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babunello [35]
2 years ago
11

When the number of atoms on the right side a chemical equation matches the number of atoms on the left side of a chemical equati

on, it is said to be… In equilibrium. Neutral. Positive. Balanced.
Physics
1 answer:
Hitman42 [59]2 years ago
7 0

Answer:

Balanced.

Explanation:

A Balanced Chemical equation is a scientific term that describes a chemical equation that has the same number of atoms on each side of the equation.

Hence, when the number of atoms on the right side of a chemical equation matches the number of atoms on the left side of a chemical equation, it is said to be BALANCED.

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A manufacturer claims its cleanser works twice as fast as any other. Could test be performed to support the claim? Explain
lora16 [44]

Yes, a test could be performed to support the claim.

 

Hypothesis: The claim that a manufacturer’s cleanser works twice as fast as any other cleanser.

 

So, based from this hypothesis, we can perform the following tests:

We assign Cleanser A to the manufacturer that claims that their cleanser works twice as fast as any other cleanser and Cleanser B to the cleanser to be compared with.

 

1.       Get two tiles and put the same amount of stain on them.

2.       Apply Cleanser A on the first tile and Cleanser B on the second tile.

3.       Apply the same amount of force in removing the stains on both tiles

4.       Record the amount of time it takes to remove the stains on each tile.

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3 years ago
A 50 n force is acting on a lever 1.5 m from the fulcrum balances an object 1m from the fulcrum on the other arm. what is the we
Angelina_Jolie [31]

Answer:

A 100 N force acting on a lever 2 m from the fulcrum balances an object 0.5 m from the fulcrum on. ... What is the weight of the object(in newtons)? What is its mass (in kg)? ... mass at the one end and effort arm is the distance between pivot and effort applied at the other end.

Explanation:

hpoe this helps you.

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2 years ago
A ball is let go and falls to the ground. It bounces a few times before
Ksenya-84 [330]

Answer:

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2 years ago
A car of mass m accelerates from speed v_1 to speed v_2 while going up a slope that makes an angle theta with the horizontal. Th
Karo-lina-s [1.5K]

Answer:

Work done by external force is given as

Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2

Explanation:

As per work energy Theorem we can say that work done by all force on the car is equal to change in kinetic energy of the car

so we will have

Work_{external} + Work_{gravity} + Work_{friction} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2

now we have

W_{gravity} = -mg(Lsin\theta)

W_{friction} = -\mu mgcos(\theta) L

so from above equation

Work_{external} - mgLsin\theta - \mu mgLcos(\theta) = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2

so from above equation work done by external force is given as

Work_{external} = mgLsin\theta + \mu mgLcos(\theta) + \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2

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3 years ago
Newtons second law of motion-skateboard
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Answer:

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