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Nataly [62]
3 years ago
9

This force on compass dials is an example of a force that _______.

Physics
2 answers:
romanna [79]3 years ago
4 0
D. Acts at a distance, because magnetic forcefield is in two dots, Earth's north and south magnetic poles, and magnet is on some distance from them.
TEA [102]3 years ago
3 0

acts at a distance

is the answer

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How is amplitude changed in an instrument or tuning fork
azamat

Amplitude is affected by the energy wave in the instrument. High energy wave means high amplitude and low energy wave means low amplitude.

<u>Explanation:</u>

The amplitude of a periodic variable is a measure of its change over a single period. There are various definitions of amplitude, which are all functions of the magnitude of the differences between the variable's extreme values.

The amount of energy carried by a wave is related to the amplitude of the wave. Amplitude of an instrument is directly affected by the wave of the energy in the instruments. High energy wave means high amplitude and low energy wave means low amplitude in the instrument.

7 0
3 years ago
How does the electric force between two charged particles change if the distance between them is increased by a factor of 2
mart [117]
Electric force will increase
6 0
3 years ago
Why are these waves called out phase waves?
ss7ja [257]

Answer:

^ What he said

Explanation:

Peace out

4 0
2 years ago
Which name is given to the type of friction that objects falling through air experience?
Pani-rosa [81]

Some call it "air resistance", and others just call it "drag".

4 0
3 years ago
Read 2 more answers
Peter designed a road with a curve of radius 30 m that is banked so that a 950 kg car traveling at 40.0 km/h can round it even i
spayn [35]

Answer:

v = 15.56 m/s

v = 56 km/h

Explanation:

When coefficient of friction is approximately zero then we have

F_ncos\theta = mg

F_n sin\theta = \frac{mv^2}{R}

tan\theta = \frac{v^2}{Rg}

here we know that

v = 40 km/h = 11.11 m/s

R = 30 m

tan\theta = \frac{11.11^2}{30\times 9.81}

\theta = 22.75 degree

now when friction coefficient is 0.30 then we have

F_n cos\theta = mg + F_f sin\theta

F_f cos\theta + F_n sin\theta = \frac{mv^2}{R}

now we have

v = \sqrt{Rg(\frac{\mu + tan\theta}{1 - \mu tan\theta})}

v = \sqrt{30(9.81)(\frac{0.30 + tan22.75}{1 - (0.30) tan22.75})}

v = 15.56 m/s

v = 56 km/h

3 0
3 years ago
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