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1 year ago

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A 15.0-gram lead ball at 25.0°C was heated with 40.5 joules of heat. Given the specific heat of lead is 0.128 J/g∙°C, what is th

e final temperature of the lead?

1 answer:

mr Goodwill [35]1 year ago

3 0

Answer:

$T=4985.5^{\circ}K$

Explanation:

The equation that relates heat Q with the temperature change $T-T_0$ of a substance of mass <em>m </em>and specific heat <em>c </em>is $Q=mc(T-T_0)$ .

We want to calculate the final temperature <em>T, </em>so we have:

$T=\frac{Q}{mc}+T_0$

Which for our values means (in this case we do not need to convert the mass to Kg since <em>c</em> is given in g also and they cancel out, but we add $273^{\circ}$ to our temperature in $^{\circ}C$ to have it in $^{\circ}K$ as it must be):

$T=\frac{Q}{mc}+T_0=\frac{40.5J}{(15g)(0.128J/g^{\circ}C)}+(298^{\circ}K)=4985.5^{\circ}K$

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zhenek [66]

Complete Question

The complete question is shown on the first uploaded image

Answer:

Force on each hand is 196.22 N

Force on each foot is 95.8 N

Explanation:

In order to get a better understanding of this question let us explain some concepts

Normal Force:

We can define normal force Fn as that type of force which makes a 90 degree angle with the surface on which it is exerted.

Torque:

We can define torque as the moment of forces that tends to produce or cause rotation

From the question we are given that

Weight of body is (W) = 584 N

The normal force on both hands (Ha) = ?

The normal force on both legs (Lg) = ?

Looking at the diagram the person is at equilibrium so

584 = Ha + Lg

an also this mean that torques acting on the body is balanced

So, 0.410 Ha = 0.840 Lg

Making Lg the subject of formula in the equation above we

Lg = 0.4881 Ha

Considering the first equation and replacing Lg with this recent equation we have

584 = Ha + 0.4881 Ha

Therefore Ha = 392.44 N

This value obtained is for both hands for each hand we divide by 2

Therefore we have for each hand $= 392.44/2$ =196.55 N

Since we have been able to get the force on both hands we can substitute it in to the equation where we made Lg the subject of formula and we have

Lg = 0.4881 × 392.44

= 191.22 N

The value above is the force on both legs to obtain the force on each leg we have

191.22/2 = 95.8 N.

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There are other forces at work here nevertheless we will imagine it is just a conservation of momentum exercise. Also the given mass of the astronaut is light astronaut.

The solution for this problem is using the formula: m1V1=m2V2 but we need to get V1:

V1= (m2/m1) V2

V1= (10/63) 12 = 1.9 m/s will be the final speed of the astronaut after throwing the tank.

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A Micro –Hydro turbine generator is accelerating uniformly from an angular velocity of 610 rpm to its operating angular velocity

Salsk061 [2.6K]

Answer:

Angular displacement of the turbine is 234.62 radian

Explanation:

initial angular speed of the turbine is

$\omega_i = 2\pi f_1$

$\omega_1 = 2\pi(\frac{610}{60})$

$\omega_1 = 63.88 rad/s$

similarly final angular speed is given as

$\omega_f = 2\pi f_2$

$\omega_2 = 2\pi(\frac{837}{60})$

$\omega_2 = 87.65 rad/s$

angular acceleration of the turbine is given as

$\alpha = 5.9 rad/s^2$

now we have to find the angular displacement is given as

$\theta = \omega t + \frac{1}{2}\alpha t^2$

$\theta = (63.88)(3.2) + (\frac{1}{2})(5.9)(3.2^2)$

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Answer:

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Explanation:

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A wave travels through a medium at 251 m/s and has a wavelength of 5.10 cm. What is its frequency? What is its angular frequency

allochka39001 [22]

Explanation:

It is given that,

Speed of a wave, v = 251 m/s

Wavelength of the wave, λ = 5.1 cm = 0.051 m

(1) The frequency of the wave is given by :

$\nu=\dfrac{v}{\lambda}$

$\nu=\dfrac{251\ m/s}{0.051\ m}$

$\nu=4921.56\ Hz$

(2) Angular frequency of the wave is given by :

$\omega=2\pi\nu$

$\omega=2\pi\times 4921.56\ Hz$

$\omega=30923.07\ rad/s$

(3) The period of oscillation is given by T as :

$T=\dfrac{1}{\nu}$

$T=\dfrac{1}{4921.56}$

T = 0.000203 seconds

or

T = 0.203 milliseconds

Hence, this is the required solution.

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