All categories

3 years ago

Compare the way light passes through convex and concave lenses.

1 answer:

3 0

When light passes through concave lenses the refracted rays diverge so that they appear to come from one point called the principal focus.
Light that passes through a convex lens can be focused on a surface like a screen.

You might be interested in

A total electric charge of 3.50 nC is distributed uniformly over the surface of a metal sphere with a radius of 24.0 cm. If the

djyliett [7]

Answer:

a, V = 65.6V

b. V = 131.0V

C. V = 131.0V

Explanation:

Given

Total electric charge $q= 3.50nC=3.50\times10^{-9}C$ distributed uniform at surface of metal sphere where radius is R = 24cm

R = 24cm = 0.024m

we have to find potential at different distance

a. r = 48cm = 0.048m

b. r = 24cm = 0.024m

c. r = 12cm = 0.012m

a. solution

Here the distance r is grater then R,we can get the potential outside of sphere

$V = \dfrac{q}{4\pi\varepsilon_0r}$

$\textrm{where is} \dfrac{1}{4\pi\varepsilon_0}=9.0\times10^{9}N.m^{2}/c^2$

now we plug the value q and r in potential v

$V=(9\times10^9N.m^2/C^2)\dfrac{3.50\times10^{-9}C}{0.48m}$

=65.6V

b.solution

Here distane r is the same from radius R, so we can get potential inside sphere

$v= \dfrac{q}{4\pi\varepsilon_0R}$

Now we plug the value of R and q

$V=(9\times10^9N.m^2/C^2)\dfrac{3.50\times10^{-9}C}{0.24m}$

$V=131V$

3.solution

Inside sphere electric feild is zero.so potential is the same every point inside sphere and equal potential on the surface

$V=131V$

5 0

3 years ago

What is the force between two negatively charged objects if q1 = 2.5x10^-6 C, q2 = 4.3x10-6 C and the distance between the charg

aliina [53]

The force between the charges is 60.47N

Explanation:

Given:

Charge, q₁ = 2.5 X 10⁻⁶ C

q₂ = 4.3 X 10⁻⁶ C

Distance, r = 0.04m

Force, F = ?

According to Coulomb's law:

$F = k\frac{q_1q_2}{r^2}$

where,

k = Coulomb's constant

and the value of k is 9 X 10⁹ Nm²/C²

On substituting the value we get:

$F = 9 X 10^9X\frac{2.5 X 10^-^6 X 4.3 X 10^-^6}{(0.04)^2} \\\\F = 60.47N$

Therefore, the force between the charges is 60.47N

6 0

3 years ago

A ball of a mass 0.3 kg is released from rest at a height of 8 m. How fast is it going when it hits the ground? Acceleration due

Vaselesa [24]

In order to solve this problem, there are two equations that you need to know to solve this problem and pretty much all of kinematics. The first is that d=0.5at^2 (d=vertical distance, a=acceleration due to gravity and t=time). The second is vf-vi=at (vf=final velocity, vi=initial velocity, a=acceleration due to gravity, t=time). So to find the time that the ball traveled, isolate the t-variable from the d=0.5at^2. Isolate the t and the equation now becomes $\sqrt{(2d)/a}$ . Solving the equation where d=8 and a=9.8 makes the time $\sqrt{(2*8)/9.8}$ =1.355 seconds. With the second equation, the vi=0 m/s, the vf is unknown, a=9.8 m/s^2 and t=1.355 sec. Substitute all these values into the equation vf-vi=at, this makes it vf-0=9.8(1.355). This means that the vf=13.28 m/s.