Answer:
No.
Explanation:
Given the following :
Velocity (V) of ball = 5m/s
Radius = 1m
Can the ball reach the highest point of the circular track
of radius 1.0 m?
The highest point in the track could be considered as the diameter of the circle :
Radius = diameter / 2;
Diameter = (2 * Radius) = (2*1) = 2
Maximum height which the ball can reach :
Using the relation :
Kinetic Energy = Potential Energy
0.5mv^2 = mgh
0.5v^2 = gh
0.5(5^2) = 9.8h
0.5 * 25 = 9.8h
12.5 = 9.8h
h = 12.5 / 9.8
h = 1.2755
h = 1.26m
Therefore maximum height which can be reached is 1.26m.
Since h < Diameter
Explanation:
Drinking water requirement in town 2.0 m^3/s of water per second
nitrate in local well nitrate per 15
of water
nitrate in distant reservoir 
Let the flow rate of well
flow rate of reservoir 
Drinking water requirement is
or 
therefore, the total flow of drinking water
Answer:

Explanation:
<u>Frictional Force
</u>
When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

The centripetal acceleration a_c is computed as

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

The normal force N is equal to the weight of the car, thus

Equating both forces

Simplifying

Substituting the values


Scalar quantities have only a magnitude. So the answer is scalar quantities.
Answer:

Explanation:
Given data
length=100mm
Diameter=5mm
Thermal conductivity=5 W/m.K
Power=50 W
Temperature=25°C
The temperature of heater surface follows from the rate equation written as:

Where S can be estimated from the conduction shape factor for a vertical cylinder in semi infinite medium

Substitute the given values
![S=\frac{2\pi (0.1m)}{ln[\frac{4*0.1m}{0.005m} ]}\\ S=0.143m](https://tex.z-dn.net/?f=S%3D%5Cfrac%7B2%5Cpi%20%280.1m%29%7D%7Bln%5B%5Cfrac%7B4%2A0.1m%7D%7B0.005m%7D%20%5D%7D%5C%5C%20S%3D0.143m)
The temperature of heater is then:

The temperature reached by the heater when dissipating 50 W with the surface of the block at a temperature of 25°C.
