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Arte-miy333 [17]
3 years ago
8

If two charged objects each have 2.5 C of charge on them and are located 100 m apart, how strong is the electrostatic force betw

een them?
Physics
1 answer:
LiRa [457]3 years ago
4 0

Answer:

5.619×10⁶ N

Explanation:

Applying,

F = kqq'/r²................... Equation 1

Where F = electrostatic force between the charges, k = coulomb's constant, q = first charge, q' = second charge, r = distance btween the charges

From the questiion,

Given: q = 2.5 C, q' = 2.5 C, r = 100 m

Constant: 8.99×10⁹ Nm²/C²

Substitute these values into equation 1

F = (2.5×2.5×8.99×10⁹)/100²

F = 56.19×10⁵

F = 5.619×10⁶ N

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A motorcycle is travelling at a constant velocity of 30ms. The motor is in high gear and emits a hum of 700Hz. The speed of soun
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a) T=1.43\times 10^{-3}\ s

b) d=0.0429\ m

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Given:

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  • original frequency of sound from the source, f_s=700\ Hz
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(a)

We know time period is inverse of frequency:

Mathematically:

T=\frac{1}{f}

T=\frac{1}{700}

T=1.43\times 10^{-3}\ s

(b)

Distance travelled by the motorcycle during one period of sound oscillation:

d=v_s.T

d=30\times 1.43\times 10^{-3}

d=0.0429\ m

(c)

The distance travelled by the sound during the period of one oscillation is its wavelength.

\lambda=\frac{S}{f}

\lambda=\frac{340}{700}

\lambda=0.4857\ m

(d)

observer frequency with respect to a stationary observer:

<u>According to the Doppler's effect:</u>

\frac{f_o}{f_s}= \frac{S+v_o}{S-v_s} ...........................(1)

where:

f_o\ \&\ v_o are the observed frequency and the velocity of observer respectively.

Here, observer is stationary.

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\frac{f_o}{700}= \frac{340+0}{340-30}

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