Answer:
okay sooo the weight is: 294 n
the normal force is 286 n
the acceleration is: -0.38 m/s²
Answer:
5525 N/C
Explanation:
Magnitude of electric field ( E ) = 3500 N/c
Direction of electric field : positive X axis
point charge ( q ) = -9.0 * 10^-9
<u>Calculate the Magnitude of the net electric field at (a) x = -0.20 m</u>
Magnitude = 5525 N/C
Electric field due to q = ( 9 * 10^9 * 9 * 10^-9 ) / ( -0.2 )^2
= 81 / 0.04 = 2025 N/c
<em>Therefore the magnitude of the net electric field </em>
= 2025 + 3500
= 5525 N/C
Answer:
0.000314 Am²
6.049*10^-7 T
Explanation:
A
From the definitions of magnetic dipole moment, we can establish that
= , where
= the magnetic dipole moment in itself
= Current, 100 A
= Area, πr² (r = diameter divided by 2). Converting to m², we have 0.000001 m²
On solving, we have
= ,
= 100 * 3.14 * 0.000001
= 0.000314 Am²
B
=
(0)/4
* 2
/
³, where
(0) = constant of permeability = 1.256*10^-6
z = 4.7 cm = 0.047 m
B = 1.256*10^-6 / 4*3.142 * [2 * 0.000314/0.047³]
B = 1*10^-7 * 0.000628/1.038*10^-4
B = 1*10^-7 * 6.049
B = 6.049*10^-7 T
Answer:

Explanation:
#First, we need to determine the actual emf required. The generator's internal resistance will cause a voltage drop inside the generator/
Internal resistance is defined using the formula:

#The bulb is rated 12.0V,25.0W
Current, 
Therefore, the voltage drop in the generator is calculated as:
Actual EMF required is thus 1.536V+12.0V=13.536V
#peak voltage is 
#For a generator, by Faraday's Law


#The rate of the generator is 25.39Hz