Answer:
11.8 m/s
Explanation:
At the top of the hill, there are two forces on the car: weight force pulling down (towards the center of the circle), and normal force pushing up (away from the center of the circle).
Sum of forces in the centripetal direction:
∑F = ma
mg − N = m v²/r
At the maximum speed, the normal force is 0.
mg = m v²/r
g = v²/r
v = √(gr)
v = √(9.8 m/s² × 14.2 m)
v = 11.8 m/s
One of the concepts to be used to solve this problem is that of thermal efficiency, that is, that coefficient or dimensionless ratio calculated as the ratio of the energy produced and the energy supplied to the machine.
From the temperature the value is given as
Where,
T_L = Cold focus temperature
T_H = Hot spot temperature
Our values are given as,
T_L = 20\° C = (20+273) K = 293 K
T_H = 440\° C = (440+273) K = 713 K
Replacing we have,
Therefore the maximum possible efficiency the car can have is 58.9%
Answer: 1. The field energy will increase
2. The energy increases, and the lines of force are denser
3. It points toward the field of earths magnetic poles
4. 1 and 2 only
5. 2, 4, 1, 3
Explanation: just took it
29.213 cm3
First, calculate the mass of the water used. You do this by subtracting the original mass of the flask from the combined mass of the water and flask, giving:60.735 g - 31.601 g = 29.134 g
So we now know we have 29.134 g of water. To calculate the volume of the flask, simply divide by the density of the water, giving:29.134 g / (0.9973 g/cm3) = 29.213 cm3
A) Work energy relation;
Work =ΔKE ; work done = Force × distance, while, Kinetic energy = 1/2 MV²
F.s = 1/2mv²
F× 4×10^-2 = 1/2 × 5 ×10^-3 × (600)²
F = 900/0.04
= 22500 N
Therefore, force is 22500 N
b) From newton's second law of motion;
F = Ma
Thus; a = F/m
= 22500/(5×10^-3)
= 4,500,000 m/s²
But v = u-at
0 = 600- 4500,000 t
t = 1.33 × 10^-4 seconds
