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solmaris [256]
3 years ago
15

An unusual lightning strike has a vertical portion with a current of –400 A downwards. The Earth’s magnetic field at that locati

on is parallel to the ground and has a magnitude of 30 μT. In units of N (Newtons), the force exerted by the Earth’s magnetic field on the 25 m-long current is
Physics
1 answer:
marusya05 [52]3 years ago
8 0

Answer:

The magnetic force is 0.3 N.        

Explanation:

Given that,

Current in an unusual lightning strike, I = -400 A

The Earth’s magnetic field at that location is parallel to the ground and has a magnitude of 30 μT, B=30\times 10^{-6}\ T

We need to find the force exerted by the Earth’s magnetic field on the 25 m-long current.

The magnetic force is given by :

F=ILB\\\\F=400\times 25\times 30\times 10^{-6}\\\\F=0.3\ N

So, the magnetic force is 0.3 N.

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There is up to 30 times more gold in a tons of old mobile phones than in a tons of gold ore. true or false
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3 years ago
Can some answer 7 a and b please
pav-90 [236]

Answer:

a.After 15 second Mr Comer's speed =  1.7 m/s

b.Distance travelled by Mr.Comer in 15 seconds  =   18.75\ m

Explanation:

a. Lets recall our first equation of motion V_{f} =V_{i} + at

Now we know that V_{i} = 0.8m/s , a = 0.06\ m/s^2 and

t = 15 \ sec

Plugging the values we have.

V_{f} =V_{i} + at

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V_{f} =1.7 m/s

Then Mr.Comer's speed after 15 sec = 1.7ms^-1

b.

Lets find the distance and recall our third equation of motion.

V_{f} ^2-V{i}^2 = 2as

So s = distance covered.

Dividing both sides with 2a we have.

\frac{V_{f} ^2-V{i}^2}{2a} = 2as

Plugging the values.

\frac{(1.7)^2-(0.8)^2}{2\times0.06} = 2as

s= 18.75\ m

So Mr.Comer will travel a distance of s= 18.75\ m.

4 0
3 years ago
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