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aleksklad [387]
4 years ago
5

How can the motion of an object that is NOT moving change?

Chemistry
1 answer:
Alenkinab [10]4 years ago
8 0
By making it change probably
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1. I am a solution with a pOH of 6. What am I?
ivolga24 [154]

Explanation:

I think 1) acidic solution

2) basic solution.

8 0
3 years ago
Two positively charged objects (N pole) are separated by a large distance. One of the positively charged objects (N pole) is rep
Eddi Din [679]

Answer:

B

Explanation:

Applying law of electrostatic which states that like charges repel each other and unlike charges attract each other

N and S are unlike charges that turn and make the former repulsive force (due to two like charges N and N)to <em>reduce</em> and attractive force between N and S to <em>increa</em><em>se</em>

6 0
3 years ago
Comare and contrast physical property and chemical property
Nimfa-mama [501]
A physical property is a quality or condition of a substance that can be observed or measured without changing the substance's composition. a chemical property is the ability of a substance to undergo a specific chemical change.
8 0
3 years ago
Pls help me. Show proof on why it’s that answer.
natulia [17]
The answer is “the sustainability of the ecosystem” because if the ecosystem isn’t sustainable enough for many different organisms to live in it then the biological diversity of the ecosystem will decrease.
6 0
3 years ago
15.0 g of cream at 10.0 ℃ are added to an insulated cup containing 150.0 g of coffee at 78.6 °C. Calculate the equilibrium tempe
elena-14-01-66 [18.8K]

Answer:

The equilibrium temperature of the coffee is 72.4 °C

Explanation:

Step 1: Data given

Mass of cream = 15.0 grams

Temperature of the cream = 10.0°C

Mass of the coffee = 150.0 grams

Temperature of the coffee = 78.6 °C

C = respective specific heat of the substances( same as water) = 4.184 J/g°C

Step 2: Calculate the equilibrium temperature

m(cream)*C*(T2-T1) = -m(coffee)*c*(T2-T1)

15.0 g* 4.184 J/g°C *(T2 - 10.0°C) = -150.0g *4.184 J/g°C*(T2-78.6°C)

62.76T2 - 627.6 = -627.6T2 + 49329.36

690.36T2 = 49956.96

T2 = 72.4 °C

The equilibrium temperature of the coffee is 72.4 °C

7 0
3 years ago
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