Answer:
b)
Explanation:
If the charge is released at rest in an electric field, it will move along the electric field, going to regions of higher electric potential if it is a negative charge (against the field direction) and towards lower potential regions if it is positive (along the field). This means that the charge will gain kinetic energy, energy that only can come from a decrease in the electric potential energy.
For a positive charge: ΔEp = q*ΔV < 0 (as ΔV < 0)
For a negative charge: ΔEp = (-q) *ΔV < 0 (as ΔV > 0)
on touching electroscope gets positively charged, so answer is B. conduction
The required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ
First, we must understand that the component of the velocity along the vertical is due to maximum height achieved and expressed as usin
θ.
The component of the velocity along the horizontal is due to the range of the object and is expressed as ucosθ.
If the <u>air resistance is ignored</u>, the velocity of the object will be constant throughout the flight and the initial velocity will be equal to the final velocity.
Hence the required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ
Learn more here; brainly.com/question/12870645
Answer:
f = 5 cm
Explanation:
using the thin lens equation, given as follows:
where,
f = focal length = ?
do = the distance of object from lens = 20 cm
di = the distance of image from lens = 6.6667 cm
Therefore,
<u>f = 5 cm</u>
