Positioning your Slinky along any direction different from its initial position will affect your reading, because there will be change in the magnetic field.
<h3>Effect of magnet on Slinky</h3>
If the Slinky is made of an iron alloy, it can be magnetized by itself. Moving the Slinky around can cause a change in the magnetic field, even if no current is flowing.
When there is a change in the magnetic field, the reading changes.
At any point, you change the orientation of the Slinky, you will need to zero the reading or adjust the Slinky back to its initial position, even if the sensor does not move.
Thus, Positioning your Slinky along any direction that is different to its initial position will affect your reading because there will be change in the magnetic field.
Learn more about magnetic field here: brainly.com/question/7802337
If you were to sit a hot cup of water out side it would frezze faster
The answer is B) <span>equilibrium
hope this helps!=-)</span>
This question is not complete.
The complete question is as follows:
One problem for humans living in outer space is that they are apparently weightless. One way around this problem is to design a space station that spins about its center at a constant rate. This creates “artificial gravity” at the outside rim of the station. (a) If the diameter of the space station is 800 m, how many revolutions per minute are needed for the “artificial gravity” acceleration to be 9.80m/s2?
Explanation:
a. Using the expression;
T = 2π√R/g
where R = radius of the space = diameter/2
R = 800/2 = 400m
g= acceleration due to gravity = 9.8m/s^2
1/T = number of revolutions per second
T = 2π√R/g
T = 2 x 3.14 x √400/9.8
T = 6.28 x 6.39 = 40.13
1/T = 1/40.13 = 0.025 x 60 = 1.5 revolution/minute
Answer:
The charge on the third object is − 21.7nC
Explanation:
From Gauss's Law
Φ = Q/ε₀
where;
Φ is the total electric flux through the shell = − 533 N⋅m²/C
Q is the total charge Q in the shell = ?
ε₀ is the permittivity of free space = 8.85 x 10⁻¹²
From this equation; Φ = Q/ε₀
Q = Φ * ε₀ = − 533 * 8.85 x 10⁻¹²
Q = −4.7 X 10⁻⁹ C = -4.7nC
Q = q₁ + q₂ + q₃
− 4.7nC = − 14.0 nC + 31.0 nC + q₃
− 4.7nC − 17nC = q₃
− 21.7nC = q₃
Therefore, the charge on the third object is − 21.7nC