Complete question:
A train has an initial velocity of 44m/s and an acceleration of -4m/s². calculate its velocity after 10s ?
Answer:
the final velocity of the train is 4 m/s.
Explanation:
Given;
initial velocity of the train, u = 44 m/s
acceleration of the train, a = -4m/s² (the negative sign shows that the train is decelerating)
time of motion, t = 10 s
let the final velocity of the train = v
The final velocity of the train is calculated using the following kinematic equation;
v = u + at
v = 44 + (-4 x 10)
v = 44 - 40
v = 4 m/s
Therefore, the final velocity of the train is 4 m/s.
Answer:
342 m/s
Explanation:
The velocity of sound in air is approximated as:
v ≈ 331.4 + 0.6 T
where v is the velocity in m/s and T is the temperature in Celsius.
At T = 18:
v ≈ 331.4 + 0.6 (18)
v ≈ 342.2
The velocity is approximately 342 m/s.
The answer would be B..
Since sand can heat up quickly, it will also cool off quickly. But water takes a long time to heat up and cool down.
Answer:
The speed of waves on this wire is 329.14 m/s
Explanation:
Given;
tension of the wire, T = 650 N
mass per unit length, μ = 0.06 g /cm = 0.006 kg/m
(convert the unit of mass per length in g/cm to kg/m by dividing by 10 = 0.06 / 10 = 0.006 kg/m)
The speed of waves on this wire is given as;
Therefore, the speed of waves on this wire is 329.14 m/s
The maximum velocity in a banked road, ignoring friction, is given by;
v = Sqrt (Rg tan ∅), where R = Radius of the curved road = 2*1000/2 = 1000 m, g = gravitational acceleration = 9.81 m/s^2, ∅ = Angle of bank.
Substituting;
30 m/s = Sqrt (1000*9.81*tan∅)
30^2 = 1000*9.81*tan∅
tan ∅ = (30^2)/(1000*9.81) = 0.0917
∅ = tan^-1(0.0917) = 5.24°
Therefore, the road has been banked at 5.24°.
