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3 years ago

11

Why the efficiency of normal light is smaller than solar light ??

1 answer:

stellarik [79]3 years ago

8 0

Answer:

part of energy is wasted in heat because of resistance in the filament (and that's how it glows)

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Which geologic feature most likely be represented by contour lines drawn far apart from one another?

andrey2020 [161]

Answer;

A plain

Explanation;

A contour line connects points of the same elevation. Contour lines are usually curves. Closed contours represent hills.

Contour lines can not cross since they represent different elevations. A contour interval is the difference in elevation between one contour and an adjacent contour.

5 0

3 years ago

In what way are mercury and venus similar ?

Neporo4naja [7]

Answer:

they are both planets they are both made of rock and metel

Explanation:

3 0

3 years ago

Read 2 more answers

What is physics? In your own words too

denis23 [38]

The study of how the world works

4 0

3 years ago

When an excited electron falls from n = 3 to n = 2, a red radiation in the Balmer series emitted. What is the frequency of this

vichka [17]

Answer:

The frequency of this photon is $4.57\times10^{14}\ Hz$

(D) is correct option.

Explanation:

Given that,

Excited states,

$n=3$

$n=2$

We need to calculate the wavelength

Using formula for energy

$E=13.6(\dfrac{1}{n^2}-\dfrac{1}{m^2})$

$E=13.6(\dfrac{1}{4}-\dfrac{1}{9})$

$E=1.888\ eV$

$E=1.888\times1.6\times10^{-19}\ J$

$E=3.0208\times10^{-19}\ J$

We need to calculate the frequency

Using formula of frequency

$E=hf$

$f=\dfrac{E}{h}$

Where, E =energy

$f=\dfrac{3.0208\times10^{-19}}{6.63\times10^{-34}}$

$f=4.57\times10^{14}\ Hz$

Hence, The frequency of this photon is $4.57\times10^{14}\ Hz$

4 0

3 years ago

Suppose the kicker launches the ball at 60∘ instead of 30∘. Assuming that the goal is 4.55 m high and 40 m away, what minimum in

Nikitich [7]

Answer:22 m/s

Explanation:

Given

launch angle $\theta =60^{\circ}$

height of goal $h=4.55\ m$

and horizontal distance $x=40\ m$

Suppose initial speed is $u$

Trajectory of a Projectile is given by

$y=x\tan \theta -\frac{1}{2}\frac{gx^2}{u^2\cos ^2\theta }$

substituting the values we get

$4.55=40\tan (60)-0.5\times \frac{9.8\times (40)^2}{u^2\cdot \cos ^260 }$

$4.55=69.28-0.5\times \frac{15,680}{u^2\cdot 0.25}$

$\frac{31,360}{u^2}=69.28-4.55$

$\frac{31,360}{64.73}=u^2$

$u^2=484.47$

$u=22.01\ m/s$

So, initial launch speed is $22\ m/s$

4 0

3 years ago