Answer:
Approximately 
(given that the magnitude of this charge is 
.)
Explanation:
If a charge of magnitude 
is placed in an electric field of magnitude 
, the magnitude of the electrostatic force on that charge would be 
.
The magnitude of this charge is 
. Apply the unit conversion 
:
.
An electric field of magnitude 
would exert on this charge a force with a magnitude of:
.
Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.
The force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is zero.
<h3>
Force required to pull one end at a constant speed</h3>
The force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is determined by applying Newton's second law of motion as shown below;
F = ma
where;
- m is mass
- a is acceleration
At a constant speed, the acceleration of the object will be zero.
F = m x 0
F = 0
Thus, the force required to pull one of the microscope sliding at a constant speed of 0.28 m/s relative to the other is zero.
Learn more about constant speed here: brainly.com/question/2681210
Answer:
8.874
Explanation:
You need to times 5.22 kg and 1.7 m/s to get 8.874.
The answer is asthenosphere
Answer:
600Hz
Explanation:
In electrical systems of alternating current, the harmonics are, as in acoustics, frequencies multiples of the fundamental working frequency of the system and whose amplitude decreases as the multiple increases. For example, if we have systems fed by the 50 Hz network, harmonics of 100, 150, 200, etc. may appear.
In our case having a fundamental wave of 100Hz, I can have harmonics of 200,300,400, ..., 600Hz
