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DaniilM [7]
3 years ago
14

A 0.500 kg block of lead is heated from 295 K to 350. K. How much heat was absorbed by the lead? (express your answer to the nea

rest joule
Physics
1 answer:
katen-ka-za [31]3 years ago
6 0

Answer:

The amount of energy needed to change the temperature of a certain supstance is given in the equation:

Q = c • m • ∆T

-Q is the amount of energy in Joules

-c is the specific heat capacity, a constant value for each supstance that shows amount of energy needed to raise the temperature of 1 g of a supstance by 1 K. Its unis is J/g•K (for lead, c=0.129J/g•K)

-m is the mass of the given supstance in grams

-∆T is the change in the temperature in Kelvins. In this case it's 350-295=55K

So, now, let's calculate:

Q = 0.129J/g•K • 500g • 55K

Q = 3,547.5 J

Rounded to the nearest whole number that is 3,548 Joules

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2 years ago
Provided following are four different ranges of stellar masses. Rank the stellar mass ranges based on how many stars in each ran
elena-s [515]

Highest to lowest number:

-less than 1 solar mass

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<h3>What is Stellar masses ?</h3>

Stellar mass is a phrase that is used by astronomers to describe the mass of a star.

  • It is usually enumerated in terms of the Sun's mass as a proportion of a solar mass ( M ☉). Hence, the bright star Sirius has around 2.02 M ☉.

  • Stellar masses are not fixed, although they change for single stars only on long periods.

Learn more about Stellar masses here:

brainly.com/question/1128503

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3 0
1 year ago
A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-
alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, q_1=-3\ nC

It is placed at a distance of 9 cm at x axis

Charge, q_2=+4\ nC

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0

Here,

r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}

So,

\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}

Squaring both sides,

3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

8 0
3 years ago
) each plate of a parallel-plate air-filled capacitor has an area of 0.0020 , and the separation of the plates is an electric fi
Dvinal [7]
I attached the full question.
We know that for a parallel-plate capacitor the surface charge density is given by the following formula:
\sigma=\varepsilon_0 \frac{V}{d}
Where V is the voltage between the plates and d is separation.
Voltage is by definition:
V=Ed
Voltage is analog to the mechanical work done by the force.
Above formula is correct only If the field is constant, and we can assume that it is since no function has been given.
The charge density would then be:
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Please note that elecric permittivity of air is very close to  elecric permittivity of vacum, it is common to use them <span>interchangeably</span>.

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noname [10]
Climate is a particular place's distance from the equator
4 0
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