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brilliants [131]
3 years ago
6

The angle between an incidentray and the mirror is 40°.1) What is the angle of reflection?​

Physics
1 answer:
Vlad1618 [11]3 years ago
5 0

Answer:

1) 50°

Explanation:

We need to find the angle of incidence first before finding the angle of reflection.

Angle of incidence = 90° - 40°

                                = 50°

Since the angle of incidence is the same as the angle of reflection, the angle of reflection here would be 50°.

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What is the wavelength (in nm) of a photon emitted during transition from the n = 3 state to the n = 1 state in the H atom?
Mkey [24]

Answer:

\lambda=103\ nm

Explanation:

E_n=-2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules

For transitions:

Energy\ Difference,\ \Delta E= E_f-E_i =-2.179\times 10^{-18}(\frac{1}{n_f^2}-\frac{1}{n_i^2})\ J=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

\Delta E=2.179\times 10^{-18}(\frac{1}{n_i^2} - \dfrac{1}{n_f^2})\ J

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

Where,  

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

So,  

\frac {h\times c}{\lambda}=2.179\times 10^{-18}(|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)\ J

\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m

So,  

\lambda=\frac {6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times (|\frac{1}{n_i^2} - \dfrac{1}{n_f^2}|)}\ m

Given, n_i=3\ and\ n_f=1

\lambda=\frac{6.626\times 10^{-34}\times 3\times 10^8}{{2.179\times 10^{-18}}\times |(\frac{1}{3^2} - \dfrac{1}{1^2})}|\ m

\lambda=\frac{10^{-26}\times \:19.878}{10^{-18}\times \:2.179|\left(\frac{1}{9}-\frac{1}{1}\right)|}\ m

\lambda=1.03\times 10^{-7}\ m

1 m = 10⁻⁹ nm

\lambda=103\ nm

4 0
2 years ago
Show that the acceleration of any object down an incline where friction behaves simply (that is, where fk=μkN ) is a=g(sinθ−μkco
11Alexandr11 [23.1K]

Answer:

a=g(sinθ-μkcosθ)

Explanation:

In an inclined plane the forces that interact with the object can be seen in the figure. The normal force, the weight w and the decomposition of the force vector of weight can be observed.

wx=m*g*sinθ

wy=m*g*cosθ

As the objects moves down an incline, acceleration in y axis is 0.

Then, by second Newton's Law:

Fy = m*ay

FN - m*g cos θ = 0,

FN=m*g cos θ

In x axis the forces that interacs are the x component of weight and friction force:

Fx = m*ax

mg sen u-FN*μk=m*a

Being friction force, Fr=FN*μk, we replace with its value in below formula:

m*g *sinθ-(m*g*cosθ*μk)=m*a

Then, isolating a:

a=(m*g sinθ-(m*g*cosθ*μk))/m

Solving, we have next equation:

a=g sinθ-(g*cosθ*μk)

Applying distributive property we have:

a=g*(sinθ-μk*cosθ)

5 0
3 years ago
If u accomplished 10,000 newton meters of work how much work did you do in units of joules
Amiraneli [1.4K]
1.0 joule= 1.0 newtons × 1.0 meter = 1.0 newton × meter 

Work = 10 newtons × 5 meters = 50 newton × meter
 
3 0
3 years ago
A 0.5-kilogram apple falls from a height of 2 meters to 1.50 meters. Ignoring frictional effects, what is the kinetic energy of
spin [16.1K]

The  final kinetic energy of the ball is 2.45 J

Explanation:

We can solve this problem by using the law of conservation of energy.

In absence of frictional effect, the mechanical energy of the apple must be conserved during the fall. So we can write:

U_i +K_i = U_f + K_f

where :

U_i is the initial potential energy, at the top

K_i is the initial kinetic energy, at the top

U_f is the final potential energy, at the bottom

K_f is the final kinetic energy, at the bottom

By explicing the potential energy, we can rewrite the equation as:

mgh_i + K_i = mgh_f + K_f

where:

m = 0.5 kg is the mass of the apple

g=9.8 m/s^2 is the acceleration of gravity

h_i = 2 m is the initial height

h_f=1.50 m is the final height

The initial kinetic energy is zero, since the ball starts from rest:

K_i = 0

Therefore we can solve the equation for K_f, the final kinetic energy of the ball:

K_f = mg(h_i-h_f)=(0.5)(9.8)(2-1.50)=2.45 J

Learn more about kinetic energy and potential energy:

brainly.com/question/6536722

brainly.com/question/1198647  

brainly.com/question/10770261  

#LearnwithBrainly

3 0
3 years ago
Read 2 more answers
Is the relationship between velocity and centripetal force a direct, linear or nonlinear square relationship?
Svet_ta [14]

Answer:

non linear square relationship

Explanation:

formula for centripetal force is given as

a = mv^2/r

here a ic centripetal acceleration , m is mass of body moving in circle of radius r and v is velocity of body . If m ,and r are constant we have

a = constant × v^2

a α v^2

hence non linear square relationship

5 0
3 years ago
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