Answer:
0.231 m/s
Explanation:
m = mass attached to the spring = 0.405 kg
k = spring constant of spring = 26.3 N/m
x₀ = initial position = 3.31 cm = 0.0331 m
x = final position = (0.5) x₀ = (0.5) (0.0331) = 0.01655 m
v₀ = initial speed = 0 m/s
v = final speed = ?
Using conservation of energy
Initial kinetic energy + initial spring energy = Final kinetic energy + final spring energy
(0.5) m v₀² + (0.5) k x₀² = (0.5) m v² + (0.5) k x²
m v₀² + k x₀² = m v² + k x²
(0.405) (0)² + (26.3) (0.0331)² = (0.405) v² + (26.3) (0.01655)²
v = 0.231 m/s
Answer:
The velocity of the plane at take off is 160 m/s.
The distance travel by the plane in that time is 3200 meter.
Explanation:
Given:
Acceleration, a = 4 m/s²
Time, t = 40 s
u = 0 i .e initial velocity
To Find:
velocity , v = ?
distance , s =?
Solution:
we have first Kinematic equation
v = u + at
∴ v = 0 + 4×40
∴ v = 160 m/s
Now by Third Kinematic equation

∴ s = 0 + 0.5 × 4× 40²
∴ s = 3200 meter
Answer:

at t = 0.001 we have

at t = 0.01

at t = infinity

Explanation:
As we know that they are in series so the voltage across all three will be sum of all individual voltages
so it is given as

now we will have

now we have

So we will have

at t = 0 we have
q = 0

also we know that
at t = 0 i = 0




so we have

at t = 0.001 we have

at t = 0.01

at t = infinity

Power is the amount of energy consumed per unit time. Having no direction, it is a scalar quantity. <span>As is implied by the equation for </span>power<span>, a unit of </span>power <span>is equivalent to a unit of work divided by a unit of time. The formula would be as follows:
P = W/t
We calculate as follows:
500 W = 15000 J / t
t = 30 s</span>