The net force on the block parallel to the incline is
∑ F = -mg sin(θ) = ma
where m is the mass of the block, g = 9.8 m/s² is the acceleration due to gravity, θ is the angle the incline makes with the horizontal, and a is the acceleration of the block. Solving for a gives
a = -g sin(θ)
so the block would slide down the incline with acceleration
a = - (9.8 m/s²) sin(30°) = -4.9 m/s²
The mass of a is smaller than the mass of b.
Answer:
The distance the train travels before coming to a (complete) stop = 40/81 km which is approximately 493.83 meters
Explanation:
The initial speed of the train u = 80 km/h = 22 2/9 m/s = 22. m/s
The magnitude of the constant acceleration with which the train slows, a = 0.5 m/s²
Therefore, we have the following suitable kinematic equation of motion;
v² = u² - 2 × a × s
Where;
v = The final velocity = 0 (The train comes to a stop)
s = The distance the train travels before coming to a stop
Substituting the values gives;
0² = 22.² - 2 × 0.5 × s
2 × 0.5 × s = 22.²
s = 22.²/1 = 493 67/81 m = 40/81 km
The distance the train travels before coming to a (complete) stop = 40/81 km ≈ 493.83 m.
Answer:
r = 5.07 m
Explanation:
given,
velocity of the man , v = 3.43 m/s
centripetal acceleration, a = 2.32 m/s²
magnitude of position of = ?
using centripetal acceleration formula
r = 5.07 m
The magnitude of the position vector relative to rotational axis is equal to 5.07 m.
Answer:
Suppose the micrometeoroid weighed 1 g = .001 kg
Suppose also the spacecraft were moving at 18,000 mph (1.5 hrs per rev)
Usually, the smaller particle would be moving but for simplicity suppose that it were stationary wrt the ground
v = 18000 miles / hr * 1500 m/mile / 3600 sec/hr = 7500 m/s
KE = 1/2 * .001 kg * (7500 m)^2 = 28,125 Joules
One can see that 28000 Joules could be damaging amount of energy