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2 years ago

If a 5 kg bowling ball is lifted to a height of 2 meters, how much potential

Physics

1 answer:

natka813 [3]2 years ago

8 0

Answer:

$\boxed {\boxed {\sf 98.1 \ Joules}}$

Explanation:

The formula for potential energy is:

$E_p=mgh$

where m is the mass, g is the gravitational acceleration, and h is the height.

The mass is 5 kilograms and the height is 2 meters. Assuming this ball is on Earth, then the gravitational acceleration is 9.81 meters per square second.

$m= 5 \ kg \\g= 9.81 \ m/s^2 \\\h= 2 \ m$

Substitute the values into the formula.

$E_p= ( 5 \ kg )(9.81 \ m/s^2)( 2 \ m)$

Multiply the first two numbers.

$E_p=49.05 \ kg*m/s^2 (2 \ m )$

Multiply again.

$E_p= 98.1 \ kg*m^2/s^2$

1 kilogram square meter per square second is equal to 1 Joule.
Our answer is equal to 98.1 Joules

$E_p= 98.1 \ J$

The ball has 98.1 Joules of potential energy.

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An electron is moving east in a uniform electric field of 1.55 N/C directed to the west. At point A, the velocity of the electro

valkas [14]

Answer:

Final velocity of electron, $v=6.45\times 10^5\ m/s$

Explanation:

It is given that,

Electric field, E = 1.55 N/C

Initial velocity at point A, $u=4.52\times 10^5\ m/s$

We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :

$v^2=u^2+2as$ ........(1)

a is the acceleration, $a=\dfrac{F}{m}$

We know that electric force, F = qE

$a=\dfrac{qE}{m}$

Use above equation in equation (1) as:

$v^2=u^2+\dfrac{2qEs}{m}$

$v^2=(4.52\times 10^5\ m/s)^2+2\times \dfrac{1.6\times 10^{-19}\ C\times 1.55\ N/C}{9.1\times 10^{-31}\ kg}\times 0.395\ m$

v = 647302.09 m/s

$v=6.45\times 10^5\ m/s$

So, the final velocity of the electron when it reaches point B is $6.45\times 10^5\ m/s$ . Hence, this is the required solution.

3 0

3 years ago

A 1.0 kg rock is thrown straight upward with an initial speed of 8.0 m/s. What is its speed

Ronch [10]

Answer:5.7m/s

Explanation:

Mass=1kg

Initial velocity=u=8m/s

height=h=1.6m

Final velocity =v

Acceleration due to gravity=g=9.8m/s^2

v^2=u^2-2xgxh

v^2=8^2-2x9.8x1.6

v^2=8x8-2x9.8x1.6

v^2=64-31.36

v^2=32.64

Take the square root of both sides

√(v^2)=√(32.64)

v=5.7

Speed at the height of 1.6m is 5.7m/s

8 0

3 years ago

Identify the number of and types of elements in this schematic diagram

Nataly_w [17]

-- The long line and short line close together at the left side
   of the diagram represent a single-cell battery.
It's the only one in this diagram.
It's a device that stores chemical energy and delivers it on demand.

-- The zig-zag lines with circles around them represent light bulbs.
There are three of them in this diagram.
They are devices used to produce light by dissipating electrical energy.

-- The zig-zag lines without circles, at the top of the diagram,
   represent resistors.
There are two of them in this diagram.
They are devices used to change or control electrical parameters
   within a circuit by dissipating electrical energy.

-- The short straight line between two small circles at the bottom
   of the diagram represents a switch.
There is only one switch in this circuit.
It's a device used to easily and quickly start or stop the flow of current
    past a certain point in a circuit.

In this circuit ...

-- When the switch is closed (as drawn), the light bulb nearest the battery
glows brightest, the light bulb in the middle glows less bright, and the light
bulb on the right side glows dimmest of all.

-- When the switch is open, the light bulb nearest the battery glows, and
neither of the other two light bulbs glows at all.

4 0

3 years ago

Read 2 more answers

to 10 Hz. Superimposed on this signal is 60-Hz noise with an amplitude of 0.1 V. It is desired to attenuate the 60-Hz signal to

givi [52]

Answer:

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

Explanation:

For this case we can use the formula for the Butterworth filter gain given by:

[tec] G = \frac{1}{\sqrt{1 +(\frac{f}{f_c})^{2n}}}[/tex]

Where:

G represent the transfer function and we want that G =0.1 since the desired signal is less than 10% of it's value

$f_c = 10 Hz$ represent the corner frequency

$f= 60 Hz$ represent the original frequency

n represent the filter order and that's the variable that we need to find

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

7 0

3 years ago

Graph the linear equation y=2x-6

My name is Ann [436]

I can't draw you the graph right now because i'm out but look up graphing calculator and click on the first one, type in your equation and it will show you the graph

4 0

3 years ago