Answer:
the maximum vertical height the person in the cart can reach is 18.42 m
Explanation:
Given;
mass of the person in cart, m₁ = 45 kg
mass of the cart, m₂ = 43 kg
acceleration due to gravity, g = 9.8 m/s²
final speed of the cart before it goes up the hill, v = 19 m/s
Apply the principle of conservation of energy;
Therefore, the maximum vertical height the person in the cart can reach is 18.42 m
Answer:
Mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 34.16 g
Explanation:
The vertical forces (with respect to the inclined plane) acting on the 100 g block include the component of the weight of the block in the direction vertical to the inclined plane and the normal reaction of the plane on the block.
And sum of upward forces = sum of downward forces.
N = mg cos θ
m = 100 g = 0.10 kg
g = acceleration due to gravity = 9.8 m/s²
θ = 15°
N = (0.1×9.8×cos 15°) = 0.946582 N
The horizontal forces (With respect to the inclined plane) include the frictional force (acting upwards for the inclined plane, opposite to the intended direction of motion), the Tension in the rope (acting downwards, away from the 100 g block) and the horizontal component (with respect to the inclined plane) of the weight of the block, F, (also acting downards).
For the body to slide down the inclined plane at constant speed, the downward sloping forces must balance the frictional force, that is, there will be no acceleration.
Frictional force = Tension + F
Frictional force = μN
where μ = coefficient of kinetic friction = 0.60
N = normal reaction = 0.9466 N
Frictional force = Fr = (0.60 × 0.9466) = 0.56796 N = 0.568 N
The horizontal component (with respect to the inclined plane) of the weight of the block (also acting downards) = mg sin θ
F = (0.10 × 9.8 × sin 15°) = 0.253624 N
Tension in the rope = T = ?
Fr = F + T
T = Fr - F = 0.568 - 0.253624 = 0.314376 N = 0.3144 N
But the balance on the rope now has the total weight on the container (weight of container + weight on the container) to be equal to 2T.
2T = mg
2 × 0.3144 = 9.8m
m = 0.06416 kg = 64.16 g.
Mass of the container = 30 g
So, mass that one should put in the container so that the 100 g block slides down the inclined plane at constant speed = 64.16 - 30 = 34.16 g
Hope this Helps!!!
The rate of fuel burning in grams per hour if the DT reaction is used is 1.08 × J/g per hour
<h3>How is the rate of fuel burning in grams per hour calculated when the D-T reaction is used?</h3>
E = 17.59 MeV
Mass = 2.014u + 3.016u
= 5.030u
Energy per Kg = (17.59××1.6×
) ÷ ( 5.030×1.66×
)
= 3.37× J/Kg
= 3.0× J/g
Rate of fuel burning in grams per hour = 3.0× × 3600
= 3.6×3.0×
= 1.08 × J/g per hour
To learn more about fusion reactor and energy production, refer
#SPJ4