Answer:
R₁ = 50.77 Ω
Explanation:
Since, we know that:
Electric Power = P = VI
but from Ohm's Law:
V = IR
(or) I = V/R
Therefore,
P = V²/R
(OR) R = V²/P
where,
V = Battery Voltage
R = Resistance of combination
FOR SERIES COMBINATION:
R = Rs = (57 V)²/48 W
Rs = 67.69 Ω
but, we know that:
Rs = R₁ + R₂
R₁ + R₂ = 67.69 Ω
R₁ = 67.69 Ω - R₂ __________ eqn (1)
FOR PARALLEL COMBINATION:
R = Rp = (57 V)²/256 W
Rp = 12.69 Ω
but, we know that:
Rp = (R₁R₂)/(R₁ + R₂) = 12.69 Ω
using eqn (1) and value of R₁ + R₂, we get
Rp = 12.69 = R₂(67.69 - R₂)/67.69
859.08 = 67.69 R₂ - R₂²
R₂² - 67.69 R₂ + 859.08 = 0
Solving this quadratic equation we get the answers:
Either, R₂ = 50.76 Ω
Either, R₂ = 16.92 Ω
Since, it is stated in the question that R₁ > R₂. Therefore, we choose the second value. So,
<u>R₂ = 16.92 Ω</u>
using this value in eqn (1), we get:
R₁ = 67.69 Ω - 16.92 Ω
<u>R₁ = 50.77 Ω</u>
Answer:
A 75.1 N and a direction of 152° to the vertical.
B 85.0 N at 0° to the vertical.
Explanation:
A) The interaction partner of this normal force has what magnitude and direction?
The interaction partner of this normal force is the component of the weight of the crate perpendicular to the ramp. <u>It has a magnitude of 85cos28° = 75.1 N and a direction of 180° - 28° = 152° to the vertical(since it is directed downwards perpendicular to the ramp).</u>
B) The normal and frictional forces are perpendicular components of the contact force exerted on the crate by the ramp. What is the magnitude and direction of the contact force?
Since this force has to balance the weight of the crate, its magnitude is 85.0 N. Its direction has to be vertically opposite to that of the weight.
Since the weight is 180° to the vertical (since it is directed downwards), this force is 0° to the vertical.
<u>So, this force has a magnitude of 85.0 N and a direction of 0° to the vertical.</u>
To solve this problem we will use a free body diagram that allows us to determine the Normal Force.
In general, the normal force would be equivalent to

Since the skier is standing on two skis, his weight will be divide by two

Pressure is given as the force applied in a given area, that is

Replacing F with N'


Our values are given as,




Replacing we have that


Therefore the pressure exerted by each ski on the snow is 776.01Pa
Answer:
I literally just got done doing something like this in my physic class. The resistance is 9.6 ohms.