Answer:
the kinetic energy lost due to friction is 22.5 J
Explanation:
Given;
mass of the block, m = 0.2 kg
initial velocity of the block, u = 25 m/s
final velocity of the block, v = 20 m/s
The kinetic energy lost due to friction is calculated as;
Therefore, the kinetic energy lost due to friction is 22.5 J
Answer:
The maximum speed that the truck can have and still be stopped by the 100m road is the speed that it can go and be stopped at exactly 100m. Since there is no friction, this problem is similar to a projectile problem. You can think of the problem as being a ball tossed into the air except here you know the highest point and you are looking for the initial velocity needed to reach that point. Also, in this problem, because there is an incline, the value of the acceleration due to gravity is not simply g; it is the component of gravity acting parallel to the incline. Since we are working parallel to the plane, also keep in mind that the highest point is given in the problem as 100m. Solving for the initial velocity needed to have the truck stop after 100m, you should find that the maximum velocity the truck can have and be stopped by the road is 18.5 m/s.
Explanation:
When t=2, the ball has fallen d(2) = 16 (2²) = 64 feet .
When t=5, the ball has fallen d(5) = 16 (5²) = 400 feet .
Distance fallen from t=2 until t=5 is (400 - 64) = 336 feet.
Time period between t=2 until t=5 is (5 - 2) = 3 seconds.
Average speed of the ball from t=2 until t=5 is
(distance covered) / (time to cover the distance)
= 336 feet / 3 seconds = 112 feet per second.
That's what choice-C says.
Answer:
(a) 5.43 x 10⁵ J
(b) 3.07 x 10⁵ J
(c) 45 °C
Explanation:
(a)
= Latent heat of fusion of ice to water = 3.33 x 10⁵ J/kg
m = mass of ice = 1.63 kg
= Energy required to melt the ice
Energy required to melt the ice is given as
= m
= (1.63) (3.33 x 10⁵)
= 5.43 x 10⁵ J
(b)
E = Total energy transferred = 8.50 x 10⁵ J
Q = Amount of energy remaining to raise the temperature
Using conservation of energy
E = + Q
8.50 x 10⁵ = 5.43 x 10⁵ + Q
Q = 3.07 x 10⁵ J
(c)
T₀ = initial temperature = 0°C
T = Final temperature
m = mass of water = 1.63 kg
c = specific heat of water = 4186 J/(kg °C)
Q = Amount of energy to raise the temperature of water = 3.07 x 10⁵ J
Using the equation
Q = m c (T - T₀)
3.07 x 10⁵ = (1.63) (4186) (T - 0)
T = 45 °C
Answer:
a) The centripetal acceleration of the car is 0.68 m/s²
b) The force that maintains circular motion is 940.03 N.
c) The minimum coefficient of static friction between the tires and the road is 0.069.
Explanation:
a) The centripetal acceleration of the car can be found using the following equation:
Where:
v: is the velocity of the car = 51.1 km/h
r: is the radius = 2.95x10² m
Hence, the centripetal acceleration of the car is 0.68 m/s².
b) The force that maintains circular motion is the centripetal force:
Where:
m: is the mass of the car
The mass is given by:
Where P is the weight of the car = 13561 N
Now, the centripetal force is:
Then, the force that maintains circular motion is 940.03 N.
c) Since the centripetal force is equal to the coefficient of static friction, this can be calculated as follows:
Therefore, the minimum coefficient of static friction between the tires and the road is 0.069.
I hope it helps you!