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alexandr402 [8]
2 years ago
7

Calculate the number of moles of water molecules in 12 dm' of watervapour at STP.​

Physics
1 answer:
Vinvika [58]2 years ago
4 0

Answer:

22.4 \:  {dm}^{3}  \: are \: occupied \: by \: 1 \: mole \\ 12 \:  {dm}^{3}  \: will \: be \: occupied \: by \: (  \frac{12}{22.4} ) \: moles \\  = 0.536 \: moles

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At a constant pressure, melting occurs by A) producing energy. B) absorbing energy. C) releasing energy. D) circulating energy.
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Realizing energy.......
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En una fundición hay un horno eléctrico con capacidad para fundir totalmente 540 kg de cobre. Si la temperatura inicial del cobr
11Alexandr11 [23.1K]

Answer:

Q = 2.95*10^5 kJ

Explanation:

In order to calculate the energy required to melt the cooper, you first calculate the energy required to reach the boiling temperature. You use the following formula:

Q_1=mc(T_b-T_1)     (1)

m: mass of cooper = 540 kg

c: specific heat of cooper = 390 J/kg°C

Tb: boiling temperature of cooper = 1080°C

T1: initial temperature of cooper = 20°C

You replace the values of the parameters in the equation (1):

Q_1=(540kg)(390\frac{J}{kg.\°C})(1080\°C-20\°C)=2.23*10^8J

Next, you calculate the energy required to melt the cooper by using the following formula:

Q_2=mL_f         (2)

Lf: melting constant of cooper = 134000J/kg

Q_2=(540kg)(134000\frac{J}{kg})=7.24*10^7J

Finally, the total amount of energy required to melt the cooper from a temperature of 20°C is the sum of Q1 and Q2:

Q=Q_1+Q_2=2.23*10^8J+7.24*10^7J=2.95*10^8J=2.95*10^5kJ

The total energy required is 2.95*10^5 kJ

3 0
3 years ago
A push broom is being pushed down across a rough floor. The broom moves to the right. What is the correct free body diagram of t
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The far right.

Fg is gravity which always acts down and since we assume the floor is flat the normal, Fn, acts opposite gravity, so straight up.

But you’re probably wondering about the pushing force, Fp, and the friction force, Ff. For the Fp, consider where the applied force is coming from. The head of the broom is on the floor and the man’s arms, where he’s applying the force from, is above and to the left, so when the man pushes the broom the force is down and to the right. The broom my not be moving down, but the applied force is still in that direction. And Ff always acts against motion so since the broom moves to the right, the friction is to the left.

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A notebook sitting on a desk has 19 j of potential energy. What is the total mechanical energy of the notebook?
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The total mechanical energy of the notebook is <u><em>19J</em></u>.

Mechanical energy is the sum of potential energy and kinetic energy.  It has no kinetic energy, because it's not moving.  So its potential energy is all the  mechanical energy it has.

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The 100-m dash can be run by the best sprinters in 10.0 s. A 66-kg sprinter accelerates uniformly for the first 45 m to reach to
irga5000 [103]

(a) 154.5 N

Let's divide the motion of the sprinter in two parts:

- In the first part, he starts with velocity u = 0 and accelerates with constant acceleration a_1 for a total time t_1 During this part of the motion, he covers a distance equal to s_1 = 45 m, until he finally reaches a velocity of v_1 = u + a_1t_1. We can use the following suvat equation:

s_1 = u t_1 + \frac{1}{2}a_1t_1^2

which reduces to

s_1 = \frac{1}{2}a_1 t_1^2 (1)

since u = 0.

- In the second part, he continues with constant speed v_1 = a_1 t_1, covering a distance of d_2 = 55 m in a time t_2. This part of the motion is a uniform motion, so we can use the equation

s_2 = v_1 t_2 = a_1 t_1 t_2 (2)

We also know that the total time is 10.0 s, so

t_1 + t_2 = 10.0 s\\t_2 = (10.0-t_1)

Therefore substituting into the 2nd equation

s_2 = a_1 t_1 (10-t_1)

From eq.(1) we find

a_1 = \frac{2s_1}{t_1^2} (3)

And substituting into (2)

s_2 = \frac{2s_1}{t_1^2}t_1 (10-t_1)=\frac{2s_1}{t_1}(10-t_1)=\frac{20 s_1}{t_1}-2s_1

Solving for t,

s_2+2s_1=\frac{20 s_1}{t_1}\\t_1 = \frac{20s_1}{s_2+2s_1}=\frac{20(45)}{55+2(45)}=6.2 s

So from (3) we find the acceleration in the first phase:

a_1 = \frac{2(45)}{(6.2)^2}=2.34 m/s^2

And so the average force exerted on the sprinter is

F=ma=(66 kg)(2.34 m/s^2)=154.5 N

b) 14.5 m/s

The speed of the sprinter remains constant during the last 55 m of motion, so we can just use the suvat equation

v_1 = u +a_1 t_1

where we have

u = 0

a_1  =2.34 m/s^2 is the acceleration

t_1 = 6.2 s is the time of the first part

Solving the equation,

v_1 = 0 +(2.34)(6.2)=14.5 m/s

3 0
3 years ago
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