Answer:
The frequency of the wave is 5 x 10⁹ Hz
Explanation:
Given;
wavelength of the radio wave, λ = 6.0 × 10⁻²m
radio wave is an example of electromagnetic wave, and electromagnetic waves travel with speed of light, which is equal to 3 x 10⁸ m/s².
Applying wave equation;
V = F λ
where;
V is the speed of the wave
F is the frequency of the wave
λ is the wavelength
Make F the subject of the formula
F = V / λ
F = (3 x 10⁸) / (6.0 × 10⁻²)
F = 5 x 10⁹ Hz
Therefore, the frequency of the wave is 5 x 10⁹ Hz
The rubber protects him from being electrocuted by the flow of current going through the plug.
Hope this helped!!
Answer:
Fx1 (6 m) sin 60 = 300 (3 m) cos 60 balancing torques about floor
Fx1 = 900 * 1/2 / 5.20 = 86.6 N this is the horizontal force that must be supplied by the wall to balance torques about the floor
This is also equal to the static force of friction that must be applied at the point of contact with the floor to balance forces in the x-direction.
Fx1 = Fx2 = 86.6 N
Answer:
Explanation:
The cannonball goes a horizontal distance of 275 m . It travels a vertical distance of 100 m
Time taken to cover vertical distance = t ,
Initial velocity u = 0
distance s = 100 m
acceleration a = 9.8 m /s²
s = ut + 1/2 g t²
100 = .5 x 9.8 x t²
t = 4.51 s
During this time it travels horizontally also uniformly so
horizontal velocity Vx = horizontal displacement / time
= 275 / 4.51 = 60.97 m /s
Vertical velocity Vy
Vy = u + gt
= 0 + 9.8 x 4.51
= 44.2 m /s
Resultant velocity
V = √ ( 44.2² + 60.97² )
= √ ( 1953.64 + 3717.34 )
= 75.3 m /s
Angle with horizontal Ф
TanФ = Vy / Vx
= 44.2 / 60.97
= .725
Ф = 36⁰ .