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Over [174]
3 years ago
14

A hot lump of 35.8 g of copper at an initial temperature of 70.7 °C is placed in 50.0 mL H2O initially at 25.0 °C and allowed to

reach thermal equilibrium. What is the final temperature of the copper and water, given that the specific heat of copper is 0.385 J/(g·°C)? Assume no heat is lost to surroundings.

Physics
1 answer:
AURORKA [14]3 years ago
4 0

I upload the answer via Image because Brainly warned me that there were inappropriate links or words.

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Interactive LearningWare 10.1 reviews the concepts involved in this problem. A spring stretches by 0.0161 m when a 3.74-kg objec
yKpoI14uk [10]

Answer:

m = 3.91 kg

Explanation:

Given that,

Mass of the object, m = 3.74 kg

Stretching in the spring, x = 0.0161 m

The frequency of vibration, f = 3.84 Hz

When the object is suspended, the gravitational force is balanced by the spring force as :

mg=kx

k=\dfrac{mg}{x}

k=\dfrac{3.74\times 9.8}{0.0161}

k = 2276.52 N/m

The frequency of vibration is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

m=\dfrac{k}{4\pi^2f^2}

m=\dfrac{2276.52}{4\pi^2\times (3.84)^2}

m = 3.91 kg

So, the mass of the object is 3.91 kg. Hence, this is the required solution.

8 0
3 years ago
What force does the water exert (in addition to that due to atmospheric pressure) on a submarine window of radius 44.0 cm at a d
Butoxors [25]
Calculate the pressure due to sea water as density*depth.
That is, 
pressure = (1025 kg/m^3)*((9400 m)*(9.8 m/s^2) = 94423000 Pa = 94423 kPa

Atmospheric pressure is  101.3 kPa
Total pressure is  94423 + 101.3 = 94524 kPa (approx)

The area of the window is π(0.44 m)^2 = 0.6082 m^2

The force on the window is
(94524 kPa)*(0.6082 m^2) = 57489.7 kN = 57.5 MN approx
3 0
3 years ago
A wire loop of radius 0.37 m lies so that an external magnetic field of magnitude 0.35 T is perpendicular to the loop. The field
Galina-37 [17]

Answer:

168.57 mV

Explanation:

Initial magnetic flux = BA , B magnetic field and A is area of loop

= .35 x 3.14 x .37²

= .15 Weber

Final magnetic flux

= - .2 x 3.14 x .37²

= -  .086 Weber

change in flux

.15 +  .086

= .236 Weber

rate of change of flux

= .236 / 1.4

= .16857 V

= 168.57 mV

5 0
3 years ago
A medical cyclotron used in the production of medical isotopes accelerates protons to 6.5 MeV. The magnetic field in the cyclotr
mart [117]

Answer:

diameter of largest orbit is 0.60 m

Explanation:

given data

isotopes accelerates KE = 6.5 MeV

magnetic field B = 1.2 T

to find out

diameter

solution

first we find velocity from kinetic energy equation

KE = 1/2 × m×v²   ........1

6.5 × 1.6 × 10^{-19} = 1/2 × 1.672 × 10^{-27} ×v²

v = 3.5 × 10^{7} m/s

so

radius will be

radius = \frac{m*v}{B*q}   ........2

radius =  \frac{1.672*10^{-27}*3.5*10^{7}}{1.2*1.6*10^{-19}}  

radius = 0.30

so diameter = 2 × 0.30

so diameter of largest orbit is 0.60 m

8 0
3 years ago
Read 2 more answers
Select the correct answer.
skelet666 [1.2K]

Answer:

B is the best answer for the question

6 0
3 years ago
Read 2 more answers
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