Complete Question
How many turns are in its secondary coil, if its input voltage is 120 V and the primary coil has 210 turns.
The output from the secondary coil is 12 V
Answer:
The value is 
Explanation:
From the equation we are told that
The input voltage is 
The number of turns of the primary coil is 
The output from the secondary is 
From the transformer equation
Here 
is the number of turns in the secondary coil
=> 
=>
=>
Answer:
1. True WA > WB > WC
Explanation:
In this exercise they give work for several different configurations and ask that we show the relationship between them, the best way to do this is to calculate each work separately.
A) Work is the product of force by distance and the cosine of the angle between them
WA = W h cos 0
WA = mg h
B) On a ramp without rubbing
Sin30 = h / L
L = h / sin 30
WB = F d cos θ
WB = F L cos 30
WB = mf (h / sin30) cos 30
WB = mg h ctan 30
C) Ramp with rubbing
W sin 30 - fr = ma
N- Wcos30 = 0
W sin 30 - μ W cos 30 = ma
F = W (sin30 - μ cos30)
WC = mg (sin30 - μ cos30) h / sin30
Wc = mg (1 - μ ctan30) h
When we review the affirmation it is the work where there is rubbing is the smallest and the work where it comes in free fall at the maximum
Let's review the claims
1. True The work of gravity is the greatest and the work where there is friction is the least
2 False. The job where there is friction is the least
3 False work with rubbing is the least
4 False work with rubbing is the least
Acceleration is a change in *speed* over time. In this case, the speed of the car increased by 90 km/hr in 6 s, giving it a rate of 90 km/hr/6s, or 15 km/hr/s. We’re asked for the acceleration in m/s^2, though, so we’ll need to do a few conversions to get our units straight.
There are 1000 m in 1 km, 60 min, or 60 * 60 = 3600 s in 1 hr, so we can change our rate to:
(15 x 1000)m/3600s/s, or (15 x 1000)m/3600 s^2
We can reduce this to:
(15 x 10)m/36 s^2 = 150 m/36 s^2
Which, dividing numerator and denominator by 36, gets us a final answer of roughly 4.17 m/s^2
