We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is
Where,
F_D = Drag Force
= Drag coefficient
A = Area
= Density
V = Velocity
Our values are given by,
(That is proper of a cone-shape)
Part A ) Replacing our values,
Part B ) To find the torque we apply the equation as follow,
Answer:
A. Doubles.
Explanation:
In an electromagnetic device such as a generator, when a wire (conductor) moves through the magnetic field between the South and North poles of a magnet, an electromotive force (e.m.f) is usually induced across a wire
The mode of operation of a generator is that a metal core with copper tightly wound to it (conductor coil) rotates rapidly between the two (2) poles of a horseshoe magnet type. Thus when the conductor coil rotates rapidly, it cuts the magnetic field existing between the poles of the horseshoe magnet and then induces the flow of current.
When a high-resistance voltmeter is connected to an electric circuit, a deflection will arise due to the flow of electricity. Moving the magnet towards the coil of wire will cause the needle of the high-resistance voltmeter to move in one direction. Also, as the magnet is moved out from the coil of wire, the needle of the high-resistance voltmeter moves in the opposite direction.
In this scenario, a magnet is moved in and out of a coil of wire connected to a high-resistance voltmeter. If the number of coils doubles, the induced voltage doubles because the number of turns (voltage) in the primary winding is directly proportional to the number of turns (voltage) in the secondary winding.
Answer:
C) 7.35*10⁶ N/C radially outward
Explanation:
- If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
- So, we can write the following equation:
- As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
- So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
- So, on the outer surface of the shell there must be a charge that be the difference between them:
- Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:
- As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.
Answer: D
All the particles must be uncharged
Explanation:
If all the particles are positively charged, then there will be force of repulsion between them which will give different directions away from each other. The same is applicable if they are all negatively charged.
If the particles are positively and negatively charged, their will be force of attraction between them which will give different directions towards each other.
For all to be experiencing forces in the same direction, We can conclude that
All the particles must be uncharged.
Answer:
The velocity of the truck after this elastic collision is 15.7 m/s
Explanation:
It is given that,
Mass of the car, 
Mass of the truck, 
Initial velocity of the car, 
Initial velocity of the truck, u₂ = 0
After the collision the velocity of the car is, v₁ = -11 m/s
Let v₂ is the velocity of the truck after this elastic collision. Using the conservation of momentum as :
So, the velocity of the truck after this elastic collision is 15.7 m/s. Hence, the correct option is (c).
