Answer:
i)-6.25m/s
ii)18 metres
iii)26.5 m/s or 95.4 km/hr
Explanation:
Firstly convert 90km/hr to m/s
90 × 1000/3600 = 25m/s
(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)
0 = (25)^2 + 2A(50)
0 = 625 + 100A....then moved the other value to one
-625 = 100A
Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)
(ii) Firstly convert 54km/hr to m/s
In which this is 54 × 1000/3600 = 15m/s
then apply the same formula as that in (i)
0 = (15)^2 + 2(-6.25)s
-225 = -12.5s
Hence the stopping distance = 18metres
(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question
0 = u^2 + 2(-6.25)(56)
u^2 = 700
Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s
In km/hr....26.5 × 3600/1000 = 95.4 km/hr
Answer:
it will take 36.12 ms to reduce the capacitor's charge to 10 μC
Explanation:
Qi= C×V
then:
Vi = Q/C = 30μ/20μ = 1.5 volts
and:
Vf = Q/C = 10μ/20μ = 0.5 volts
then:
v = v₀e^(–t/τ)
v₀ is the initial voltage on the cap
v is the voltage after time t
R is resistance in ohms,
C is capacitance in farads
t is time in seconds
RC = τ = time constant
τ = 20µ x 1.5k = 30 ms
v = v₀e^(t/τ)
0.5 = 1.5e^(t/30ms)
e^(t/30ms) = 10/3
t/30ms = 1.20397
t = (30ms)(1.20397) = 36.12 ms
Therefore, it will take 36.12 ms to reduce the capacitor's charge to 10 μC.
The final velocity of the train at the end of the given distance is 7.81 m/s.
The given parameters;
- initial velocity of the train, u = 6.4 m/s
- acceleration of the train, a = 0.1 m/s²
- distance traveled, s = 100 m
The final velocity of the train at the end of the given distance is calculated using the following kinematic equation;
v² = u² + 2as
v² = (6.4)² + (2 x 0.1 x 100)
v² = 60.96
v = √60.96
v = 7.81 m/s
Thus, the final velocity of the train at the end of the given distance is 7.81 m/s.
Learn more here:brainly.com/question/21180604
Answer:
Explanation:
a ) starting from rest , so u = o and initial kinetic energy = 0 .
Let mass of the skier = m
Kinetic energy gained = potential energy lost
= mgh = mg l sinθ
= m x 9.8 x 70 x sin 30
= 343 m
Total kinetic energy at the base = 343 m + 0 = 343 m .
b )
In this case initial kinetic energy = 1/2 m v²
= .5 x m x 2.5²
= 3.125 m
Total kinetic energy at the base
= 3.125 m + 343 m
= 346.125 m
c ) It is not surprising as energy gained due to gravitational force by the earth is enormous . So component of energy gained due to gravitational force far exceeds the initial kinetic energy . Still in a competitive event , the fractional initial kinetic energy may be the deciding factor .
