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2 years ago

3.Cuanto Calor pierden 514 ml de agua si su temperatura desciende de 12°C a 11°C. Expresa el resultado en calorias.

Physics

1 answer:

Scrat [10]2 years ago

3 0

Answer:

514 cal

Explanation:

In order to calculate the lost heat by the amount of water you first take into account the following formula:

$Q=mc(T_2-T_1)$ (1)

Q: heat lost by the amount of water = ?

m: mass of the water

c: specific heat of water = 1cal/g°C

T2: final temperature of water = 11°C

T1: initial temperature = 12°C

The amount of water is calculated by using the information about the density of water (1g/ml):

$m=\rho V=(1g/ml)(514ml)=514g$

Then, you replace the values of all parameters in the equation (1):

$Q=(514g)(1cal/g\°C)(11\°C-12\°C)=-514cal$

The amount of water losses a heat of 514 cal

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Given:
Ca = 3Cb (1)
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Ca = heat capacity of object A
Cb = heat capacity f object B

Also,
Ta = 2Tb (2)
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Ta = initial temperature of object A
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Let
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Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
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Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
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Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb

Answer:
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where
$k= \frac{M_{a}}{M_{b}}$

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2 years ago

Two forces F1 and F2 act on a 5.00 kg object. Taking F1=20.0N and F2=15.00N, find the acceleration of the object for the configu

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A) mass m with F1 acting in the positive x direction and F2 acting perpendicular in the positive y direction

m = 5.00 kg
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F2=15.00N ... y direction

Net force ^2 = F1^2 + F2^2 = (20N)^2 + (15n)^2 = 625N^2 =>

Net force = √625 = 25N

F = m*a => a = F/m = 25.0 N /5.00 kg = 5 m/s^2

Answer: 5.00 m/s^2

b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal. 

m = 5.00 kg
F1=20.0N ... x direction
F2=15.00N ... 60 degress above x direction

Components of F2

F2,x = F2*cos(60) = 15N / 2 = 7.5N

F2, y = F2*sin(60) = 15N* 0.866 = 12.99 N ≈ 13 N

Total force in x = F1 + F2,x = 20.0 N + 7.5 N = 27.5 N

Total force in y = F2,y = 13.0 N

Net force^2 = (27.5N)^2 + (13.0N)^2 = 925.25 N^2 = Net force = √(925.25N^2) =

= 30.42N

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Answer: 6.08 m/s^2

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1 year ago

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The battery capacity of a lithium ion battery in a digital music player is 750 mA-h. The manufacturer claims that the player can

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Answer:

The number of electrons is $6.3\times10^{21}\ electrons$

(D) is correct option.

Explanation:

Given that,

Battery capacity = 750 mA-h

Time t= 8 hours

Time t'=3 hours

We need to calculate the battery capacity

$Battery\ capacity=750\times10^{-3}\times3600$

$Battery\ capacity=2700\ A-s$

We need to calculate the number of electrons in 1 C Li

Using formula for number of electron

$n=\dfrac{1}{e}$

$n=\dfrac{1}{1.6\times10^{-19}}$

$n=6.25\times10^{18}\ electrons$

We need to calculate the number of electron in 2700 C

$2700\ C=2700\times6.25\times10^{18}=1.68\times10^{22}\ electrons$

The total number of electrons battery can deliver in 8 hours

$n=1.68\times10^{22}\ electrons$

We need to calculate the number of electron in 3 hours

Using formula of number of electrons

$n=\dfrac{n\times t'}{t}$

Put the value into the formula

$n=\dfrac{1.68\times10^{22}\times3}{8}$

$n=6.3\times10^{21}\ electrons$

Hence, The number of electrons is $6.3\times10^{21}\ electrons$

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2 years ago

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The Lamborghini Huracan has an initial acceleration of 0.85g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode al

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Answer:

7.9 $\frac{m}{s^{2} }$

Explanation:

Take the fact that mass is inversely proportional to accelertation:

m ∝ a

Therefore m = a, but because we are finding the change in acceleration, we would set our problem up to look more like this:

$\frac{m_{1} }{m_{2} } = \frac{a_{2} }{a_{1} } \\$

Using algebra, we can rearrange our equation to find the final acceleration, $a_{2}$ :

$a_{2} = \frac{a_{1}*m_{1} }{m_{2} } \\$

Before plugging everything in, since you are being asked to find acceleration, you will want to convert 0.85g to m/s^2. To do this, multiply by g, which is equal to 9.8 m/s^2:

0.85g * 9.8 $\frac{m }{s^{2} }$ = 8.33 $\frac{m }{s^{2} }$

Plug everything in:

7.9 $\frac{m }{s^{2} }$ = $\frac{ 8.33\frac{m}{s^{2} }*1510kg }{1590kg}$

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1 year ago

Light with a wavelength of 495 nm is falling on a surface and electrons with a maximum kinetic energy of 0.5 eV are ejected. Wha

devlian [24]

Answer:

To increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

Explanation:

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$W+K_{max}=\frac{hc}{\lambda}\\\lambda=\frac{hc}{W+K_{max}}\\\lambda=\frac{(4.14*10^{-15}eV\cdot s)(3*10^8\frac{m}{s})}{2.01eV+1.5eV}\\\lambda=3.54*10^{-7}m=354*10^{-9}m=354nm$

This wavelength corresponds to ultraviolet radiation. So, to increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.

8 0

2 years ago