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OverLord2011 [107]
3 years ago
10

What is an example of frequency

Physics
2 answers:
emmasim [6.3K]3 years ago
6 0

'Frequency' is a word that often confuses some people ... for no good reason.
It just means "frequent-ness" or "often-ness" ... how often something happens.

The SI unit of frequency is the Hertz (Hz).  Hz means 'per second'.
So  " 13 Hz "  means  13 per second.

Here are examples of frequency:

-- 780 kilohertz (on your AM radio dial)
-- 98.7 Megahertz (on your FM dial)
-- 5.8 Gigahertz
-- twice a day
-- three per week
-- every 6 months

vaieri [72.5K]3 years ago
3 0
For example 75 Hertz means that there are 75 wave tops per second in for example a radio signal.
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A rod 16.0 cm long is uniformly charged and has a total charge of -25.0 µC. Determine the magnitude and direction of the electri
Vlada [557]

Answer:

-1.4x10^6N/C

Explanation:

Pls see attached file

8 0
3 years ago
Which of the following are true for acceleration?
Fittoniya [83]

The SI unit for acceleration is m/s2 ( D)

6 0
3 years ago
If at 10m above the ground an object has 50J of Kinetic Energy, with 50J of Potential Energy. How high can the object travel?
steposvetlana [31]

Hi there!

\large\boxed{\text{B) 20 meters}}

We know that:

E_T = U + K

U = Potential Energy (J)

K = Kinetic Energy (J)

E = Total Energy (J)

At 10m, the total amount of energy is equivalent to:

U + K = 50 + 50 = 100 J

To find the highest point the object can travel, K = 0 J and U is at a maximum of 100 J, so:

100J = mgh

We know at 10m U = 50J, so we can solve for mass. Let g = 10 m/s².

50J = 10(10)m

m = 1/2 kg

Now, solve for height given that E = 100 J:

100J = 1/2(10)h

100J = 5h

<u>h = 20 meters</u>

3 0
2 years ago
The parachute on a drag racing car deploys at the end of a run. If the car has a mass of 820 kg and the car is moving 36 m/s, wh
Lelechka [254]

In order to determine the required force to stop the car, proceed as follow:

Calculate the deceleration of the car, by using the following formula:

v^2=v^2_o-2ax

where,

v: final speed = 0m/s (the car stops)

vo: initial speed = 36m/s

x: distance traveled = 980m

a: deceleration of the car= ?

Solve the equation above for a, replace the values of the other parameters and simplify:

\begin{gathered} a=\frac{v^2_o-v^2}{2x} \\ a=\frac{(36\frac{m}{s})^2-(0\frac{m}{s})^2}{2(980m)}=0.66\frac{m}{s^2} \end{gathered}

Next, consider that the formula for the force is:

F=ma

where,

m: mass of the car = 820 kg

a: deceleration of the car = 0.66m/s^2

Replace the previous values and simplify:

F=(820kg)(0.66\frac{m}{s^2})=542.20N

Hence, the required force to stop the car is 542.20N

4 0
1 year ago
If a ball has a velocity of 18 m/s, how far does it travel in 15 seconds
Stolb23 [73]

Answer:

The answer is

<h2>270 m</h2>

Explanation:

To find the distance when given the velocity and time we use the formula

<h3>distance = velocity × time</h3>

From the question

velocity of the ball = 18 m/s

time = 15 s

So the distance is

distance = 18 × 15

We have the final answer as

<h3>270 m</h3>

Hope this helps you

7 0
3 years ago
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