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3 years ago

12

6 jam jars of volume 0.37m3 are filled with Jam. An empty jam jar has a weight 200g each. The weight of all 6 jars filled with j

am 5400g. Use this information to calculate the density of Jam?

1 answer:

marta [7]3 years ago

6 0

Answer:

0.14594 g/cm³

Explanation:

Density = Mass / Volume

Mass = 5400 g

Volume = 0.37 m³

Density = M /V

volume = 0.37 m³ = 37000 cm³

Density = $\frac{5400}{37000}$

= 0.14594 g/cm³

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During an adiabatic process an object does 100 J of work and its temperature decreases by 5 K. During another process it does 25

Liono4ka [1.6K]

Answer:

The heat capacity for the second process is 15 J/K.

Explanation:

Given that,

Work = 100 J

Change temperature = 5 k

For adiabatic process,

The heat energy always same.

$dQ=0$

$dU=-dW$

We need to calculate the number of moles and specific heat

Using formula of heat

$dU=nC_{v}dT$

$nC_{v}=\dfrac{dU}{dT}$

Put the value into the formula

$nC_{v}=\dfrac{-100}{5}$

$nC_{v}=-20\ J/K$

We need to calculate the heat

Using formula of heat

$dQ=nC_{v}(dT_{1})+dW_{1}$

Put the value into the formula

$dQ=-20\times5+25$

$dQ=-75\ J$

We need to calculate the heat capacity for the second process

Using formula of heat

$dQ=nC_{v}(dT_{1})$

Put the value into the formula

$-75=nC_{v}\times(-5)$

$nC_{v}=\dfrac{-75}{-5}$

$nC_{v}=15\ J/K$

Hence, The heat capacity for the second process is 15 J/K.

5 0

3 years ago

One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction

Sophie [7]

Answer:

F = 156.3 N

Explanation:

Let's start with the top block, apply Newton's second law

F - fr = 0

F = fr

fr = 52.1 N

Now we can work with the bottom block

In this case we have two friction forces, one between the two blocks and the other between the block and the surface. In the exercise, indicate that the two friction coefficients are equal

we apply Newton's second law

Y axis

N - W₁ -W₂ = 0

N = W₁ + W₂

as the two blocks are identical

N = 2W

X axis

F - fr₁ - fr₂ = 0

F = fr₁ + fr₂

indicates that the lower block is moving below block 1, therefore the upper friction force is

fr₁ = 52.1 N

fr₁ = μ N

a

s the normal in the lower block of twice the friction force is

fr₂ = μ 2N

fr₂ = 2 μ N

fr₂ = 2 fr₁

we substitute

F = fr₁ + 2 fr₁

F = 3 fr₁

F = 3 52.1

F = 156.3 N

7 0

3 years ago

The bar graph shows energy data taken from a roller coaster at a theme park. Analyze the data and assess its validity. In 3–5 se

alexdok [17]

The data given in the bar graph is valid because it follows the law of conservation of energy, since the GPE at top of 2nd hill plus KE at top of 2nd hill equals KE at bottom of 1st hill.

<h3>What is law of conservation of energy?</h3>

The law of conservation of energy states that energy can neither be created nor destroyed but can be transformed from one form to another.

Based on the law of conservation of energy, kinetic energy of a roller coaster can be converted into potential energy of the roller coaster and vice versa.

ΔK.E = ΔP.E

where;

ΔK.E is change in kinetic energy
ΔP.E is change in potential energy

The kinetic energy of the coaster is greatest at the bottom of the hill, as the coaster moves upward, the kinetic energy decreases and will be converted into potential energy. The potential energy of the coaster increases as the coaster moves up the hill and will become maximum at the highest point of the hill.

From the given data;

GPE at top of 2nd hill + KE at top of 2nd hill = KE at bottom of 1st hill

Learn more about conservation of energy here: brainly.com/question/166559

#SPJ1

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1 year ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

<u>Distance of the center from each corners</u> $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

<u>For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:</u>

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

<u>Force by each of the charges at the remaining corners:</u>

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

<u> Now, net electric field in the vertical direction:</u>

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

<u>Now, net electric field in the horizontal direction:</u>

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

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3 years ago

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kotykmax [81]

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