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3 years ago

What is the density of an object that has a mass of 10 g and a volume

Physics

1 answer:

Korolek [52]3 years ago

7 0

2g/mL

Explanation:

Given parameters :

Mass = 10g

Volume = 5mL

Unknown:

Density of the object = ?

Solution:

Density is defined as the mass per unit volume of a body.

Density = $\frac{mass}{Volume}$

Now, input the values in the equation:

Density = $\frac{10}{5}$ = 2g/mL

Learn more:

Density brainly.com/question/5055270

#learnwithBrainly

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A frog falls from its rainforest tree. If we ignore wind resistance, (a) how much time does it take the frog to fall a distance

gtnhenbr [62]

Answer:

Explanation:

a) Using the equation of motion

S = ut + 1/2gt²

S is the distance of fall

g is the acceleration due to gravity

t is the time taken

Given S = 12.0m, g = 9.81m/s^2, un= 0m/s

12 = 0+1/2(9.81)t²

12 = 4.905t²²²

t² = 12/4.905

t² = 2.446

t = √2.446

t = 1.56secs

b) To determine how fast is the frog falling at this point, we need to calculate the speed of the frog. Using the equaton v = u+gt

v = 0+9.81(1.56)

v = 15.34m/s

Hence the frog is falling at the rate of 15.34m/s

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3 years ago

If Siobhan hits a 0.25 kg volleyball with 0.5 N of force, what is the acceleration of the ball?

Alekssandra [29.7K]

Answer:

2 meters per second²

Explanation:

8 0

3 years ago

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A heat engine accepts 200,000 Btu of heat from a source at 1500 R and rejects 100,000 Btu of heat to a sink at 600 R. Calculate

diamong [38]

To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.

By definition we know that the change in entropy is given by

$\Delta S = \frac{Q}{T}$

Where,

Q = Heat transfer

T = Temperature

On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

$W = Q_{source}-Q_{sink}$

According to the data given we have to,

$Q_{source} = 200000Btu$

$T_{source} = 1500R$

$Q_{sink} = 100000Btu$

$T_{sink} = 600R$

PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is

$\Delta S_{sink} = \frac{Q_{sink}}{T_{sink}}$

$\Delta S_{sink} = \frac{100000}{600}$

$\Delta S_{sink} = 166.67Btu/R$

On the other hand,

$\Delta S_{source} = \frac{Q_{source}}{T_{source}}$

$\Delta S_{source} = \frac{-200000}{1500}$

$\Delta S_{source} = -133.33Btu/R$

The total change of entropy would be,

$S = \Delta S_{source}+\Delta S_{sink}$

$S = -133.33+166.67$

$S = 33.34Btu/R$

Since $S\neq 0$ the heat engine is not reversible.

PART B)

Work done by heat engine is given by

$W=Q_{source}-Q_{sink}$

$W = 200000-100000$

$W = 100000 Btu$

Therefore the work in the system is 100000Btu

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3 years ago

1. Someone fires a 0.04 kg bullet at a block of wood that has a mass of 0.5 kg. (The block of wood is sitting on a frictionless

Maslowich

Since momentum is a vector quantity, take any direction as positive and other as negative. Answer won't change.

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3 years ago

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Does anyone know these four?! I really need help and fast pls

Mama L [17]

Evidence: Data gathered

Experiment: Looking through a telescope

Observations: Testing what happens

Reasoning: Thinking a problem through

I believe that these should be correct.

Hoping you pass!

3 0

3 years ago