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VikaD [51]
3 years ago
7

If you see a rainbow near the time of sunset, where in the sky will the rainbow be?

Physics
1 answer:
Igoryamba3 years ago
8 0
Closest to the sun f thats a choice or closest to the atmoshpere
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What did newton's law of universal gravitation tell us about how gravity works?
vazorg [7]
The force of gravity form the Sun will be stronger on a n object with more mass
6 0
3 years ago
One mole of titanium contains how many atoms?
Vladimir [108]

Answer:

\boxed {\boxed {\sf D. \ 6.02*10^{23} \ atoms}}

Explanation:

One mole of a substance contains the same amount of representative particles. These particles can be atoms, molecules, ions, or formula units. In this case, the particles are atoms of titanium.

Regardless of the particles, there will always be <u>6.02*10²³</u> (also known as Avogadro's Number) particles in one mole of a substance.

Therefore, the best answer for 1 mole of titanium is D. 6.02*10²³ atoms.

3 0
3 years ago
Read 2 more answers
A snail crawls 300 cm in 1 hour. Calculate the snail's speed in each of the following units. A) cm/h B) cm/min C) m/h
Crazy boy [7]
A) 300cm/h
B)1 hr=60 min
300/60=5
5cm/min
C)1m=100cm
300/100=3
3m/h
7 0
3 years ago
Which elements have similar behavior
Nana76 [90]

The elements which have similar behavior are Barium, strontium and beryllium.

Explanation:

4 0
3 years ago
How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is 3.33××10âˆ
AlekseyPX

Answer:

There are 756.25 electrons present on each sphere.

Explanation:

Given that,

The force of repression between electrons, F=3.33\times 10^{-21}\ N

Let the distance between charges, d = 0.2 m

The electric force of repulsion between the electrons is given by :

F=k\dfrac{q^2}{r^2}

q=\sqrt{\dfrac{Fr^2}{k}}

q=\sqrt{\dfrac{3.33\times 10^{-21}\times (0.2)^2}{9\times 10^9}}

q=1.21\times 10^{-16}\ C

Let n are the number of excess electrons present on each sphere. It can be calculated using quantization of charges. It is given by :

q = ne

n=\dfrac{q}{e}

n=\dfrac{1.21\times 10^{-16}}{1.6\times 10^{-19}}

n = 756.25 electrons

So, there are 756.25 electrons present on each sphere. Hence, this is the required solution.

8 0
3 years ago
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