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3 years ago

11

The big bang theory suggests that our universe formed as the result of a huge explosion that sent all existing matter flying out

ward from a single point. The big bang theory is supported by which of these observations?.

1 answer:

marin [14]3 years ago

4 0

Answer:

All galaxies appear to be moving away from all other galaxies

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In certain ranges of a piano keyboard, more than one string istuned to the same note to provide extra loudness. For example, the

steposvetlana [31]

Answer:

The beat frequency is 5.5 Hz.

Explanation:

f = 110 Hz, T = 600 N , T' = 540 N

let the frequency is f'.

$\frac{f'}{f}=\sqrt\frac{T'}{T}\\\\\frac{f'}{f}=\sqrt\frac{540}{600}\\\\\frac{f'}{f}=0.95\\\\f'= 104.5 Hz$

So, the beat frequency is f - f' = 110 - 104.5 = 5.5 Hz

6 0

3 years ago

A two-liter bottle of your favorite beverage has just been removed from the trunk of your car. The temperature of the beverage i

Ksivusya [100]

Answer:

a) 209.3 kilojoules must be removed from two liter of beverage, b) A rate of heat removal of 1.163 kilowatts is required to cool down 10 2-liter bottles, c) Cooling 10 2-L bottles during 30 minutes costs 4.9 cents.

Explanation:

a) <em>How much heat energy must be removed from your two liters of beverage?</em>

At first we suppose that the beverage has the mass and specific heat of water and that there are no energy interactions between the bottle and its surroundings.

From the First Law of Thermodynamics and definition of sensible heat, we get that amount of removed heat ( $Q$ ), measured in kilojoules, is represented by the following formula:

$Q = \rho \cdot V\cdot c\cdot (T_{o}-T_{f})$ (Eq. 1)

Where:

$\rho$ - Density of the beverage, measured in kilograms per cubic meter.

$V$ - Volume of the bottle, measured in cubic meters.

$c$ - Specific heat of water, measured in kilojoules per kilogram-Celsius.

$T_{o}$ , $T_{f}$ - Initial and final temperatures, measured in Celsius.

If we know that $\rho = 1000\,\frac{kg}{m^{3}}$ , $V = 2\times 10^{-3}\,m^{3}$ , $c = 4.186\,\frac{kJ}{kg\cdot ^{\circ}C}$ , $T_{o} = 35\,^{\circ}C$ and $T_{f} = 10\,^{\circ}C$ , then:

$Q = \left(1000\,\frac{kg}{m^{3}}\right)\cdot (2\times 10^{-3}\,m^{3})\cdot \left(4.186\,\frac{kJ}{kg\cdot ^{\circ}C} \right) \cdot (35\,^{\circ}C-10\,^{\circ}C)$

$Q = 209.3\,kJ$

209.3 kilojoules must be removed from two liter of beverage.

b) <em>You are having a party and need to cool 10 of these two-liter bottles in one-half hour. What rate of heat removal, in kW, is required?</em>

The total amount of heat that must be removed from 10 2-L bottles is:

$Q_{T} = 10\cdot (209.3\,kJ)$

$Q_{T} = 2093\,kJ$

If we suppose that bottles are cooled at constant rate, then, rate of heat removal is determined by this formula:

$\dot Q = \frac{Q_{T}}{\Delta t}$ (Eq. 2)

Where:

$Q_{T}$ - Total heat, measured in kilojoules.

$\Delta t$ - Time, measured in seconds.

$\dot Q$ - Rate of heat removal, measured in kilowatts.

If we know that $Q_{T} = 2093\,kJ$ and $\Delta t = 1800\,s$ , we find that rate of heat removal is:

$\dot Q = \frac{2093\,kJ}{1800\,s}$

$\dot Q = 1.163\,kW$

A rate of heat removal of 1.163 kilowatts is required to cool down 10 2-liter bottles.

c) <em>Assuming that your refrigerator can accomplish this and that electricity costs 8.5 cents per kW-hr, how much will it cost to cool these 10 bottles (in $)?</em>

A kilowatt-hour equals 3600 kilojoules. The electricity cost is equal to the removal heat of 10 bottles ( $Q_{T}$ ), measured in kilojoules, and unit electricity cost ( $c$ ), measured in US dollars per kilowatt-hour. That is:

$C = c\cdot Q_{T}$

If we know that $c = 0.085\,\frac{USD}{kWh}$ and $Q_{T} = 2093\,kJ$ , the total cost of cooling 10 bottles is:

$C = \left(0.085\,\frac{USD}{kWh}\right)\cdot \left(2093\,kJ\right)\cdot \left(\frac{1}{3600}\,\frac{kWh}{kJ} \right)$

$C = 0.049\,USD$

Cooling 10 2-L bottles during 30 minutes costs 4.9 cents.

3 0

4 years ago

Two identical balls each have a mass of 35.0 grams and a charge of . The balls are released from rest when they are separated by

OlgaM077 [116]

The given question is incomplete. The complete question is as follows.

Two identical balls each have a mass of 35.0 grams and a charge of q = 3.50 \times 10^{-6}C[/tex]. The balls are released from rest when they are separated by a distance of 6.00 cm. What is the speed of each ball when the distance between them has tripled? Use k = $9.00 \times 10^{9} Nm^{2}/C^{2}$ .

Explanation:

According to the conservation of energy, the formula will be as follows.

$\frac{kq_{1}q_{2}}{r_{1}} = \frac{kq_{1}q_{2}}{(3r_{1})} + \frac{1}{2}mv^{2} + \frac{1}{2}mv^{2}$

or, $\frac{kq_{1}q_{2}}{r_{1}}[1 - \frac{1}{3}] = mv^{2}$

Putting the given values into the above formula as follows.

$\frac{kq_{1}q_{2}}{r_{1}}[1 - \frac{1}{3}] = mv^{2}$

$\frac{9 \times 10^{9} \times (3.5 \times 10^{-6})^{2}}{0.09} \times \frac{2}{3} = \frac{35}{1000}v^{2}$

$v^{2}$ = 23.333

v = 4.83 m/s

Thus, we can conclude that speed of each ball when the distance between them has tripled is 4.83 m/s.

7 0

4 years ago

A group of students put a ford focus in neutral and push the car with 500 N of force to the right, causing the car to travel a d

SOVA2 [1]

50 meters is the answer

4 0

3 years ago

A spring hangs from the ceiling with an unstretched length of x0=0.45 m . A m1=7.9 kg block is hung from the spring, causing the

andrezito [222]

Answer:0.126 meters

Explanation:see attached photo

6 0

3 years ago