A non-uniform rod 60cm long and weighs 32N is balanced at the 45cm mark. A load of 2N is hung on the zinc rod at the 25cm mark.
1 answer:
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Answer:
Explanation:
Given
Em wave is in the form of
where
Wave constant for EM wave k is
Wavelength of wave
Answer:
v_wind = 101.46 km / h , θ = 61.8
Explanation:
This is a velocity composition exercise.
Let's do the problem in parts. Let's start by knowing the speed of the plane without air.
v = d / t
v = 750 / 3.14
v = 238.85 km / h
This is the speed of the plane relative to the Earth and it does not change.
In the second part, when there is wind, the travel time is greater than when there is no wind, therefore the wind delays the plane. To be more general, suppose that the wind has two components vₓ and
Let's use trigonometry to find the components of the plane's speed
cos θ = v_N / v
sin θ = v_W / v
v_N = v cos θ
v_W = v sin θ
let's calculate
V _N = 238.85 cos 22 = 221.46 km / h
v_W = -238.85 sin 22 = -89.47
the negative sign is because the plane is going west and the positive sign is the east direction.
As it indicates that the destination of the avine is towards the north, the x component of the wind must be
vₓ - v_W = 0
vₓ = v-w
vₓ = 89.47 km / h
in the direction to the East.
Now let's analyze the component of the wind in the Nort-South direction,
Indicate the travel time, let's calculate the speed that the component must have the speed of the plane
v_total = d / t
v_total = 750 / 4.32
v_total = 173.61 km / h
This is the final speed of the plane, which can be written
v_total = v_n - vy
vy = v_n - v_total
vy = 221.46 - 173.61
vy = 47.85 km
this component is directed towards the south
Let's use the Pythagorean Theorem, to find the magnitude
v_wind² = vₓ² + vy²
v_wind = √ (89.47² + 47.85²)
v_wind = 101.46 km / h
the address will then be found using trigonometry
θ = Vy / vx
θ = tan⁻¹ (vy / vx)
θ = tan⁻¹1 (47.85 / 89.47)
θ = 28.14
Therefore, the magnitude of the wind speed is 101.5 km / h and its direction is 28º south of the East, to give this value
90- θtea = 90- 28.2
θ = 61.8
East of South
Answer:
2.7x10⁻⁸ N/m²
Explanation:
Since the piece of cardboard absorbs totally the light, the radiation pressure can be found using the following equation:
<u>Where:</u>
: is the radiation pressure
I: is the intensity of the light = 8.1 W/m²
c: is the speed of light = 3.00x10⁸ m/s
Hence, the radiation pressure is:
Therefore, the radiation pressure that is produced on the cardboard by the light is 2.7x10⁻⁸ N/m².
I hope it helps you!