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lianna [129]
2 years ago
12

What particular appropriate program/application is installed on the system where in the user can utilize and browse educational

websites?​
Physics
1 answer:
Bond [772]2 years ago
3 0

Answer:educational software

Explanation:

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Examples of angular motion<br>​
ohaa [14]

Answer:

A figure skater doing a double axle

The swing of a baseball bat

The leverage on a hockey stick

hope it helps

3 0
2 years ago
Read 2 more answers
What is the acceleration of a 4,000 kg car pushed with a<br> force of 12,000 N?
jek_recluse [69]

Answer:

3 m/s

Explanation:

A= F/m

12,000/ 4000 = 3

8 0
2 years ago
Read 2 more answers
Will the positioning of your Slinky along the east-west, north-south, or some other orientation affect your readings? Why?
Oxana [17]

Positioning your Slinky along any direction different from its initial position will affect your reading, because there will be change in the magnetic field.

<h3>Effect of magnet on Slinky</h3>

If the Slinky is made of an iron alloy, it can be magnetized by itself. Moving the Slinky around can cause a change in the magnetic field, even if no current is flowing.

When there is a change in the magnetic field, the reading changes.

At any point, you change the orientation of the Slinky, you will need to zero the reading or adjust the Slinky back to its initial position, even if the sensor does not move.

Thus, Positioning your Slinky along any direction that is different to its initial position will affect your reading because there will be change in the magnetic field.

Learn more about magnetic field here: brainly.com/question/7802337

5 0
2 years ago
A LASIK vision-correction system uses a laser that emits 10-ns-long pulses of light, each with 3.0 mJ of energy. The laser beam
lyudmila [28]
 <span>P = energy/t = 0.0025/1E-8 = 250000 W 
I(ave) = P/A = 250000/(pi*0.425E-3^2) = 4.4056732E11 W/m^2 
I(peak) = 2I(ave) = 8.8113463E11 W/m^2 
Electric field E = sqrt(I(peak)*Z0) = 1.8219499E7 V/m, where 
free-space impedance Z0 = sqrt(µ0/e0) = 376.73031 ohms</span>
7 0
2 years ago
When I wave a charged golf tube at the front of the classroom with a frequency of two oscillations per second, I produce an elec
borishaifa [10]

To solve the exercise it is necessary to take into account the concepts of wavelength as a function of speed.

From the definition we know that the wavelength is described under the equation,

\lambda = \frac{c}{f}

Where,

c = Speed of light (vacuum)

f = frequency

Our values are,

f = 2Hz

c = 3*10^8km/s

Replacing we have,

\lambda = \frac{c}{f}

\lambda = \frac{3*10^8km/s}{2Hz}

\lambda = 1.5*10^8m

<em>Therefore the wavelength of this wave is 1.5*10^{8}m</em>

8 0
2 years ago
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