Use Hooke's Law to determine the variable force in the spring problem. A force of 450 newtons stretches a spring 30 centimeters.
1 answer:
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Answer:
Explanation:
Given that,
Frequency of a radio antenna is 1 MHz
Power, P = 21 kW
We need to find the the waves intensity 25 km from the antenna . The object emits intenisty evenly in all direction. It can be given by :
So, the wave intensity 25 km from the antenna is .
Answer:
Explanation:
There is electric field between the plates whose value is given by the following expression
electric field E = V /d where V is potential between the plates and d is distance between them
E = 300 / 5 x 10⁻³
= 60 x 10³ N/c
Force on electron = q E where q is charge on the electron
F = 1.6 X 10⁻¹⁹ X 60 X 10³ = 96 X 10⁻¹⁶ N.
Acceleration a = force / mass
a = 96 x 10⁻¹⁶/ mass = 96 x 10⁻¹⁶ / 9.1 x 10⁻³¹
= 10.55 x 10¹⁵ m / s²
For midway , distance travelled
s = 2.5 x 10⁻³ m
s = 1\2 a t²
t =
=
t = .474 x 10⁻¹⁸ s
For striking the plate time is calculated as follows
t = [/tex]
t = 0.67 x 10⁻¹⁸ s
Answer:
<h3>The answer is option B</h3>Explanation:
The wavelength of a wave can be found by using the formula
where
c is the speed of the wave
f is the frequency
From the question
c = 343 m/s
f = 466 Hz
We have
We have the final answer as
<h3>0.74 m</h3>Hope this helps you