All categories

3 years ago

1.Mention two uses of the concave mirror.

Physics

1 answer:

Anuta_ua [19.1K]3 years ago

8 0

Answer:

1. telescope

$f = \frac{r}{2}$

$f = \frac{r}{2}$ f- focal length

$f = \frac{r}{2}$ f- focal length r- the radius of curvature of the mirror

$\frac{1}{f} = \frac{1}{p} + \frac{1}{l}$

p-the distance of the object from the vertex of the mirror

l-the distance of the figure from the vertex of the mirror

You might be interested in

Particles q1 = +8.0 UC, 92 = +3.5 uc, and

Olin [163]

Answer:

F_total = 29.4 N, directed to the right of particle 2

Explanation:

We must solve this problem in parts, first we calculate each force and then we apply Newton's law to add the forces.

Let's use Coulomb's law to calculate each force

F = $k \frac{q_1q_2}{r_{12}^2}$

particles 1 and 2

q₁ = 8.0 10⁻⁶ C, q₂ = 3.5 10⁻⁶ C x₁₂ = 0.10 m

F₁₂ = 9 10⁹ 8.0 3.5 10⁻¹² / 0.1²

F₁₂ = 2.59 10¹ N

Since the two charges are of the same sign, this force is repulsive and is directed towards the positive side of the x axis.

particles 2 and 3

q₂ = 3.6 10⁻⁶ C, q₃ = 2.5 10⁻⁶ C, x₂₃ = 0.15 m

we calculate

F₂₃ = 9 10⁹ 3.5 2.5 10⁻¹²/ 0.15²

F₂₃ = 3.5 N

as the charge is of different sign, the force is attractive, therefore it is directed to the right of the load 2

Now we add the forces as vectors

F_total = ∑ F = F₁₂ + F₂₃

F_total = 25.2 +3.5

F_total = 29.4 N

directed to the right of particle 2

5 0

3 years ago

Read 2 more answers

What is the biggest star in our universe?

oksano4ka [1.4K]

<span>VY Canis Majoris is the largest star ever discovered.</span>

7 0

3 years ago

A 20~\mu F20 μF capacitor has previously charged up to contain a total charge of Q = 100~\mu CQ=100 μC on it. The capacitor is t

sertanlavr [38]

Explanation:

The given data is as follows.

C = $20 \times 10^{-6} F$

R = $100 \times 10^{3}$ ohm

$Q_{o} = 100 \times 10^{-6}$ C

Q = $13.5 \times 10^{-6} C$

Formula to calculate the time is as follows.

$Q_{t} = Q_{o} [e^{\frac{-t}{\tau}]$

$13.5 \times 10^{-6} = 100 \times 10^{-6} [e^{\frac{-t}{2}}]$

0.135 = $e^{\frac{-t}{2}}$

$e^{\frac{t}{2}} = \frac{1}{0.135}$

= 7.407

$\frac{t}{2} = ln (7.407)$

t = 4.00 s

Therefore, we can conclude that time after the resistor is connected will the capacitor is 4.0 sec.

4 0

3 years ago

Does any one know the answer

valina [46]

<span>Fossils are the preserved remains or traces of animals, plants, and other organisms from the remote past. The totality of fossils, both discovered and undiscovered, and their placement in fossiliferous (fossil-containing) rock formations and sedimentary layers (strata) is known as the fossil record</span>

6 0

3 years ago

a current of 180 mini amphere passes through a conductor for 5minute calculate the quantity of electricity transported

oksano4ka [1.4K]

Answer:

Explanation:

You can calculate the total electric charge that passes through the conductor as $q=It=(180\times 10^{-3})(5\times 60)= 54 C$ . It means that the number of electron that passes through the conductor is:

$n=\frac{q}{e}=\frac{54}{1.6\times 10^{-19}}=33.75\times 10^{19}$

8 0

2 years ago