Answer:
The minimum frequency of the coil is 7.1 Hz
Explanation:
Given;
number of turns, N = 200 turns
cross sectional area, A = 300 cm² = 300 x 10⁻⁴ m²
magnitude of magnetic field strength, B = 30 x 10⁻³ T
maximum value of the induced emf, E = 8 V
Maximum induced emf is given as;
E = NBAω
where
ω is angular velocity (ω = 2πf)
E = NBA2πf
where;
f is the minimum frequency, measured in hertz (Hz)
f = E / (NBA2π)
f = 8 / (200 x 30 x 10⁻³ x 300 x 10⁻⁴ x 2 x 3.142)
f = 7.073 Hz
f = 7.1 Hz
Therefore, the minimum frequency of the coil is 7.1 Hz
Answer: true
Explanation: this is were it is located
A. 0.5kg
To get this answer you need to follow the equation of KE=0.5*mv^2
But we don't have the m part in the equation. So just plug in the numbers to see which works best, though I can tell you before we do that the answer would be a.
As you may know, gravity, is a force of 9.8 m/s. And we want to get 9.8 Joules. So if we take a half a kg stone, release it at one meter, we get half of the normal gravity pull, 4.90 Joules. That means if we take half a kg stone and drop it at a doubled height, we get 9.8 Joules.
That is also to say that if we have a 1kg stone and drop it at one meter you will get the normal pull of gravity in Joules, 9.8J.
Be careful though, this does not mean if you drop a 1kg stone and a .5 kg stone the 1kg will hit first. This simply means that the 1kg stone will have twice the Joules that the .5kg stone has.
Answer:
<h2>0.67 m/s²</h2>
Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
f is the force
m is the mass
From the question we have
We have the final answer as
<h3>0.67 m/s²</h3>
Hope this helps you
firstly you get your acceleration with the formula, a=v-u/t. Then you use the formula for kinetic energy 1/2mv^2
then you can finally get the answer for power by dividing your previous answer by the time
