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grin007 [14]
3 years ago
9

How much force is needed to accelerate a 1000-kg car at a rate of 3 m/s2? a 1003 N b 0.003 N c 3000 N d 333.3 N

Chemistry
1 answer:
Bogdan [553]3 years ago
7 0

Answer:

Option c.

Explanation:

According to Newton's law, formula to solve this problem is:

F = m . a

Mass → 1000 kg

a → 3 m/s²

F = 1000 kg . 3m/s² →  3000 N

1 N = 1 kg .  m/s²  according to SI, Internation System of units.

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What is the iupac name of ch3ch2ch2ch2ch2ch3
Alborosie

Answer:

hexane

Explanation:

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3 years ago
A piece of wood has a mass of 36g and measures 3cm X 6cm X 4cm. What is the density of the wood? Would the piece of wood float o
Fudgin [204]

Answer:

Explanation:

The density is  

1.1 g c m 3 .

4 0
3 years ago
Please help me!! 70 points!!
ICE Princess25 [194]
Do all substances dissolve in water? Kids explore the varying levels of solubility of common household substances in this fun-filled experiment!

Materials Needed:
4 clear, glass jars filled with plain tap water
Flour
Salt
Talcum or baby powder
Granulated sugar
Stirrer
Step 1: Help your child form a big question before starting the experiment.

Step 2: Make a hypothesis for each substance. Perhaps the salt will dissolve because your child has watched you dissolve salt or sugar in water when cooking. Maybe the baby powder will not dissolve because of its powdery texture. Help your child write down his or her predictions.

Step 3: Scoop a teaspoon of each substance in the jars, only adding one substance per jar. Stir it up!

Step 4: Observe whether or not each substance dissolves and record the findings!

Your child will likely note that that sugar and salt dissolve, while the flour will partially dissolve, and the baby powder will remain intact. The grainy crystals of the sugar and salt are easily dissolved in water, but the dry, powdery substances are likely to clump up or remain at the bottom of the jar.

As you can see, the scientific method is easy to work into your child’s scientific experiments. Not only does it increase your child’s scientific learning and critical thinking skills, but it sparks curiosity and motivates kids as they learn to ask questions and prove their ideas! Get started today with the above ideas, and bring the scientific method home to your child during your next exciting science experiment
7 0
2 years ago
10. What effect does temperature have on molecular motion? Using this explanation, explain why both pressure and volume can decr
vovangra [49]

Answer:

If the temperature increases the molecular movement as well, and if it increases the same it will happen with the molecular movement.

Pressure, volume and temperature are three factors that are closely related since they increase the temperature, the pressure usually decreases due to the dispersion of the molecules that can be generated, so the volume also increases.

If the temperature drops, the material becomes denser, its molecules do not collide with each other, their volume and pressure increases.

Explanation:

The pressure is related to the molecular density and the movement that these molecules have.

The movement is regulated by temperature, since if it increases, the friction and collision of the molecules also.

On the other hand, the higher the volume, the less pressure there will be on the molecules, since they are more dispersed among themselves.

(in the opposite case that the volume decreases, the pressure increases)

5 0
3 years ago
The standard enthalpy of formation for glucose [c6h12o6(s)] is −1273.3 kj/mol. what is the correct formation equation correspond
balu736 [363]
The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is 
     C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is 
     6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)

Using the equation for the standard enthalpy change of formation 
     ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
     ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}

C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
     ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
                           = -1273.3 - (0 + 0 + 0)
                           = -1273.3
8 0
3 years ago
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