Answer:
t=67.7s
Explanation:
From this question we know that:
Vo = 6m/s
a = 1.8 m/s2
D = 1500m
And we also know that:
Replacing the known values:
Solving for t we get 2 possible answers:
t1 = -44.3s and t2 = 67.7s Since negative time represents an instant before the beginning of the movement, t1 is discarded. So, the final answer is:
t = 67.7s
Answer:
4.9 minutes
Explanation:
Given; T(t) = Ce^-kt + Ts
Now;
T(t) = 190 degrees Fahrenheit
Ts = 60 degrees
To obtain C;
190 = Ce^0 + 60
190 - 60 = C
C = 130
Hence, to find k when t=11
172 = 130 e^-11k + 60
172 -60/130 = e^-k
e^-k = 0.86
ln(e^-k) = ln( 0.86)
-k = -0.15
k = 0.15
Hence at 122 degrees, t is;
T(t) = Ce^-kt + Ts
122 = 130e^-0.15t + 60
122 - 60/130 = e^-0.15t
0.477 = e^-0.15t
ln (e^-0.15t) = ln (0.477)
-0.15t = -0.74
t = 0.74/0.15
t = 4.9 minutes
Answer:
(a) t = 0 s
(b) t = 0 s, 30 s, 55 s
(c) t = 40 s to t = 60 s
(d) t = 10 s to t = 15 s
(e) a = 6 m/s^2
Explanation:
(a) The car is at starting position at t = 0 s and v = 0 m/s.
(b) The velocity of car is zero when the time is t = 0 s, 30 s and 55 s.
(c) from t = 40 s to 60 s the car is moving in the negative direction.
(d) The fastest speed is 60m/s from t = 10 s to t = 15 s.
(e) The slope of the velocity time graph gives acceleration.
a = (60 - 0) / (10 - 0) = 6 m/s^2
Answer:
- the Thevenin equivalent circuit for the battery is uploaded below
- the battery delivered 117 watts of power to the starter motor
Explanation:
Given the data in the equation
diagram of the Thevenin equivalent circuit for the battery is uploaded below.
Current I = 10 A
Voltage 1 = 12 V
voltage 2 = 11.7 v
R = (V1 - V2) / I
R = (12-11.7)/10
R = 0.3 / 10
R = 0.03Ω
Thevenin equivalent circuit
= V2 / I = 11.7 / 10
= 1.17Ω
so, power delivered to the starter motor will be;
p = (V2)² /
P = ( 11.7 V )² / 1.17Ω
p = 136.89 / 1.17
p = 117 watts
Therefore, the battery delivered 117 watts of power to the starter motor
