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Vlad [161]
3 years ago
9

What are the features of image formed by convex mirror.​

Physics
2 answers:
faltersainse [42]3 years ago
5 0

Explanation:

the image is virtual and errect

All the distances are measured from the pole

Convex mirrors reflect light outwards (diverging light rays) and therefore they are not used to focus light. The image is virtual, erect and smaller in size than the object, but gets larger (maximum upto the size of the object) as the object comes towards the mirror.

STatiana [176]3 years ago
4 0
The image is always virtual and erect. The image is highly diminished or point sized. It is always formed between F and P.
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If a train travels 500 kilometers from Stockholm to
o-na [289]

Hello,

Average speed is total distance divided by total time. From the problem, our total distance is given as 500 kilometers and given time is 5 hours. Therefore, the average speed is:

\displaystyle{v_\text{average}=\sum_{i=1}^n \dfrac{s_i}{t_i}}\\\\\displaystyle{v=\dfrac{500\ \text{km}}{5 \ \text{h}}}\\\\\displaystyle{v=100 \ \text{km/h}}

Therefore, the average speed is 100 km/h. Please let me know if you have any questions!

4 0
11 months ago
A skier descends a mountain at an angle of 35.0º to the horizontal. If the mountain is 235 m long, what are the horizontal and v
notsponge [240]

d=235m

Alpha=35⁰

h=d×sin(alpha)=134.8m

L=d×cos(alpha)=192.5m

8 0
3 years ago
A 500-Ω resistor, an uncharged 1.50-μF capacitor, and a 6.16-V emf are connected in series. (a) What is the initial current? (b)
Alexeev081 [22]

Answer:

a) 0.01232 A

b) 0.00075 s = 0.75 ms

c) 0.0045323 A = 4.532 mA

d) 3.894 V

Explanation:

R = 500 Ω

V = 6.16 V

C = 1.50 μF

Let Vs be the voltage of the emf source

Let Vc be the voltage across the capacitor at any time

a) Current flows as a result of potential difference between two points. So, the current flows according to difference in voltage between the emf source and the capacitor.

At time t = 0,

There is no voltage on the capacitor; Vc = 0 V

Current in the circuit is given by

I = (Vs - Vc)/R

I = (6.16 - 0)/500

I = 0.01232 A

b) Time constant for an RC circuit is given by τ

τ = RC = (500) (1.5 × 10⁻⁶) = 0.00075 s

c) The current decay in an RC circuit (called decay because the current in the circuit starts to fall as the capacitor's voltage rises as the capacitor charges) is given by

I = I₀ e⁻ᵏᵗ

where k = (1/τ)

I₀ = Current in the circuit at t = 0 s; I₀ = 0.01232 A

At t = τ = 0.00075 s, kt = (τ/τ) = 1

I = 0.01232 e⁻¹ = 0.0045323 A = 4.532 mA

d) The voltage for a charging capacitor is given by

Vc = Vs (1 - e⁻ᵏᵗ)

where k = (1/τ)

At t = τ = 0.00075 s, Vc = ?, Vs = 6.16 V, kt = 1

Vc = 6.16 (1 - e⁻¹) = 6.16 (0.6321)

Vc = 3.894 V

4 0
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