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9966 [12]
3 years ago
10

A smith needs to melt 0.0500 kg of gold at 21.0°C. How much heat must be added? (Remember, she has to heat it to the melting poi

nt first.) (Unit=J)

Physics
1 answer:
baherus [9]3 years ago
4 0

Answer:

9704.6 J

Explanation:

Total Thermal Energy

= Energy required to bring gold to melting point + Energy required to change the state of gold from solid to liquid

= mcT + <em>m</em><em>l</em><em>f</em><em> </em><em> </em>[bolded is energy used to bring gold to melting point, <em>i</em><em>t</em><em>a</em><em>l</em><em>i</em><em>c</em><em>i</em><em>s</em><em>e</em><em>d</em><em> </em><em>i</em><em>s</em><em> </em><em>s</em><em>t</em><em>a</em><em>t</em><em>e</em><em> </em><em>c</em><em>h</em><em>a</em><em>n</em><em>g</em><em>e</em><em> </em><em>o</em><em>f</em><em> </em><em>g</em><em>o</em><em>l</em><em>d</em><em> </em><em>f</em><em>r</em><em>o</em><em>m</em><em> </em><em>s</em><em>o</em><em>l</em><em>d</em><em> </em><em>t</em><em>o</em><em> </em><em>l</em><em>i</em><em>q</em><em>u</em><em>i</em><em>d</em><em>]</em>

= (0.0500)(126)(1063 - 21) + <em>(</em><em>0</em><em>.</em><em>0</em><em>5</em><em>0</em><em>0</em><em>)</em><em>(</em><em>6</em><em>.</em><em>2</em><em>8</em><em> </em><em>×</em><em> </em><em>1</em><em>0</em><em>^</em><em>4</em><em>)</em>

= <u>9704.6</u><u> </u><u>J</u>

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imagine that a tank is filled with water the hight of the liquid column is 7 meters and the area is 1.5 sq meters (m™). what's t
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Answer - 12,900 Newtons

Explanation

First, we find the volume of the water
Volume = Area * Heinght
= 1.5 m² x 7 m
=  10.5 m³
 
Covert the volume to liters
1 m³ of water = 1000  liters
10.5 m³ of water = 10.5 m³ * 1000 liters liter/m³
= 10,500 liters
 
Use the volume of water to calculate the mass
1 liter of water weighs 1 kg
10,500 liters of water = 10,500 * 1 kg/liter
= 10,500 kg
 
Now, we can calculate the force of gravity on the water
Force of gravity on the water = Weight of the water
Weight = Mass * Acceleration
 
Mass = 10,500kg
Acceleration (due to gravity) = 9.8 m/s²

Force of gravity on the water
= Weight of the water
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8 0
3 years ago
A fluid is flowing through a circulat tube at 0.4 kg/s. Tube inner surface is smooth with a diameter 0.014 m. Fluid density is 9
ElenaW [278]

Answer:

The convection coefficient is 15456.48\ W/m^{2}K

Solution:

Mass flow rate, \dot{m} = 0.4\ kg

Inner diameter of the tube, d = 0.014 m

Fluid density, \rho_{f} = 990\ kg/m^{3}

Specific Heat, C = 3845 J/K

Thermal Conductivity, K = 0.74

Prandtl Number, P_{r} = 8.6

Heat flux, \dot{q} = 71,297\ W/m^{2}

Viscosity, \mu = 0.00079\ Ns/m^{2}

Now,

To calculate the convection heat coefficient, h:

Determine the cross sectional area of the circular tube:

A_{c} = \frac{\pi}{4}d^{2} = \frac{\pi}{4}\times (0.014)^{2} = 1.54\time 10^{- 4}\ m^{2}

Determine the velocity of the fluid inside the tube by mass flow rate:

\dot{m} = \rho_{f}A_{c}v

0.4 = 990\times 1.54\time 10^{- 4}v

v = 2.624 m/s

Determine the Reynold's Number, R_{e}:

R_{e} = \frac{\rho_{f}dv}{\mu}

R_{e} = \frac{990\times 0.014\times 2.624}{0.00079} = 46036.253

Thus it is clear that R_{e} > 10,000 hence flow is turbulent.

Now,

Determine the Nusselt Number:

N_{u} = 0.023R_{e}^{0.8}P_{r}^{0.4}

N_{u} = 0.023\times 46036.253^{0.8}\times 8.6^{0.4} = 292.42

Also,

N_{u} = \frac{dh}{K}

where

h = convection coefficient

Now,

292.42 = \frac{0.014\times h}{0.74}

h = 15456.48\ W/m^{2}K

7 0
3 years ago
Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg
vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

\omega = \frac{v}{R}

Where,

\omega =Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

\alpha = \frac{\omega}{t}

Where

\alpha =Angular acceleration

\omega = Angular velocity

t = Time

Our values are

v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})

v = 16.67m/s

r = 0.25m

t=6s

Replacing at the previous equation we have that the angular velocity is

\omega = \frac{v}{R}

\omega = \frac{ 16.67}{0.25}

\omega = 66.67rad/s

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

\alpha = \frac{\omega}{t}

\alpha = \frac{66.67}{6}

\alpha = 11.11rad/s^2

Therefore the angular acceleration of a point on the outer edge of the tires is 11.11rad/s^2

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