Answer:
The initial acceleration of the 59g particle is
Explanation:
Newton's second laws relates acceleration (a), net force(F) and mass (m) in the next way:
(1)
We already know the mass of the particle so we should find the electric force on it to use on (1), the magnitude of the electric force between two charged objects by Columb's law is:
with q1 and q2 the charge of the particles, r the distance between them and k the constant . So:
Using that value on (1) and solving for a
Answer:
(a) 0.38 m
(b) 2.78 m/s
(c) 0.11 watt
Explanation:
mass, m = 0.3 kg
spring constant, K = 160 N/m
initial compression, d = 12 cm = 01.2 m
initial speed, u = 3 m/s
(a) Let the maximum stretch is y.
Use conservation of energy
Initial potential energy + initial kinetic energy = final potential energy
0.5 x K x d² + 0.5 x m x u² = 0.5 x K x y²
160 x 0.12 x 0.12 + 0.3 x 0.12 x 0.12 = 160 x y²
2.304 + 0.00432 = 160 y²
y = 0.38 m
y = 38 cm
(b) Let v is the maximum speed.
The speed is maximum when the stretch in the spring is zero, so by use of conservation of energy
Initial potential energy + initial kinetic energy = final kinetic energy
0.5 x K x d² + 0.5 x m x u² = 0.5 x m x v²
160 x 0.12 x 0.12 + 0.3 x 0.12 x 0.12 = 0.3 x v²
2.304 + 0.00432 = 0.3 v²
v = 2.78 m/s
(c) The time period of the spring mass system is given by
T = 0.272 second
Energy dissipated per cycle = 0.03 J
Power, P = 0.03 / 0.272 = 0.11 Watt
Answer:
C) 3,000 kg m/s
Explanation:
We can consider the horizontal velocity of the motorcycle to be zero, since it rolls off the edge of the cliff very slowly. So, we only need to find the vertical velocity at the time of the impact with the ground.
The vertical velocity of the motorcycle at time t is given by (free-fall motion):
where
is the initial vertical velocity (zero, since the motorcycle is not moving)
g = 9.8 m/s^2 is the acceleration due to gravity
t is the time
Since the motorcycle hits the ground after t = 3 seconds, we have
And since we know its mass, m=100 kg, we can find its momentum:
and the negative sign simply means downward direction.
Which of the following are characteristics of noble gases?
✅
✅