Answer:
350 miles
Explanation:
When the car starts 2 hours later, the train would have a head start of
50 * 2 = 100 miles
The speed of the car relative to the train is
70 - 50 = 20 mi/hr
For the car to catch up with the train, it must cover the 100 miles difference at the rate of 20mi/hr. So the time it would need to cover this difference is
100 / 20 = 5 hours
After 5 hours, the car would have traveled a distance of
5 * 70 = 350 miles which is also the distance from the station to where the car catches up
<span>Matter is c) something that has mass and occupies space. Everything in our known universe takes up space, and everything is made up of matter. Matter as a concept is not something that can have any unit of measurement put upon it. </span>
Answer:
B. 6 cm
Explanation:
First, we calculate the spring constant of a single spring:

where,
k = spring constant of single spring = ?
F = Force Applied = 10 N
Δx = extension = 4 cm = 0.04 m
Therefore,

Now, the equivalent resistance of two springs connected in parallel, as shown in the diagram, will be:

For a load of 30 N, applying Hooke's Law:

Hence, the correct option is:
<u>B. 6 cm</u>
Answer
given,
force per unit length = 350 µN/m
current, I = 22.5 A
y = y = 0.420 m



I₂ = 32.67 A
distance where the magnetic field is zero


there the distance at which the magnetic field is zero in the two wire is at 0.248 m.