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2 years ago

An object that looks white when exposed to sunlight reflects all colors of light. What

Physics

1 answer:

Ierofanga [76]2 years ago

6 0

It looks blue as it is only reflecting blue light

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Consider a current carrying a wire coming out of your computer screen towards you. Which statement below correctly describes the

Ugo [173]

Answer:

1. The magnetic field encircles the wire in a counterclockwise direction

Explanation:

When we have a current carrying wire perpendicular to the screen in which the current flows out of the screen then by the Maxwell's right-hand thumb rule we place the thumb of our right hand in the direction of the current and curl the remaining fingers around the wire, these curled fingers denote the direction of the magnetic field which is in the counter-clock wise direction.

Ever current carrying conductor produces a magnetic field around it.

5 0

3 years ago

The temperature at one of the Viking sites on Mars was found to vary daily from -90.OF to -5.0 C. Convert these temperatures to

kykrilka [37]

Answer:

I don't know about this, this is which class question

6 0

2 years ago

What force must act on a 50.0-kg mass to give it an ancceleration of 0.30 m/s^2?

ratelena [41]

Answer:

15.0 N

Explanation:

see pic

3 0

2 years ago

What is the magnitude of the electric force of attraction between two point charges + 2.8 mc and -1.2 mc if the distance

Anit [1.1K]

Answer:

Explanation:

it just a

8 0

2 years ago

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

3 0

3 years ago