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Firdavs [7]
3 years ago
11

Particles in a __________ wave move in an elliptical or circular motion. question 4 options: sound longitudinal transverse surfa

ce
Physics
2 answers:
liubo4ka [24]3 years ago
7 0

Answer: The correct answer is "surface".

Explanation:

The particles oscillate to the parallel to the direction of the energy transportation in the longitudinal wave. For example, sound wave.

Transverse wave: The particles oscillate to the perpendicular to the direction of the energy transportation. For example, light wave.

Surface wave: This wave is neither transverse wave or longitudinal wave. It moves both parallel and perpendicular to the direction of the propagation of the wave. It moves in a circular motion to the direction of the energy transportation.

Therefore, Particles in a surface wave move in an elliptical or circular motion.

VikaD [51]3 years ago
3 0
The sentence can be completed as follows:
"Particles in a SURFACE wave move in an elliptical or circular motion". So Answer D is the correct one.

<span>Explanation: surface waves are waves that move in circular or elliptical motion at the interface between two mediums (for example, between water and air). An example of surface wave is the wave obtained when you throw a stone in the water: the wave produced when the stone touches the water is a surface wave, that moves with concentrical circles around the source.</span>
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In a series circuit, each circuit element has the same current. The answer is letter A. This is because the electrons flowing in a current has only a single path to follow. There are no other path that the current moves along.

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4 years ago
If you plan to exercise every day what is something you can do to make it more convenient so that you exercise more regularly?
jolli1 [7]

Answer: exercise doing something you like. for example if you like dancing incorporate dancing in your workouts so you will be more likely to do it regularly

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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
A solenoid with 3,000.0 turns is 70.0 cm long. If its self-inductance is 25.0 mH, what is 5) its radius? (The value of o is 4 ×
Elan Coil [88]
The self-inductance of a solenoid is given by:
L= \frac{\mu_0 N^2 A}{l}
where
\mu_0 is the vacuum permeability
N is the number of turns
A is the cross-sectional area of the solenoid
l is the length of the solenoid

For the solenoid in our problem, N=3000, l=70.0 cm=0.70 m and the self-inductance is L=25.0 mH=0.025 H, therefore the cross-sectional area is
A= \frac{Ll}{\mu N^2}= \frac{(0.025 H)(0.70 m)}{(4\pi \cdot 10^{-7}N/A^2)(3000)^2}= 1.55 \cdot 10^{-3}m^2
And since the area is related to the radius by
A=\pi r^2
The radius of the solenoid is
r= \sqrt{ \frac{A}{\pi} } = \sqrt{ \frac{1.55 \cdot 10^{-3} m^2}{\pi} } =0.022 m=2.2 cm

6 0
3 years ago
Which of the following describes inertia?
Brrunno [24]
<span>a resistance to change in motion</span>
5 0
3 years ago
Read 2 more answers
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