What happens to end a of the rod when the ball approaches it closely this first time is; It is strongly attracted.
<h3>Electrostatics</h3>
I have attached the image of the rod.
We are told that the ball is much closer to the end of the rod than the length of the rod. Thus, if we point down the rod several times, the distance of approach will experience no electric field and as such the charge on end point A of the rod must be comparable in magnitude to the charge on the ball.
This means that their fields will cancel.
Finally, we can conclude that when a charge is brought close to a conductor, the opposite charges will all navigate to the point that is closest to the charge and as a result, a strong attraction will be created.
This also applies to a strong conducting rod and therefore it is strongly attracted.
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Answer:
The electric field due to the right ring at a location midway between the two rings is 
Explanation:
Given that,
Radius of first ring = 5 cm
Radius of second ring = 20 cm
Charge on the left of the ring = +30 nC
Charge on the right of the ring = -30 nC
We need to calculate the electric field due to the right ring at a location midway between the two rings
Using formula of electric field
Put the value into the formula
Hence, The electric field due to the right ring at a location midway between the two rings is
B. Parabolic
This is because an object that is in projectile motion has gravity acting on it.
Answer:
Explanation:
Maximum vertical distance or height = h = 35.4 m
let's consider the initial speed at the top is zero.
As the roller coaster is coming from top to bottom there is the conversion of gravitational potential energy into kinetic energy. So we will consider the law of conservation of energy.
As in this case,
Loss in potential energy = Gain in Kinetic energy
mgh = 1/2mv²
mass will cancel out will mass.
gh = 1/2 v²
v = √2gh
v = √2×9.8×35.4
v =√693.84
v = 26.34 m/s
The rollar coaster will have the maximum speed of 26.34 m/s when it reaches the bottom if we ignore the frictional forces.
Answer:
4.92°
Explanation:
The banking angle θ = tan⁻¹(v²/rg) where v = designated speed of ramp = 30 mph = 30 × 1609 m/3600 s = 13.41 m/s, r = radius of curve = 700 ft = 700 × 0.3048 m = 213.36 m and g = acceleration due to gravity = 9.8 m/s²
Substituting the variables into the equation, we have
θ = tan⁻¹(v²/rg)
= tan⁻¹((13.41 m/s)²/[213.36 m × 9.8 m/s²])
= tan⁻¹((179.8281 m²/s)²/[2090.928 m²/s²])
= tan⁻¹(0.086)
= 4.92°
