Answer:
Force of friction, f = 751.97 N
Explanation:
it is given that,
Mass of the car, m = 1100 kg
It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.
From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.


f = 751.97 N
So, the force of friction on the car is 751.97 N. Hence, this is the required solution.
Answer:
current is 13.88 A
Explanation:
given data
no of loops = 50 loops
resistance = 40 Ω
R = 15.0 Ω
I = 5.0 amp
N = 18
to find out
rms current
solution
we will apply here transformed principle that is
v(s) / v(p) = I(p) / I(s) = N(s) / N(p) ....................1
here p is primary and s is secondary
put here all these value
5 / I(s) = 18 / 50
I(s) = 250 /18
I(s) = 13.88
so current is 13.88 A
Answer:
Rp = 3.04×10² Ω.
Explanation:
From the question given:
1/Rp = 1/4.5×10² Ω + 1/ 9.4×10² Ω
Rp =?
We can obtain the value of Rp as follow:
1/Rp = 1/4.5×10² + 1/ 9.4×10²
Find the least common multiple (lcm) of 4.5×10² and 9.4×10².
The result is 4.5×10² × 9.4×10²
Next, divide the result of the lcm by each denominator and multiply the result obtained with the numerator as shown below:
1/Rp = (9.4×10² + 4.5×10²) /(4.5×10²) (9.4×10²)
1/Rp = 13.9×10²/4.23×10⁵
Cross multiply
Rp × 13.9×10² = 4.23×10⁵
Divide both side by 13.9×10²
Rp = 4.23×10⁵ / 13.9×10²
Rp = 3.04×10² Ω.