First do 1.6 m (how far he jumps) 9.8 m/s (what gravity is measured at) then times 2
= 31.36
Sq root = 5.6
Answer:
0.54 A
Explanation:
Parameters given:
Number of turns, N = 15
Area of coil, A = 40 cm² = 0.004 m²
Change in magnetic field, ΔB = 5.1 - 1.5 = 3.6 T
Time interval, Δt = 2 secs
Resistance of the coil, R = 0.2 ohms
To get the magnitude of the current, we have to first find the magnitude of the EMF induced in the coil:
|V| = |(-N * ΔB * A) /Δt)
|V| = | (-15 * 3.6 * 0.004) / 2 |
|V| = 0.108 V
According to Ohm's law:
|V| = |I| * R
|I| = |V| / R
|I| = 0.108 / 0.2
|I| = 0.54 A
The magnitude of the current in the coil of wire is 0.54 A
Answer:
A. 
B. t = 50 s
Explanation:
A. The vectorial equation of the person who is getting closer to the other person is:

r: position vector
v: speed vector = 6m/s i (if you consider the motion as a horizontal motion)
Then, you replace and obtain:

B. The time is:

d: distance to the observer = 300m
v: speed of the person on the car = 6.00 m/s

Answer:0.27
Explanation:
Given
One worker Pushes with force 
other Pulls it with a rope of rope 
mass of crate 
both forces are horizontal and crate slides with a constant speed
Both forces are in the same direction so Friction will oppose the forces and will be equal in magnitude of sum of two forces because crate is moving with constant speed i.e. net force is zero on it

where
is the friction force



where
is the coefficient of static friction



Answer:
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Explanation: