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Usimov [2.4K]
3 years ago
9

A school bus takes 0.53 hours to reach the school from your house. If the average speed of the bus is 19km/h, what is the displa

cement of the bus during the trip?
Physics
2 answers:
Artist 52 [7]3 years ago
5 0
D = v*t= 19*0,53 = 10.07 km
Harlamova29_29 [7]3 years ago
4 0

Answer:

The displacement of the bus during the trip is 10.07 km.

Explanation:

Given that,

Time = 0.53 h

Average speed of bus = 19 km/h

We need to calculate the displacement of the bus

Using formula of average velocity

v_{av}=\dfrac{D}{t}

D = v_{avg}\times t

Where, D = displacement

t = time

Put the value into the formula

D=19\times0.53

D=10.07\ km

Hence, The displacement of the bus during the trip is 10.07 km.

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A flat, 181 ‑turn, current‑carrying loop is immersed in a uniform magnetic field. The area of the loop is 4.97 cm2 and the angle
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Explanation:

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A: area of the loop = 4.97cm^2 = 4.97(10^-2m)^2=4.97*10^-4m^2

N: turns = 181

θ: angle between B and the magnetic dipole (same as the direction  of the normal to the plane)

You replace the values of the parameters in (1). Furthermore you do B the subject of the formula:

B=\frac{\tau}{NIAsin\tetha}=\frac{1.51*10^{-5}Nm}{(181)(2.47*10^{-3}A)(4.97*10^{-4}m^2)(sin30.1\°)}=0.135T

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A negative charge of -6.0x10-6 C exerts an
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The magnitude of the second charge given that the first is –6×10¯⁶ C and is located 0.05 m away is +3.0×10¯⁶ C

<h3>Coulomb's law equation </h3>

F = Kq₁q₂ / r²

Where

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  • K is the electrical constant
  • q₁ and q₂ are two point charges
  • r is the distance apart

<h3>How to determine the second charge </h3>
  • Charge 1 (q₁) = –6×10¯⁶ C
  • Electric constant (K) = 9×10⁹ Nm²/C²
  • Distance apart (r) = 0.05 m
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F = Kq₁q₂ / r²

Cross multiply

Fr² = Kq₁q₂

Divide both side by Kq₁

q₂ = Fr² / Kq₁

q₂ = (65 × 0.05²) / (9×10⁹ × 6×10¯⁶)

q₂ = +3.0×10¯⁶ C (since the force is attractive)

Learn more about Coulomb's law:

brainly.com/question/506926

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