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3 years ago

During the new moon phase, why is the Moon not visible in the sky?

Physics

1 answer:

katrin [286]3 years ago

4 0

Answer:

During the new moon phase, the moon is not visible because Sometimes it can be detected by noting the visible absence of the stars that it blocks. Additionally, during the new moon, sometimes enough light is reflected off the surface of the Earth that the disk of the moon is faintly visible

Explanation:

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Ultraviolet radiation is dangerous because it has a high enough energy to damage skin cells. UV radiation is called ultraviolet

BabaBlast [244]

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Ultraviolet radiation has broad range of wavelengths, higher number means greater risk of exposure to UV rays that can be dangerous to skin cells. Sunlight is the main source of electromagnetic radiation and it is transmitted in different wavelengths known as electromagnetic spectrum. This spectrum is divided into several regions in order of decreasing wavelength and increasing energy and frequency. UV radiation has frequency and energy that is higher than purple light or violet radiation and the wavelength of ultraviolet radiation is smaller than violet radiation.

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3 years ago

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Unbalanced forces cause a change in motion. How do you know this?

ololo11 [35]

Answer:

This happens in two ways. If an object is at rest and an unbalanced force pushes or pulls the object, it will move. Unbalanced forces can also change the speed or direction of an object that is already in motion.

Explanation:

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2 years ago

A bus travels east for 3 km, then north for 4 km. What is its final displacement?

ikadub [295]

5 km northeast. Left and up would make northeast

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2 years ago

List 3 quantities of waves.

VARVARA [1.3K]

If you mean like electromagnetic waves then, Mico waves, UV rays, and infrared waves

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3 years ago

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A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge a

makkiz [27]

Answer:

E = (-3.61^i+1.02^j) N/C

magnitude E = 3.75N/C

Explanation:

In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:

$\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}]$ (1)

Where the minus sign means that the electric field point to the charge.

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

q = -4.28 pC = -4.28*10^-12C

r: distance to the charge from the point P

The point P is at the point (0,9.83mm)

θ: angle between the electric field vector and the x-axis

The angle is calculated as follow:

$\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°$

The distance r is:

$r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m$

You replace the values of all parameters in the equation (1):

$\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}$

The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C

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2 years ago