Explanation:
(a) A circuit has 50 Ω, a 100Ω and a 150 Ω resistor are connected in series. We know that in series combination, current through each resistor is same. So, current through a 50 Ω, a 100Ω and a 150 Ω resistor is same.
(b) Ohm's law of given by :
V = I R
V is potential difference. As I is same, so, the resistor having highest resistance will have highest potential difference. So,
.
Hence, this is the required solution.
Consider the motion of the car before brakes are applied:
v₀ = maximum initial velocity of the car before the brakes are applied
t = reaction time = 0.50 s
x₀ = distance traveled by the car before brakes are applied
since car moves at constant speed before brakes are applied
Using the equation
x₀ = v₀ t
x₀ = v₀ (0.50)
Consider the motion after brakes are applied :
v₀ = initial velocity of the car before the brakes are applied
a = acceleration = - 10 m/s²
v = final velocity of the car after it comes to stop = 0 m/s
x = stopping distance = initial distance - distance traveled before applying the brakes = 38 - x₀ = 38 - v₀ (0.50)
Using the equation
v² = v²₀ + 2 a x
inserting the values
0² = v²₀ + 2 (- 10) (38 - v₀ (0.50))
v²₀ = 20 (38 - v₀ (0.50))
v₀ = 23 m/s
Answer:
R (120) = 940Ω
Explanation:
The variation in resistance with temperature is linear in metals
ΔR (T) = R₀ α ΔT
where α is the coefficient of variation of resistance with temperature, in this case α = -0,0005 / ºC
let's calculate
ΔR = 1000 (-0,0005) (120-0)
ΔR = -60
Ω
ΔR = R (120) + R (0) = -60
R (120) = -60 + R (0)
R (120) = -60 + 1000
R (120) = 940Ω
To solve this problem we will apply the concepts related to the Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. Mathematically it can be described as,

Here,
= Frequency of Source
= Speed of sound
f = Frequency heard before slowing down
f' = Frequency heard after slowing down
v = Speed of the train before slowing down
So if the speed of the train after slowing down will be v/2, we can do a system equation of 2x2 at the two moments, then,
The first equation is,



Now the second expression will be,



Dividing the two expression we have,

Solving for v, we have,

Therefore the speed of the train before and after slowing down is 22.12m/s
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