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Jlenok [28]
3 years ago
7

When a 57-gram piece of aluminum at 100oC is placed in water, it loses 735 calories of heat while cooling to 30oC. Calculate the

specific health capacity of the aluminum
Physics
2 answers:
Archy [21]3 years ago
7 0

Answer:

specific heat capacity of aluminium is 0.184 Calroie/ gram-K.

Explanation:

As we know that heat transferred to a given material to change its temperature is given as

Q = ms\Delta T

here we know

Q = heat given = 735 calorie

m = mass = 57 gram

initial temperature = 100 degree C

final temperature = 30 degree C

now by above formula we will have

735 = 57(s)(100 - 30)

s = 0.184 \frac{cal}{g-K}

Tomtit [17]3 years ago
4 0
Are you still in need of help with this question

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3 years ago
An electron is released from rest at the negative plate of a parallel plate capacitor. The charge per unit area on each plate is
Andreas93 [3]

Answer:

v=1.08\times 10^7\ m/s

Explanation:

Initial speed of the electron, u = 0

The charge per unit area of each plate, \dfrac{Q}{A}=1.69\times 10^{-7}\ C/m^2

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An electron is released from rest, u = 0

Using equation of kinematics,

v^2-u^2=2ad..........(1)

Acceleration of the electron in electric field, a=\dfrac{qE}{m}............(2)

Electric field, E=\dfrac{\sigma}{\epsilon_o}............(3)

From equation (1), (2) and (3) :

v=\sqrt{\dfrac{2q\sigma d}{m\epsilon_o}}

v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 1.69\times 10^{-7}\times 1.75\times 10^{-2}}{9.1\times 10^{-31}\times 8.85\times 10^{-12}}}

v = 10840393.1799 m/s

or

v=1.08\times 10^7\ m/s

So, the electron is moving with a speed of 1.08\times 10^7\ m/s before it reaches the positive plate. Hence, this is the required solution.

3 0
3 years ago
Which statements correctly describe matter? Select all that apply.
strojnjashka [21]

Answer:

D

I hope these is correct

3 0
3 years ago
I don't understand how to get 50000 ohms, please can someone help? I would have just left my answer at 25,000 ohms!
pochemuha
I think this is shown in the fine print on the second sheet . . .


The question starts out by saying that the scale uses <u>two</u> 3V cells <u>in series</u>.
That makes the total battery voltage 6 volts.

0.12 milliamperes = 0.00012 Amperes.

Resistance = (voltage) / (current in Amperes).

Resistance = 6 / 0.00012 = 50,000 Ω






8 0
3 years ago
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