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Firdavs [7]
3 years ago
5

At 20 degrees C, how much sodium chloride could be dissolved into 2 L of water?

Physics
1 answer:
sashaice [31]3 years ago
3 0
<span>35.89g of sodium chloride dissolves in 100g of water at 20 degrees celsius.
35.89 divided by 100 multiplied by 2000 = 718 g</span>
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What is the acceleration of an object with a constant velocity
valina [46]

Zero.

Acceleration is defined as the change in velocity over time.

Since in your case there is no change, there is no acceleration, so it is zero:

Or in formula: <span>a=<span><span>Δv</span>t</span></span>

Where a=acceleration, <span>Δv</span>=change in velocity and t=time

6 0
3 years ago
Read 2 more answers
Two balls of equal size are dropped from the same height from the roof of a building. the mass of ball a is twice that of ball b
Sindrei [870]
The mass of ball a is twice the mass of ball b:
m_a = 2 m_b
This means that the initial potential energy of ball a (U_a = m_a gh=2 m_b gh) is twice the potential energy of ball b (U_b = m_b  g h):
U_a = 2 U_b
When the two balls reach the ground, the potential energy of each ball has converted into kinetic energy (since now their altitude is h=0), because the total mechanical energy of each ball must be  conserved. Therefore:
K_a = U_a
K_b = U_b
and so the kinetic energy of ball a must be twice the kinetic energy of ball b:
K_a = 2 K_b
8 0
3 years ago
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Which type of power plants generate the least amount of pollution? A) Biomass power plants B) Geothermal power plants C) Nuclear
sladkih [1.3K]

Answer:

C

Explanation:

6 0
3 years ago
% = Wo/Wi x 100 Solve for Wo<br> % = Wo/Wi x 100 Solve for Wi
irga5000 [103]
1) % = (Wo /Wi) * 100

Solve for Wo => Wo = (% / 100) * Wi

For example, % =30% and Wi = 250 => Wo = (30 /100) * 250 = 0.30 * 250 = 75

Wo = 75

2) % = (Wo / Wi) * 100

Solve for Wi

=> Wi = Wo * (%/100)

For example, Wo = 125 and % = 40%

=> Wi = 125 * (40 / 100) = 125 * 0.40 = 50

Wi = 50
8 0
3 years ago
A square loop of wire, with sides of length a, lies in the first quadrant of the xy plane, with one corner at the origin. In thi
Ainat [17]

Answer:

emf=-\dfrac{1}{2}kta^5

Explanation:

Given that

B(y, t) = k y ³t²

To find the total flux over the loop we have to integrate over the loop

\phi =\int B.dS

Given that loop is square,so

\phi =\int B.dS

B(y, t) = k y ³t²

\phi =kt^2\int_{0}^{a}dx\int_{0}^{a}y^3dy

\phi =\dfrac{1}{4}kt^2a^5

We know that emf given as

emf=-\dfrac{d\phi }{dt}

\phi =\dfrac{1}{4}kt^2a^5

So

emf=-\dfrac{1}{2}kta^5

5 0
3 years ago
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