The north and North Pole or the south and the South Pole are facing each other which is causing them to repel
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!!
I think it is <span>Alpha rays.</span>
Answer:
3420.39 N
Explanation:
Applying,
Fd = 1/2(mv²-mu²)................. Equation 1
Where F = force on the bumber, d = distance, m = mass of the car, v = final velocity, u = initial velocity.
make F the subject of the equation
F = (mv²-mu²)/2d............... Equation 2
From the question,
Given: m = 890 kg, v = 0 m/s (to rest), u = 1.4 m/s, d = 0.255 m
Substitute these values into equation 2
F = [(890×0²)-(890×1.4²)]/(2×0.255)
F = -1744.4/0.51
F = -3420.39 N
The negative sign denotes that the force in opposite direction to the motion of the car.
Answer:
Hz
Explanation:
In alternating current (AC) circuits, voltage (V) oscillates in a sine wave pattern and has a general equation as a function of time (t) as follows;
V(t) = V sin (ωt + Ф) -----------------(i)
Where;
V = amplitude value of the voltage
ω = angular frequency = 2 π f [f = cyclic frequency or simply, frequency]
Ф = phase difference between voltage and current.
<u><em>Now,</em></u>
From the question,
V(t) = 230 sin (100t) ---------------(ii)
<em><u>By comparing equations (i) and (ii) the following holds;</u></em>
V = 230
ω = 100
Ф = 0
<em><u>But;</u></em>
ω = 2 π f = 100
2 π f = 100 [divide both sides by 2]
π f = 50
f = Hz
Therefore, the frequency of the voltage is Hz
