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pickupchik [31]

1 year ago

8

An aluminum block of mass 12.00 kg is heated from 20 c to 118 c. if the specific heat of aluminum is 913 j-1 kg k-1 then how muc

h energy is required?

1 answer:

Jobisdone [24]1 year ago

6 0

Answer:

107 kJ

Explanation:

12.00 kg * 913 J /kg-C * (118-20)°C ( I corrected spec heat units)

( a C degree change is the same as a kelvin change)

= 1 073 688 J = 107 kJ ( three sig digits)

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A thin nonconducting rod with a uniform distribution of positive charge Q is bent into a complete circle of radius R. The centra

Dmitriy789 [7]

Answer:

(a). If z = 0, The electric field due to the rod is zero.

(b). If z = ∞, The electric field due to the rod is $E\propto\dfrac{1}{z^2}$ .

(c). The positive distance is $\dfrac{R}{\sqrt{2}}$

(d). The maximum magnitude of electric field is $1.54\times10^{4}\ N/C$

Explanation:

Given that,

Radius = 2.00 cm

Charge = 4.00 mC

(a). If the radius and charge are R and Q.

We need to calculate the electric field due to the rod

Using formula of electric field

$E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qz}{(z^2+R^2)^{\frac{2}{3}}}$

Where, Q = charge

z = distance

If z = 0,

Then, The electric field is

$E=0$

(b). If z = ∞, z>>R

So, R = 0

Then, the electric field is

$E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Q}{z^2}$

$E\propto\dfrac{1}{z^2}$

(c). In terms of R,

We need to calculate the positive distance

If $E\rightarrow E_{max}$

Then, $\dfrac{dE}{dz}=0$

$\dfrac{Q}{4\pi\epsilon_{0}}(\dfrac{(z^2+R^2)^\frac{3}{2}-\dfrac{3z}{2}(z^2+R^2)^\dfrac{1}{2}}{(z^2+R^2)^2})=0$

Taking only positive distance

$z=\dfrac{R}{\sqrt{2}}$

(d). If R = 2.00 and Q = 4.00 mC

We need to calculate the maximum magnitude of electric field

Using formula of electric field

$E_{max}=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{Qz}{(z^2+R^2)^{\frac{2}{3}}}$

$E_{max}=9\times10^{9}\times\dfrac{4.0\times10^{-6}\times\dfrac{2.00}{\sqrt{2}}}{((\dfrac{2.00}{\sqrt{2}})^2+(2.00)^2)^{\frac{2}{3}}}$

$E_{max}=15418.7\ N/C$

$E_{max}=1.54\times10^{4}\ N/C$

Hence, (a). If z = 0, The electric field due to the rod is zero.

(b). If z = ∞, The electric field due to the rod is $E\propto\dfrac{1}{z^2}$ .

(c). The positive distance is $\dfrac{R}{\sqrt{2}}$

(d). The maximum magnitude of electric field is $1.54\times10^{4}\ N/C$

6 0

2 years ago

The magnitude of the momentum of an object is 64 kilogram meters per second. If the magnitude of te velocity is doubled, the mag

dsp73

Answer:

$128 \frac{kg*m}{s}$

Explanation:

$P=m*v\\if double velocity:\\P=m*2v=2m*v\\$

7 0

3 years ago

When a fixed amount of ideal gas goes through an isobaric expansion A) its internal (thermal) energy does not change.B) the gas

Bingel [31]

<h2>Answer: its temperature must increase.</h2>

Explanation:

In an isobaric process the pressure remains constant, which means the initial pressure and the final pressure will be the same.

In addition, during this thermodynamic process, the volume of the ideal gas expands or contracts in such a way that the variation of pressure $\Delta P$ is neutralized.

Now, according to the First law of Thermodynamics that establishes the conservation of energy:

$\Delta U=\Delta Q-\Delta W$ (1)

Where:

$\Delta U$ is the internal energy

$\Delta Q$ is the heat transferred

$\Delta W$ is the work

Now, for an isobaric process:

$\Delta W=P\Delta V$ (2)

Where:

$P$ is the pressure (<u>always positive</u>)

$\Delta V$ is the volume variation of the gas

<u />

<u>Here we have two possible results:</u>

-If the gas expands (positive $\Delta V$ ), the work is positive.

-If the gas compresses (negative $\Delta V$ ), the work is negative.

In this case we are talking about the first result (work is positive).

Then, according to the above, equation (1) can be written as follows:

$\Delta U=\Delta Q - P\Delta V$ (3)

Clearing $\Delta Q$ :

$\Delta Q=\Delta U+P \Delta V$ (4)

Then, for an ideal gas in an isobaric process, part of the heat ( $Q$ ) added to the system will be used to do work (positive in this case) and the other part <u>will increase the internal energy</u>, hence <u>the temperature will increase as well.</u>

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3 years ago

Which is an example of projectile motion?

stiv31 [10]

A person throwing a rock

5 0

3 years ago

Read 2 more answers

A 12.7 L box is filled with 6.0 moles of Argon gas and cooled to a temperature of 210 K. What is the pressure?

sp2606 [1]

Answer:

Gases are easily compressed. We can see evidence of this in Table 1 in Thermal Expansion of Solids and Liquids, where you will note that gases have the largest coefficients of volume expansion. The large coefficients mean that gases expand and contract very rapidly with temperature changes. In addition, you will note that most gases expand at the same rate, or have the same β. This raises the question as to why gases should all act in nearly the same way, when liquids and solids have widely varying expansion rates.

Explanation:

3 0

2 years ago