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goldfiish [28.3K]
3 years ago
12

Which one of the following is not equivalent to 2.50 miles?

Physics
1 answer:
lisabon 2012 [21]3 years ago
6 0

Answer:

Choice C is not equivalent to 2.50 miles.

Explanation:

The given data is now converted into feet, inches, kilometers, yards and centimeters:

mi - ft

x = (2.50\,mi)\cdot \left(5280\,\frac{ft}{mi} \right)

x = 13200\,ft

x = 1.320\times 10^{4}\,ft (Choice A)

mi - in

x = (2.50\,mi)\cdot \left(5280\,\frac{ft}{mi} \right)\cdot \left(12\,\frac{in}{ft} \right)

x = 158400\,in

x = 1.584\times 10^{5} (Choice B)

mi - km

x = (2.50\,mi)\cdot \left(1.61\,\frac{km}{mi} \right)

x = 4.025\,km (Different from Choice C)

mi - yd

x = (2.50\,mi)\cdot \left(5280\,\frac{ft}{mi}\right) \cdot \left(\frac{1}{3}\,\frac{yd}{ft}  \right)

x = 4400\,yd

x = 4.40\times 10^{3}\,yd (Choice D)

mi - cm

x = (2.50\,mi)\cdot \left(1.61\,\frac{km}{mi} \right)\cdot \left(100000\,\frac{cm}{km}\right)

x = 402500\,cm

x = 4.025\times 10^{5}\,cm (Choice E)

Choice C is not equivalent to 2.50 miles.

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Answer:

 

D)  

The energy of the wave moves perpendicular to the direction the wave travels.

Explanation:

I had this question on USA test prep and this was the correct answer

7 0
3 years ago
For an object like a planet, with a typical temperature of a few hundred kelvin, what kind of blackbody radiation would it princ
navik [9.2K]

Answer:

Low-temperature blackbody

Explanation:

There are 3 types of blackbody temperatures.

Low-temperature blackbody

High temperature extended area blackbody

High-temperature cavity blackbody

A Low-temperature blackbody is a type of black body radiation that has the range of -40° C to 175° C, typically between 233 K and 448 K. A perfect fit for the temperature range mentioned in the question, "a few hundred Kelvin". Therefore, it's the kind of blackbody temperature that the object would emit.

6 0
3 years ago
Un recipiente contiene 224 dm3 de Ozono de masa 4.561 Kg a 51.09 grados celsius. Calcula la presión del Ozono
kari74 [83]

Answer:

Por lo tanto, la presión del ozono es:

P=0.011\: atm  

Explanation:

Podemos usar la ecuacion de los gases ideales;

PV=nRT (1)

Tenemos:

El volumen V = 224 dm³ = 224 L

La temperatura T = 51.09 C = 324.09 K

La masa es m = 4.561 kg

Lo necesitamos ahora es calvular n que es el numero de moles;

recordemos que el peso molecular del ozono M = 48 g/mol.

n=\frac{m}{M}=\frac{4.561}{48}=0.095\: mol

Finalmente, usando la ecuacion 1 despejamos la presion P

P=\frac{nRT}{V}

P=\frac{0.095*0.082*324.09}{224}  

Por lo tanto, la presion del ozono es:

P=0.011\: atm  

Espero te haya ayudado!

5 0
3 years ago
A sound wave propagates through a region filled with an ideal gas at constant temperature T. It approaches an acoustically perme
nataly862011 [7]

Answer:

The answer is <em>e.2</em>

Explanation:

We should make use of Snell's refractive law. The arriving wave has a certain velocity at T in a medium, then instantly it reaches a medium (same composition) at T' where velocity would either decrease or increase.

When the incidence angle is 30 °, and we want to make the refraction angle 90 ° such that no sound passes through the barrier (this would be named total internal refraction), so we want the second medium to be "faster" than in the first.

<em>The steps are in the image attached:</em>

5 0
3 years ago
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Answer: clay

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8 0
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