Question: Initially, the car travels along a straight road with a speed of 35 m/s. If the brakes are applied and the speed of the car is reduced to 13 m/s in 17 s, determine the constant deceleration of the car.
Answer:
1.29 m/s²
Explanation:
From the question,
a = (v-u)/t............................ Equation 1
Where a = deceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.
Given: v = 13 m/s, u = 35 m/s, t = 17 s.
a = (13-35)/17
a = -22/17
a = -1.29 m/s²
Hence the deceleration of the car is 1.29 m/s²
20m/s, honestly I have know idea what the asnwer is but if i had to guess, this would be my answer.
The resistance needed to be added is R
The Current is 2 ma
The voltage reading is a maximum of 50 volts.
The ma meter has an internal resistance of 40 ohms.
Formula
E = I * R
Givens
E = 50
I = 2 ms
R = R + 40
Solution
E = I * R
I = 2 ma [ 1 amp / 1000 ma] = 0.002 amp
50 = 0.002 * (R + 40) Divide by 0.002
50/0.002 = R + 40
25000 = R + 40 Subtract 40 from both sides.
R = 25000 - 40
R = 24960 Answer
Answer:
7.4 cm
Explanation:
K = 2.17 x 10^3 N/m
m = 4.71 kg
v = 1.78 m/s (It is maximum velocity)
The angular velocity
ω = 24 rad/s
Maximum velocity, v = ω x A
Where, A be the maximum displacement
1.78 = 24 x A
A = 0.074 m = 7.4 cm
Absorber, it absorbs the light, like a black shirt is an absorber..
