3 years ago

PLEASE HELP ME!!! THIS IS DO TODAY. AND NO LINKS PLEASE!!!!!

Physics

1 answer:

AnnZ [28]3 years ago

3 0

Answer:

to late

Explanation:

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The 50N group of force would be greater
So whichever object is being pulled will be pulled towards the 50N force

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A heat lamp produces 32.8 watts of power at a wavelength of 6.5 μm. How many photons are emitted per second?

Svet_ta [14]

Answer:

$Photon/sec=1.07\times 10^{21}$

Explanation:

The expression for the power is:-

$Power=\frac{Energy}{Time}$

Also, $E=n\times \frac{h\times c}{\lambda}$

Where,

n is the number of photons

h is Plank's constant having value $6.626\times 10^{-34}\ Js$

c is the speed of light having value $3\times 10^8\ m/s$

$\lambda$ is the wavelength of the light

So,

$Power=\frac{n}{Time}\times \frac{h\times c}{\lambda}$

Thus, the expression for photons per second is:-

$\frac{n}{Time}=\frac{Power\times \lambda}{h\times c}$

Given that:-

Power = 32.8 Watt

$\lambda=6.5\ \mu m=6.5\times 10^{-6}\ m$

So,

$Photon/sec=\frac{32.8\times 6.5\times 10^{-6}}{6.626\times 10^{-34}\times 3\times 10^8}=1.07\times 10^{21}$

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Hi there!

We can begin by solving for the pendulum's velocity at the bottom of its trajectory using the work-energy theorem.

Recall:
$E_i = E_f$

Initially, we just have Potential Energy. At the bottom, there is just Kinetic Energy.

$PE = KE\\\\$

Working equation:
$\large\boxed{mgh = \frac{1}{2}mv^2}$

Rearrange to solve for velocity:
$gh = \frac{1}{2}v^2\\\\v = \sqrt{2gh}\\\\v = \sqrt{2(9.8)(1.5)} = 5.42 \frac{m}{s}$

Now, we can do a summation of forces:
$\Sigma F = T - W$

The net force is the centripetal force, so:
$\frac{mv^2}{r} = T - W$

Rearrange to solve for tension:
$T = \frac{mv^2}{r} + W\\\\T = \frac{5(5.42^2)}{4} + 5(9.8) = \boxed{85.75 N}$

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