Ikes teacher tells him that this process has caused each strip of tape to gain a negative electric charge. What could ike do nex
1 answer:
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Answer:
0.107 m
Explanation:
The magnetic field at the center of a current-carrying loop is given by
where
is the vacuum permeability
I is the current
r is the radius of the loop
In this problem we have
I = 3.40 A is the current in the loop
is the magnetic field at the centre of the loop
So, solving the formula for r we find
To find the magnitude and direction of the electric field, let us find the horizontal and vertical components of the field separately, then we will use those values to calculate the total magnitude and direction.
The tension in the thread is 6.57×10⁻²N and the thread is aligned horizontally, so the tension force is directed entirely horizontally. The sphere is in static equilibrium, therefore the horizontal component of the electrostatic force acting on the sphere, Fx, must act in the opposite direction of the tension and have a magnitude of 6.57×10⁻²N. We know this equation relating a charge, an electric field, and the force that the field exerts on the charge:
F = Eq
F is the electric force, E is the electric field, and q is the charge
Let us adjust the equation for only the horizontal components of the above quantities:
Fx = (Ex)(q)
Fx is the horizontal component of the electric force and Ex is the horizontal component of the electric field.
Given values:
F = 6.57×10⁻²N
q = 6.80×10³C
Plug in these values and solve for Ex:
6.57x10⁻² = Ex(6.80×10³)
<u>Ex = 9.66×10⁻⁶N/C</u>
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Since the sphere is in static equilibrium, the vertical component of the electrostatic force acting on the sphere, Fy, must have the same magnitude and act in the opposite direction of the sphere's weight. If we assume the weight to act downwards, then Fy must act upward.
We know the weight of the sphere is given by:
W = mg
W is the weight, m is the mass, and g is the acceleration of objects due to earth's gravity field near its surface.
We also know this equation:
F = Eq
Let us adjust for the vertical components:
Fy = (Ey)(q)
Set Fy equal to W and we get:
(Ey)(q) = mg
Given values:
q = 6.80×10³C
m = 0.018kg
g = 9.81m/s²
Plug in the values and solve for Ey:
(Ey)(6.80×10³) = 0.018(9.81)
<u>Ey = 2.60×10⁻⁵N/C</u>
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Let's now use the Pythagorean theorem to find the total magnitude of the electric field:
E =
E = 2.77×10⁻⁵N/C
The direction of the electric field is given by:
θ = tan⁻¹(Ey/Ex)
θ = 20.4° off the horizontal
Answer:
The answer to the question is;
The required torque that it would take to cause the gyroscopes to precess through an angle of 1.0×10−6 degree during a 5.0-hour exposure of a galaxy is 2.44 ×10⁻¹² N·m
Explanation:
To solve the question we first resolve the units of the given quantities as follows
The gyroscopes spin at 19,200 rpm that is 19,200 revolutions per minute
1 revolution = 2π rad and
1 minute = 60 seconds
Therefore 19,200 revolutions per minute = 2π×19,200÷60 rad/s
= 2010.619 rad/s
The angle of precess is given as 1.0×10⁻⁶ °. We convert the angle to radians as follows
360 ° = 2π radians
1 ° = radians and
1.0×10⁻⁶ ° = radians × 1.0×10⁻⁶ ° = 1.745×10⁻⁸ rad
To find the torque we note that the torque is given by
Precession angular speed × The moment of inertia × angular velocity
The precession angular speed is given by
The precession angle was determined in rad as 1.745×10⁻⁸ rad
The precession time is 5 hours which is equal to 5×60×60 = 18000 s
Therefore the precession velocity = = 9.696×10⁻¹³ rad/s
The moment of inertia is given by
Formula for the moment of inertia of a thin walled cylinder I = m·r²
Where:
r = Radius of the gyroscope = Diameter/2 = 5.0 cm/2 = 2.5 cm = 0.025 m
m = Mass of each gyroscopes = 2.0 kg
Therefore I = m·r² = 2.0 kg × (0.025 m)² = 0.00125 kg·m²
Torque, τ = Ω·I·ω
Where:
Ω = Precession velocity = 9.696×10⁻¹³ rad/s
I = Moment of inertia = 0.00125 kg·m²
ω = Angular speed = 2010.619 rad/s
τ = 9.696×10⁻¹³ rad/s × 0.00125 kg·m² × 2010.619 rad/s =
2.44 ×10⁻¹² kg·m²/ s² = 2.44 ×10⁻¹² N·m
The required torque is 2.44 ×10⁻¹² N·m.