A ball dropped by melissa is falling downward toward the floor
1 answer:
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Answer:
D) Image cannot be seen
Explanation:
As we know by lens formula
now we know that
object is placed at the focal point of lens
so we will have
so the image will form at infinite
so image will be at infinite distance from the lens
so it cannot be seen
Answer:
<em>The car will be moving at 5.48 m/s at the bottom of the hill</em>
Explanation:
<u>Principle of Conservation of Mechanical Energy</u>
In the absence of friction, the total mechanical energy is conserved. That means that
is constant, being U the potential energy and K the kinetic energy
U=mgh
When the car is at the top of the hill, its speed is 0, but its height h should be enough to produce the needed speed v down the hill.
The Kinetic energy is then, zero. When the car gets enough speed we assume it is achieved at ground level, so the potential energy runs out to zero but the Kinetic is at max. So the initial potential energy is transformed into kinetic energy.
We are given the initial potential energy U=45 J. It all is transformed to kinetic energy at the bottom of the hill, thus:
Multiplying by 2:
Dividing by m:
Taking square roots:
v = 5.48 m/s
The car will be moving at 5.48 m/s at the bottom of the hill
Answer:
(a) Charge density σ=6.6375×10²nC/m²
(b) Total charge Q=1.47×10²nC
Explanation:
Given Data
A=47.0 cm =0.47 m
Electric field E=75.0 kN/C
To find
(a) Charge density σ
(b)Total Charge Q
Solution
For (a) charge density σ
From Gauss Law we know that
Φ=Q/ε₀.......eq(i)
Where
Φ is electric flux
Q is charge
ε₀ is permittivity of space
And from the definition of flux
Φ = EA
The flux is electric field passing perpendicularly through the surface
Put the this Φ in equation(i)
EA =Q/ε₀
where Q(charge)=σA
EA=(σA)/ε₀
E=σ/ε₀
σ=ε₀E
σ=6.6375×10²nC/m²
For (b) total charge Q
Q=σA