Answer:
Explanation:
Given that,
Emf, V = 22 mV
Number of turns in the coil us 519
Rate of change of current is 10 A/s.
We need to find the magnetic flux through each turn of the coil at an instant when the current is 3.70 A.
Let's find the inductance first. So,
We have,
, 
is magnetic flux
So, the magnetic flux is 
.
Yes that is correct. We know this because 4.00 x 10 4 Pa is constant. If you have 2.00×10−3m3 then you do the following: (2.00×10^−3)(4.00×10^<span> 4) = </span>8.00×10^−3. That is how you get your answer
Answer:
480
Explanation:
resistance equals to potential difference divide by electric current
120÷0.25
=480
Answer:
41.74 m/s
Explanation:
The energy used to draw the bowstring = the kinetic energy of the arrow.
Fd = 1/2mv²................................ Equation 1
Where F = force, d = distance move string, m = mass of the arrow, v = speed of the arrow.
make v the subject of the equation
v = √(2Fd/m)...................... Equation 2
Given: F = 201 N, m = 0.3 kg, d = 1.3 m.
Substitute into equation 2
v = √(2×201×1.3/0.3)
v = √(1742)
v = 41.74 m/s.
Hence the arrow leave the bow with a speed of 41.74 m/s
Answer:
The only parameter that changes is mass m
It is only necessary to calculate the ratio Eh/Ee
The kinetic energy of the heavy paricle is three times the kinetic energy of an electron
