has a magnitude of 4.30x 104 V/

hodyreva [135]

A) The work done by the electric field is zero

B) The work done by the electric field is $9.1\cdot 10^{-4} J$

C) The work done by the electric field is $-2.4\cdot 10^{-3} J$

Explanation:

A)

The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).

The work done by a force is given by the equation

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the particle

$\theta$ is the angle between the direction of the force and the direction of the displacement

In this problem, we have:

The force is directed vertically upward (because the field is directed vertically upward)
The charge moves to the right, so its displacement is to the right

This means that force and displacement are perpendicular to each other, so

$\theta=90^{\circ}$

and $cos 90^{\circ}=0$ : therefore, the work done on the charge by the electric field is zero.

B)

In this case, the charge move upward (same direction as the electric field), so

$\theta=0^{\circ}$

and

$cos 0^{\circ}=1$

Therefore, the work done by the electric force is

$W=Fd$

and we have:

$F=qE$ is the magnitude of the electric force. Since

$E=4.30\cdot 10^4 V/m$ is the magnitude of the electric field

$q=32.0 nC = 32.0\cdot 10^{-9}C$ is the charge

The electric force is

$F=(32.0\cdot 10^{-9})(4.30\cdot 10^4)=1.38\cdot 10^{-3} N$

The displacement of the particle is

d = 0.660 m

Therefore, the work done is

$W=Fd=(1.38\cdot 10^{-3})(0.660)=9.1\cdot 10^{-4} J$

C)

In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is

$\theta=90^{\circ}+45^{\circ}=135^{\circ}$

Moreover, we have:

$F=1.38\cdot 10^{-3} N$ (electric force calculated in part b)

While the displacement of the charge is

d = 2.50 m

Therefore, we can now calculate the work done by the electric force:

$W=Fdcos \theta = (1.38\cdot 10^{-3})(2.50)(cos 135.0^{\circ})=-2.4\cdot 10^{-3} J$

And the work is negative because the electric force is opposite direction to the displacement of the charge.

Learn more about work and electric force:

brainly.com/question/6763771

brainly.com/question/6443626

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0

3 years ago

Two aluminum soda cans are charged and repel each other, hanging motionless at an angle. which of the forces on the left can has

Sliva [168]

Assuming that the can is motionless, we can then assume that the vertical component of T = mg and that Fe = the horizontal component of T.
<span> Since T itself is larger than it's vertical or horizontal components separately, then T is greater than all the forces.</span>

7 0

3 years ago

Read 2 more answers

Name the material used to transfer of charges from one body to other

Evgesh-ka [11]

Answer:

conductor

Explanation:

A "conductor" is a material that allows the charges to pass freely from one body to the other. This causes a movement among the electrons and this means that<em> the charge will be passed entirely to the object receiving it.</em> This is also called <em>"conductive material."</em>

Examples of conductors are: <em>copper, aluminum, gold, silver, seawater, etc.</em>

The opposite of conductors are called "insulators." These do not allow the free movement of charges from one object to the other.

Examples of insulators: <em>plastic, rubber, paper, glass, wool, dry air, etc.</em>

6 0

3 years ago

when a person uses an iron to remove the wrinkles from a shirt, why does heat travel from the iron to the shirt?

artcher [175]

I think the answer to your question is that:
There is chemical energy being produced (by the water) and that causes the base of iron to absorb the heat. Then when the heated base of the iron is placed onto the shirt, the heat flattens the wrinkles on it.

3 0

3 years ago

At a rock concert, the sound intensity 1.0 m in front of the bank of loudspeakers is 0.10 W/m². A fan is 30 m from the loudspeak

Klio2033 [76]

To solve this problem we will apply the concepts related to the Area, the power and the proportionality relationships between intensity and distance.

The expression for sound power is,

$P = AI$

Here,

A = Area

I = Intensity

P = Power

At the same time the area can be written as,

$A = \frac{\pi d^2}{4}$

Now the intensity is inversely proportional to the square of the distance from the source, then

$I \propto \frac{1}{r^2}$

The expression for the intensity at different distance is

$\frac{I_1}{I_2}= \frac{r^2_2}{r_1^2}$

Here,

$I_1$ = Intensity at distance 1

$I_2$ = Intensity at distance 2

$r_1$ = Distance 1 from light source

$r_2$ = Distance 2 from the light source

If we rearrange the expression to find the intensity at second position we have,

$I_2 = I_1 (\frac{r_1^2}{r_2^2})$

If we replace with our values at this equation we have,

$I_2 = (0.10W/m^2)(\frac{1.0m^2}{30.0m^2})$

$I_2 = 1.11*10^{-4} W/m^2$

Now using the equation to find the area we have that

$A = \frac{\pi (8.4*10^{-3}m)^2}{4}$

$A = 5.5*10^{-5}m^2$

Finally with the intensity and the area we can find the sound power, which is

$P = AI$

$P = (5.5*10^{-5}m^2)(1.11*10^{-4}W/m^2)$

$P = 6.1*10^{-9}J/s$

Power is defined as the quantity of Energy per second, then

$E = 6.1*10^{-9}J$

8 0

3 years ago