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kompoz [17]
3 years ago
8

Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.650 µC charge and flies due west at a speed of

540 m/s over the Earth's south magnetic pole, where the 8.00 ✕ 10−5 T magnetic field points straight up. What are the magnitude (in N) and direction of the magnetic force on the plane?
Physics
1 answer:
Ganezh [65]3 years ago
4 0

Answer : F=2808\times 10^{-11}\ N

Explanation :

It is given that,

Charge, q=0.650\ \mu C=0.65\times 10^{-6}\ C

Velocity of Aircraft, v=540\ m/s  (in west)

Magnetic field, B=8 \times 10^{-5}\ T ( in north )

Using the relation as :

F=q(v\times B)

Magnetic force is ,

F=0.65\times 10^{-6}\ C\times 540\ m/s\times 8 \times 10^{-5}\ T

F=2808\times 10^{-11}\ N

Using Right hand thumb rule, the direction of force is in the plane perpendicular to the velocity and the magnetic field i.e. -\hat{k}.

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A motorboat traveling 4 m/s, East encounters a current traveling 3.0 m/s, North. a) What is the resultant velocity of the motorb
Nostrana [21]

Explanation:

It is given that,

Velocity in East, v_1=4\ m/s

Velocity in North, v_2=3\ m/s

(a) The resultant velocity is given by :

v=\sqrt{4^2+3^2}=5\ m/s

(b) The width of the river is, d = 80 m

Let t is the time taken by the boat to travel shore to shore. So,

t=\dfrac{d}{v}

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t = 16 seconds

(c) Let x is the distance covered by the boat to reach the opposite shore. So,

x=v_2\times t

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4 0
3 years ago
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3 years ago
A candy-filled piñata is hung from a tree for Elia's birthday. During an unsuccessful attempt to break the 4.4-kg piñata, Tonja
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Answer: v = 0.6 m/s

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Momentum is calculated as Q = m.v

For the piñata problem:

Q_{i}=Q_{f}

m_{p}v_{p}_{i}+m_{s}v_{s}_{i}=m_{p}v_{p}_{f}+m_{s}v_{s}_{f}

Before the collision, the piñata is not moving, so v_{p}_{i}=0.

After the collision, the stick stops, so v_{s}_{f}=0.

Rearraging, we have:

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v_{p}_{f}=\frac{m_{s}v_{s}_{i}}{m_{p}}

Substituting:

v_{p}_{f}=\frac{(0.54)(4.8)}{(4.4)}

v_{p}_{f}= 0.6

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