Answer:
Equal to 5000N
Explanation:
The stress on the material is defined by force per unit of cross-sectional area. So it depends on the force and the diameter of the wire, which is the same for both wires. The material that defines the breaking point, is also the same. Therefore, both wires have their breaking point the same at 5000N. The wire length plays no role in here.
Answer:

Given:
Initial velocity (u) = 30 m/s
Final speed (v) = 0 m/s
Acceleration (a) = - 1.5 m/,s²
To Find:
Time in which train will come to rest (t).
Explanation:

So,
Time in which train will come to rest = 20 seconds
Answer:
W = 46 J
Explanation:
We need to find the angle between the two vectors Force vector and displacement vector.
First we will find the angle α of the force vector

Then we find the angle β of the displacement vector

With these two angles we can find the angle between the two vectors
∅ = α + β = 25.56 deg
The definition of work is given by the expression

The absolute value of F will be:

The absolute value of d will be:

Now we have:

Answer:
It is very rare to see a solar eclipse from your home, because the Earth, Sun, and the moon need to align just right. Not everyone in the world can view a solar eclipse, only some area can. A solar eclipse is where the moon blocks out the sun. If you think about it: Let's say you live in Florida, U.S.A. You may see the moon coming in front of the sun, but if you lived in California or sumthin', the moon and the sun wouldn't be aligned to form a solar eclipse. It all depends on location... so it is rare to see one.
Answer:

Explanation:
From the question we are told that:
Mass 
Charge 
Velocity 
Length of Wire 
Current 
Generally the equation for Magnetic Field of Wire B is mathematically given by



Generally the equation for Force on the plane F is mathematically given by

Therefore




Therefore in Terms of g's

