If It took 40 Seconds for a vehicle of weight 40,000 Newton to move round a 7 metres, What time will it achieve same feat for an
other circular path Of radius 21 metres given the Condition of the vehicle remains the same.
1 answer:
Answer:
t = 376.99 s
Explanation:
We must solve this problem with the equations and kinematics, let's start by looking for the speed of the vehicle,
v = d / t
v = 7/40
v = 0.175 m / s
Since the speed e remains constant, we must find the length of the circle is
L = 2π r
L = 2π 21
L = 131.95 m
In the problem it does not specify clearly, but in general the curves of the road correspond to half a circle, so the length of the road is
L ’= L / 2
L ’= 131.95 / 2 = 65.97 m
as the speed is constant
t = L ’/ v
t = 65.97 / 0.175
t = 376.99 s
You might be interested in
Yes that answer is correct because sometimes tides cause waves to happen
B. If you press that into a calculator it comes up with 153.6. You then shift the decimal point 2 times forward and you end up getting 1.5 x 10^2 V.
F=nmv
where;
n=no. of bullets = 1
m=mass of bullets=2g *10^-3
V=velocity of bullets200m/sec
F=1
loss in Kinetic energy=gain in heat energy
1/2MV^2=MS∆t
let M council M
=1/2V^2=S∆t
M=2g
K.E=MV^2/2
=(2*10^-3)(200)^2/2
2 councils 2
2*10^-3*4*10/2
K.E=40Js
H=mv∆t
(40/4.2)
40Js=40/4.2=mc∆t
40/4.2=2*0.03*∆t
=158.73°C
Maximum height
= (Usinα)^2/2g
(50*0.5)^2/20
25^2/20
625/20
=31.25metres
horizontal distance = Range= [U^2 * sin2α]/g
[50^2 * sin60]/10
2500 * 0.8660/10
2165/10=216.5metres
Answer:
Distance travelled is 7 meters and the displacement is 3 meters
