(B.) I know it’s not that hard or easy so don’t be made if wrong
21+10=31 because you can see that 21 and 10 are in metres while 12 is in seconds so 21+10=31 is the answer.
Part 1
If water does not spill at the top point of the circular motion then for the minimum speed condition we can say normal force will be zero at the top position
given that
R = 1 m
g = 9.8 m/s^2
now from above equation we have
Part b)
for minimum value of angular speed we will have
Answer:
The ratio of the potential difference across a metallic conductor to the current in the conductor is known as.
B. Resistance.
Explanation:
According to ohms law " the current passing through a conductor is directly proportional to the potential difference between the ends provided the temperature of the wire remains constant".
What is resistance ?
Resistance is a measure of the opposition to current flow in an electrical circuit
what is a resistor ?
a resistor is a n electric conductor which forms resistance to free flow of electric current, the resistance is measured in Ω
22. a - (vf^2 - vi^2)/(2d)
a = (0 - 23^2)/(170)
a = -3.1 m/s^2
23. Find the time (t) to reach 33 m/s at 3 m/s^2
33-0/t = 3
33 = 3t
t = 11 sec to reach 33 m/s^2
Find the av velocuty: 33+0/2 = 16.5 m/s
Dist = 16.5 * 11 = 181.5 meters to each 33m/s speed. Runway has to be at least this long.
24. The sprinter starts from rest. The average acceleration is found from:
(Vf)^2 = (Vi)^2 = 2as ---> a = (Vf)^2 - (Vi)^2/2s = (11.5m/s)^2-0/2(15.0m) = 4.408m/s^2 estimated: 4.41m/s^2
The elapsed time is found by solving
Vf = Vi + at ----> t = vf-vi/a = 11.5m/s-0/4.408m/s^2 = 2.61s
25. Acceleration of car = v-u/t = 0ms^-1-21.0ms^-1/6.00s = -3.50ms^-2
S = v^2 - u^2/2a = (0ms^-1)^2-(21.0ms^-1)^2/2*-3.50ms^-2 = 63.0m
26. Assuming a constant deceleration of 7.00 m/s^2
final velocity, v = 0m/s
acceleration, a = -7.00m/s^2
displacement, s - 92m
Using v^2 = u^2 - 2as
0^2 - u^2 + 2 (-7.00) (92)
initial velocity, u = sqrt (1288) = 35.9 m/s
This is the speed pf the car just bore braking.
I hope this helps!!
