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ozzi
3 years ago
12

Noise pollution can lead to the disruption of speech communication in children true or false

Physics
2 answers:
LenaWriter [7]3 years ago
7 0
False, because I doesn’t matter if there is noise pollution the child will still able to learn the way words work
balu736 [363]3 years ago
7 0

The correct answer is False. Glad to be of help.

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Which is not a characteristic of an ideal fluid?
Lena [83]
The answer would be B
7 0
3 years ago
A rocket initially at rest accelerates at a rate of 99.0 meters/second2. Calculate the distance covered by the rocket if it atta
Jobisdone [24]

Answer: The correct answer is option b.

Explanation: We are given that the rocket is at rest initially final velocity is 445m/s.

The acceleration of the rocket is 99.0m/s^2

To calculate the distance of rocket, we use third equation of motion, which is:

v^2-u^2=2as

where, v = final velocity = 445m/s

u = initial velocity = 0m/s

a = acceleration = 99m/s^2

s = distance = ? m

Putting values in above equation, we get:

(445)^2-(0)^2=2(99)s\\\\s=1\times 10^3meters

3 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
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Answer: 0.000346 Nm

Explanation:

T = u X B

u = i x A = magnetic moment

T = i x A x B x sin(30)

T = 0..48 x 0.049^2 x 0.6 x 0.5 = 0.000346 Nm

6 0
4 years ago
¿Cuál es la velocidad del sonido en el aíre a -25°C?
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La velocidad del sonido en el aire (a una temperatura de 20 ºC) es de 343 m/s. La ecuación creada por Newton y posteriormente modificada por Laplace que permite obtener la velocidad del sonido en el aire teniendo en cuenta la variable de la temperatura es "331+(0,6 x Temperatura)".

               

8 0
3 years ago
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