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anygoal [31]
3 years ago
5

One piece of copper jewelry at 111°C has exactly twice the mass of another piece, which is at 28°C. Both pieces are placed insid

e a calorimeter whose heat capacity is negligible. What is the final temperature inside the calorimeter (c of copper = 0.387 J/g·K)?
Physics
1 answer:
vladimir2022 [97]3 years ago
3 0

Answer:

83.33 C

Explanation:

T1 = 111 C, m1 = 2m

T2 = 28 C, m2 = m

c  = 0.387 J/gK

Let the final temperature inside the calorimeter of T.

Use the principle of calorimetery

heat lost by hot body = heat gained by cold body

m1 x c x (T1 - T) = m2 x c x (T - T2)

2m x c X (111 - T) = m x c x (T - 28)

2 (111 - T) = (T - 28)

222 - 2T = T - 28

3T = 250

T = 83.33 C

Thus, the final temperature inside calorimeter is 83.33 C.

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3 years ago
Someone please help? :)
Gnoma [55]

Answer:

(A) F_N - mg\cos\theta

Explanation:

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8 0
4 years ago
A radio signal takes 1.28 s to travel from a transmitter on the Moon to the surface of Earth. The radio waves travel at 3.00108
yulyashka [42]

Answer:

s = 3.84 x 10⁸ m

Explanation:

The distance traveled by an object, while in uniform motion, is given by the following equation:

s = vt

where,

s = distance covered

v = speed

t = time interval

In this case:

s = distance between Moon and Earth = ?

v = speed of radio waves = 3 x 10⁸ m/s

t = time taken to travel = 1.28 s

Therefore,

s = (3 x 10⁸ m/s)(1.28 s)

<u>s = 3.84 x 10⁸ m</u>

3 0
3 years ago
A surface will be an equipotential surface if (there may be more than one correct choice).A. the electric field is zero at all p
aev [14]

Answer:

c. A, C

Explanation:

On equi-potential surface, fields are equal in magnitude at all points . If field is zero at all points , they will be equal so first option is correct.

Field is perpendicular to  equipotential surface at all points.

6 0
3 years ago
Assume it takes 8.00 min to fill a 50.0-gal gasoline tank. (1 U.S. gal = 231 in.3) (a) Calculate the rate at which the tank is f
vagabundo [1.1K]

The volumetric rate or flow rate of a fluid is defined as the amount of the volume of a fluid circulating on a surface per unit of time. In this case we have units given initially: Gallons and minutes. For the first part we will convert the minutes to seconds, and we will obtain the flow rate under that measure. For the second case we will convert the gallons to cubic meters and obtain the desired value. Recall the following conversion rates,

1 min = 60s

1 U.S Gal = 0.00378541178 m^3

If the flow rate is defined as the volume by time, the flow rate with the given values is

Q = \frac{V}{t}

Q = \frac{50Gal}{8min}

Q = 6.25 Gal/min

PART A ) Converting to Gal/seconds, we have,

Q = 6.25 \frac{Gal}{min}(\frac{1min}{60s})

Q = 0.10416Gal/s

PART B) Converting Gal/seconds to m^3/s

Q = 0.104116\frac{Gal}{s} (\frac{0.00378541178 m^3}{1 Gal})

Q = 3.941*10^{-4}m^3/s

4 0
3 years ago
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