Ask question

Ask question

All categories

2 years ago

14

Which of these objects has kinetic energy?

2 answers:

balandron [24]2 years ago

4 0

I cant tell what you’re talking about but here are some examples of kinetic energy;

Walking, jogging, and running

A driving car

a baseball being thrown

Damm [24]2 years ago

3 0

Answer:

A ball moving through the air.

Explanation:

The ball has momentum which is a form of kinetic energy.

I don't know if that is correct, but I hope it helps!!!!

You might be interested in

Convert 56 kilometers to inches

svlad2 [7]

The conversion for km to inches is:

1km=39370.1in

Now we can solve for 56 km..

56km=39370.1*56
56km=<span> 2204725.6in

Answer=2,204,725.6in</span>

7 0

2 years ago

An ultrasonic tape measure uses frequencies above 20 MHz todetermine dimensions of structures such as buildings. It does so byem

ohaa [14]

Answer:

a) 1m

b) 2μs

c) 3mm

Explanation:

3 0

2 years ago

Argue as to why drivint on icy or snowy roads can be dangerous

Vika [28.1K]

Icy/Snowy roads have less friction than normal roads. This means that the wheels are less likely to stay positioned because of traction, and you will spin out of control

6 0

2 years ago

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

Please help me fast please help

inysia [295]

Answer:

the net energy Gained per hour equals 30Kcal/h

4 0

2 years ago