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Vlad [161]
3 years ago
6

Please help me out. Convert 28 pints to cups. Using one step conversion.

Physics
1 answer:
KengaRu [80]3 years ago
6 0

Answer:

56

Explanation:

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3 years ago
A Chef is looking for a new frying pan that will allow her to cook food quickly at temperatures above 500°C. Use the data provid
makvit [3.9K]

Answer:

Explanation:

1st one

What is your evidence?

Very heavy professional or restaurant pans will have iron handles, while those for home use will be made of brass or stainless steel. All are perfectly safe for oven use.

6 0
2 years ago
Someone help please by providing work and answers please :)
Nastasia [14]
First we gotta use an equation of motion:

d = ut + \frac{1}{2} a {t}^{2}

Our vertical distance d= 100 m, initial vertical speed u = 0 m/s (because velocity is fully horizontal), and vertical acceleration a = 9.8 m/s2 because of gravity. Let's plug it all in!

100 = 0 + \frac{1}{2} (9.8) {t}^{2}

Now we just need to solve for t:

{t}^{2} = \frac{2(100)}{9.8} \\ \\ t = \sqrt{\frac{2(100)}{9.8}}

Hit the calculators, and you'll get 4.5 seconds!
5 0
3 years ago
Pls help ASAP this questions are science questions ( biology chemistry physics)
Vilka [71]

Answer:

It would be A.

Explanation:

The scale goes from 0 to 14, With 0 being acidic and 14 being basic.

So if 7 is neutral, then anything less than 7 is moving more towards being more acidic.  Anything higher than 7 is moving towards being more basic.

4 0
3 years ago
Read 2 more answers
In a charging process, 4 × 1013 electrons are removed from one small metal sphere and placed on a second identical sphere. Initi
Alina [70]

Answer:

The distance between the two spheres is 914.41 X 10³ m

Explanation:

Given;

4 X 10¹³ electrons, and its equivalent in coulomb's is calculated as follows;

1 e = 1.602 X 10⁻¹⁹ C

4 X 10¹³ e = 4 X 10¹³ X 1.602 X 10⁻¹⁹ C = 6.408 X 10⁻⁶ C

V = Ed

where;

V is the electrical potential energy between two spheres, J

E is the electric field potential between the two spheres N/C

d is the distance between two charged bodies, m

V = \frac{K*q}{d^2}*d = \frac{K*q}{d}

d = \frac{K*q}{V}

where;

K is coulomb's constant = 8.99 X 10⁹ Nm²/C²

d = (8.99 X 10⁹ X 6.408 X 10⁻⁶)/0.063

d = 914.41 X 10³ m

Therefore, the distance between the two spheres is 914.41 X 10³ m

3 0
3 years ago
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