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pishuonlain [190]
3 years ago
8

The Light incident on a smooth reflecting surface undergoes specular reflection

Physics
1 answer:
In-s [12.5K]3 years ago
8 0
According to the laws of reflection for a a surface undergoing specular reflection the angle of reflection is equal to the angle of incidence therefore the angle of incidence will also be 55°.
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A -0.00325 C charge q1 is placed 5.62 m from a second charge q2. The first charge is repelled with a 48900 N force. What is the
blagie [28]

Answer: q2 = -0.05286

Explanation:

Given that

Charge q1 = - 0.00325C

Electric force F = 48900N

The electric field strength experienced by the charge will be force per unit charge. That is

E = F/q

Substitute F and q into the formula

E = 48900/0.00325

E = 15046153.85 N/C

The value of the repelled second charge will be achieved by using the formula

E = kq/d^2

Where the value of constant

k = 8.99×10^9Nm^2/C^2

d = 5.62m

Substitutes E, d and k into the formula

15046153.85 = 8.99×10^9q/5.62^2

15046153.85 = 284634186.5q

Make q the subject of formula

q2 = 15046153.85/ 28463416.5

q2 = 0.05286

Since they repelled each other, q2 will be negative. Therefore,

q2 = -0.05286

6 0
3 years ago
ILL ADD BRAINLIEST
Marina86 [1]

This is kinda confusing. I wish u just to a screenshot of the problem but here goes...

Forest at highest latitudes- Hardwood trees/deer, squirrel, foxes

Praries/temperate climate- Mostly small mammals/scrubs/steppes

High humidity/rainfall near equator- Abundant thick vegatation/manny species

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6 0
3 years ago
How are eras and periods of the geologic time scale named?
Natalka [10]

Answer:

mark me brainliest plz

Explanation:

In the early 1800's a system for naming geologic time periods was devised using four periods of geologic time. They were named using Latin root words, Primary, Secondary, Tertiary and Quaternary. ... Keep in mind that this chart is focused on geologic time periods. There are also geologic Eons, Eras, and epochs.

3 0
3 years ago
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A hydrogen atom in a galaxy moving with a speed of 6.65×106 m/???? away from the Earth emits light with a wavelength of 5.13×10−
Mumz [18]

Answer:

The observed wavelength on Earth from that hydrogen atom is 5.24\times 10^{-7}\ m.

Explanation:

Given that,

The actual wavelength of the hydrogen atom, \lambda_a=5.13\times 10^{-7}\ m

A hydrogen atom in a galaxy moving with a speed of, v=6.65\times 10^6\ m/s

We need to find the observed wavelength on Earth from that hydrogen atom. The speed of galaxy is given by :

v=c\times \dfrac{\lambda_o-\lambda_a}{\lambda_a}

\lambda_o is the observed wavelength

\lambda_o=\dfrac{v\lambda_a}{c}+\lambda_a\\\\\lambda_o=\dfrac{6.65\times 10^6\times 5.13\times 10^{-7}}{3\times 10^8}+5.13\times 10^{-7}\\\\\lambda_o=5.24\times 10^{-7}\ m

So, the observed wavelength on Earth from that hydrogen atom is 5.24\times 10^{-7}\ m. Hence, this is the required solution.

8 0
3 years ago
Volume is a measurement of the height of the wave known as what? <br><br><br> this is science
Cerrena [4.2K]
I’m gonna have to say “Ocean waves” as the answer
6 0
3 years ago
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