Answer:
Take the measurement of the distance (d) with a meter rule (in meters) and also measure the time (t) of the travel in seconds with a stopwatch.
question: What is the speed of the cart?
Explanation:
The speed of an object in motion is the distance covered by the object with respect to time, that is, the ratio of distance covered to the time taken to reach that distance.
Speed = distance / time
= d (in meters m) / t (in seconds s) = m/s
The speed of the runner is 300 m /38 seconds. You can simplify this answer to be about 7.9 m/s
Answer:
The fraction of its volume inside liquid is increased .
Explanation:
According to principle pf floatation , an object floats on the surface of water
when the weight of liquid displaced by it becomes equal to weight of the object . weight of the liquid depends upon the density of the liquid .
In the second case , when the body is dipped into liquid of lesser density , in order to balance the weight of body , more volume of liquid will be displaced so that weight of displaced liquid becomes equal to object's weight . So the body floats with greater depth inside liquid . The fraction of its volume inside liquid is increased .
Answer:
v = 12.12 m/s
Explanation:
It is given that,
Radius of circle, r = 30 m
The coefficient friction between tires and road is 0.5,
The centripetal force is balanced by the force of friction such that,
v = 12.12 m/s
So, the maximum speed with which this car can round this curve is 12.12 m/s. Hence, this is the required solution.
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°.
