Answer:
A)s = 104.16 m
b)s= 104.16 m
Explanation:
Given that
u = 25 m/s
μ = 0.3
The friction force will act opposite to the direction of motion.
Fr= μ m g
Fr= - m a
a=acceleration
μ m g = - m a
a= - μ g
a= - 0.3 x 10 m/s² ( take g= 10 m/s²)
a= - 3 m/s²
The final speed of the mass is zero ,v= 0
We know that
v² = u² +2 a s
s=distance
0² = 25² - 2 x 3 x s
625 = 6 s
s = 104.16 m
By using energy conservation
Work done by all the forces =Change in the kinetic energy
Negative sign because force act opposite to the displacement.
- 3 x 2 x s = - 625
s= 104.16 m
Answer:
The potential energy at point A is 17.1675 J
Explanation:
The capillary potential is the work expended to bring up a unit mass of liquid to a point in a capillary region from a level liquid surface. It is the capillary potential that facilitates the movement of moisture within soil capillaries
In meteorology it is used to describe the level of saturated soil above the water table
Potential energy is the energy inherent in a body by virtue of its position, therefore the potentials of both point A and B are
Point A, elevation = 75 cm capillary potential = -100 cm
Point B, elevation = 25 cm capillary potential = -200 cm
The total potential energy at point A is
Elevation above reference - capillary potential =75-(-100) = 175 cm
which gives per unit mass
PE = m × g × h = 1 kg × 9.81 m/s ² × 1.75 m = 17.1675 kg·m²/s² = 17.1675 J
Answer:
No, if a car is going faster. The RPM is obviously higher. If that is higher, you can burn through gas and energy much faster. A car going at 15mph would be cruising and wouldn't have to worry too much about burning our your vehicle.
Explanation:
Answer:
The value is 
Explanation:
From the question we are told that
The mass of the car is 
The acceleration is 
Generally the net force applied on the rope is mathematically represented as
Here W is the weight of the car which is evaluated as
=> 
=> 
Generally the net force can also be mathematically represented as
So
=> 
=>
Answer:
D. A zinc Zn contains 30 protons inside the nucleus and 30 electron outside the nucleus
