Answer:
the maximum angular speed (in radians per second) of a Blu-ray disc as it rotates is 57.6 m/s
Explanation:
Given information:
diameter of the disc, d = 11 cm, r = 5.5 cm = 0.055 m
angular speed ω = 10000 rev/min = (10000 rev/min)(2π rad/rev)(1/60 min/s)
= 1000π/3 rad/s
to calculate the maximum angular speed we can use the following formula
ω = v/r
v = ωr
= (1000π/3)(0.055)
= 57.6 m/s
Given :
We, have to find frequency :
Hope Helps!
Answer:
t = 23.255 s, x = 2298.98 m, v_y = - 227.90 m / s
Explanation:
After reading your extensive writing, we are going to solve the approach.
The initial speed of the plane is 250 miles / h and it is at an altitude of 2650 m; In general, planes fly horizontally for launch, therefore this is the initial horizontal speed.
As there is a mixture of units in different systems we are going to reduce everything to the SI system.
v₀ₓ = 250 miles h (1609.34 m / 1 mile) (1 h / 3600 s) = 111.76 m / s
y₀ = 2650 m
Let's set a reference system with the x-axis parallel to the ground, the y-axis is vertical. As time is a scalar it is the same for vertical and horizontal movement
Y axis
y = y₀ + v₀ t - ½ g t²
the initial vertical velocity when the cargo is dropped is zero and when it reaches the floor the height is zero
0 = y₀ + 0 - ½ g t²
t =
t = √(2 2650/ 9.8)
t = 23.255 s
Therefore, for the cargo to reach the desired point, it must be launched from a distance of
x = v₀ₓ t
x = 111.76 23.255
x = 2298.98 m
at the point and arrival the speed is
vₓ = v₀ₓ = 111.76
vertical speed is
v_y = v_{oy} - gt
v_y = 0 - gt
v_y = - 9.8 23.25 555
v_y = - 227.90 m / s
the negative sign indicates that the speed is down
in the attachment we have a diagram of the movement
Actually Welcome to the Concept of the Kinematics.
so here we get as,
V^2 = U^2 + 2as
so here, a = -0.2 m/s^2
(0.1)^2 = (0.3)^2 + (-0.2)(s)
=> 0.01 = 0.09 - 0.2s
=> 0.2s = 0.08
=> s = 0.08/0.2
=> s = 0.4 m
Answer:
The cost of energy is $ 0.34.
Explanation:
The energy is the capacity to do work.
The energy is a scalar quantity and its SI unit is Joule.
The commercial unit of energy is kWh.
Cost of 1 kWh energy = $ 0.17
energy loss by standard window is 2 kWh .
So, the cost of lost of energy is
Cost = $ 0.17 x 2 = $ 0.34