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3 years ago

7

Microwave ovens emit microwave energy with a wavelength of 12.1 cm. What is the energy of exactly one photon of this microwave r

adiation?

1 answer:

EastWind [94]3 years ago

3 0

Answer:

$1.64\cdot 10^{-24}J$

Explanation:

The energy of a photon is given by:

$E=\frac{hc}{\lambda}$

where

$h=6.63\cdot 10^{-34} Js$ is the Planck constant

$c=3\cdot 10^8 m/s$ is the speed of light

$\lambda$ is the wavelenght of the photon

For the microwave photons in this problem,

$\lambda=12.1 cm=0.121 m$

so their energy is

$E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{0.121 m}=1.64\cdot 10^{-24}J$

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Lady bird [3.3K]

I have all the answers here so take this

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2 years ago

If an object is to rest on an incline without slipping, then friction must equal the component of the weight of the object paral

leonid [27]

Answer:

$\theta = tan^{-1}\mu$

Explanation:

As we know that if the object is placed on the inclined plane then the force of friction on the object is counterbalanced by the component of the weight of the object along the inclined plane.

So we can say

$F_f = mgsin\theta$

now if we increase the inclination of the plane then the component of the weight weight along the inclined plane will increase and hence the friction force will also increase.

As we know that the limiting value or the maximum value of friction force at the static condition is given by

$F_f = \mu N$

$N = mg cos\theta$

so we have

$F_f = \mu (mg cos\theta)$

so we will have

$mg sin\theta = \mu (mg cos\theta)$

so now we have

$tan\theta = \mu$

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3 0

3 years ago

Try this out. Use a calculator!

Maslowich

Just subsitute and easy
v=55m/s
m=100kg
KE=(0.5)(100kg)(55m/s)^2
KE=(50kg)(3025 m^2/s^2)
KE=151250 J
2nd option

6 0

3 years ago

A block is pulled along a horizontal surface at a constant speed by a force (14.1i 0 j 5.1k). The direction k is perpendicular t

Pepsi [2]

Answer:

$W =84.6\ Nm$

Explanation:

given,

F = 14.1 i + 0 j + 5.1 k

displacement = 6 m

Assuming block is moving in x- direction

we know,

dW = F dx

$\int dW = F\int dx$

$W = F\int_0^6 dx$

$W = F[x]_0^6$

$W = 14.1 \times 6$

$W =84.6\ Nm$

hence, work done by the force is equal to $W =84.6\ Nm$

7 0

3 years ago

Nucleus A decays into the stable nucleus B with a half-life of 22.07 s. At t=0 s there are 1,293 A nuclei and no B nuclei. At wh

Alexeev081 [22]

Answer:

29.38 seconds

Explanation:

Half life, T = 22.07 s

No = 1293

Let N be the number of atoms left after time t

N = 1293 - 779 = 514

By the use of law of radioactivity

$N=N_{0}e^{-\lambda t}$

Where, λ is the decay constant

λ = 0.6931 / T = 0.6931 / 22.07 = 0.0314 decay per second

so,

$514=1293e^{-0.0314t}$

$2.5155=e^{0.0314t}$

take natural log on both the sides

0.9225 = 0.0314 t

t = 29.38 seconds

5 0

3 years ago