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Orlov [11]
2 years ago
15

Water is boiled in a pan on a stove at sea level. During 10 min of boiling, it is observed that 200 g of water has evaporated. D

etermine the rate of heat transfer to the water. Solve this using appropriate software. Use data from the tables.
Physics
1 answer:
Ymorist [56]2 years ago
4 0

Answer : The rate of heat transfer to the water is, 37.92 kJ/min

Explanation : Given,

Time = 10 min

Mass of water = 200 g

Latent heat of fusion of water = 334 J/g

Latent heat of vaporization of water = 2230 J/g

Now we have to calculate the rate of heat transfer to the water.

Q=\frac{m\times (L_v-L_f)}{t}

Now put all the given values in the above formula, we get:

Q=\frac{200g\times (2230-334)J/g}{10min}

Q=37920J/min=37.92kJ/min

Thus, the rate of heat transfer to the water is, 37.92 kJ/min

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Dvinal [7]
At the rear.

PWC stands for personal watercraft, and it is a small powerboat. The main components of a PWC are the hull (body of the boat), deck (surface where people walk/stand), throttle (controls speed), steering nozzle and water intake.
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2 years ago
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Two electrodes, separated by a distance d, in a vacuum are maintained at a constant potential difference. An electron, accelerat
Alja [10]

Answer:

Explanation:

Given that, the distance between the electrode is d.

The electron kinetic energy is Ek when the electrode are at distance "d" apart.

So, we want to find the K.E when that are at d/3 distance apart.

K.E = ½mv²

Note: the mass doesn't change, it is only the velocity that change.

Also,

K.E = Work done by the electron

K.E = F × d

K.E = W = ma × d

Let assume that if is constant acceleration

Then, m and a is constant,

Then,

K.E is directly proportional to d

So, as d increase K.E increase and as d decreases K.E decreases.

So,

K.E_1 / d_1 = K.E_2 / d_2

K.E_1 = E_k

d_1 = d

d_2 = d/3

K.E_2 = K.E_1 / d_1 × d_2

K.E_2 = E_k × ⅓d / d

Then,

K.E_2 = ⅓E_k

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7 0
2 years ago
What is the binding energy (in J/mol or kJ/mol) of an electron in a metal whose threshold frequency for photoelectrons is 2.50 u
Aleksandr-060686 [28]

Answer:

binding energy is 99771 J/mol

Exlanation:

given data

threshold frequency = 2.50 ×  10^{14} Hz

solution

we get here binding energy using threshold frequency of the metal that is express as

E=h\nu_{o}    ..................1

here E is the energy of electron per atom E=\frac{x}{N}  and h is plank constant i.e. 6.626\times10^{-34} Js  and x is  binding energy

and here N is the Avogadro constant = 6.023\times10^{23}

so E will \frac{x}{6.023\times10^{23}}  

E = 3.19\times10^{-19}  J  

so put value in equation 1 we get

\frac{x}{6.023\times10^{23}} = 2.50 ×  10^{14} × 6.626\times10^{-34} Js  

solve it we get

x = 99770.99

so  binding energy is 99771 J/mol

4 0
2 years ago
Solve A and B using energy considerations.
Alisiya [41]
 <span>Use the kinematic equation vf^2 = vi^2 + 2ad where; 
vf = ? 
vi = 0 m/s 
a = 9.8 m/s^2 
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final velocity at the ground (d1): vf = sqrt(2)(9.8)(10) = 14 m/s 
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7 0
2 years ago
A 0.600-mm diameter wire stretches 0.500% of its length when it is stretched with a tension of 20.0 n. what is the young's modul
Rashid [163]
The Young modulus is given by:
E= \frac{F /A}{\Delta L / L_0}
where
F is the force applied
L_0 is the initial length of the wire
A is the cross-sectional area of the wire
\Delta L is the stretch of the wire

The wire in the problem stretches by 0.5% of its length, this means 
\frac{\Delta L}{L_0}  = 0.005

We can also calculate the area of the wire; its radius is in fact half the diameter:
r= \frac{d}{2}= \frac{0.600 mm}{2}=0.300 mm=0.3 \cdot 10^{-3} m
and so the area is
A=\pi r^2 = \pi (0.3 \cdot 10^{-3} m)^2 = 2.83 \cdot 10^{-7} m^2

We know the force applied to the wire, F=20 N, so now we have everything to calculate the Young modulus:

E=  \frac{F/A}{\Delta L / L_0} = \frac{20 N/(2.83 \cdot 10^{-7} m^2)}{0.005}=1.42 \cdot 10^{10} N/m^2
3 0
1 year ago
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