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3 years ago

What is cos-^1(0.34)? A. 19.9 B. 44.2° C. 70.1° D. 18.8°

1 answer:

3 0

A calculator must be used. To put your calculator in degree mode, press the MODE button and select degree, the press the 2nd button then MODE again. For most TI calculators, press the 2nd button then press the cos button then enter the value 0.34. This will give you an answer of 70.123 (when you round to 3 decimal places).

The answer is C

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Melting, freezing, and boiling are______ changes

avanturin [10]

Melting, boiling, and freezing are state changes!
Hope this helps! Any questions please just ask! Thank you so much!

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3 years ago

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Which of the following statements best describe destructive forces? A. Forces that build up, create, landmasses. B. Forces that

MissTica

An event that breaks objects into smaller objects or pieces is called destructive force
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3 years ago

A spherically spreading EM wave comes from an 1800-W source. At a distance of 5.0 m, what is the intensity, and what is the rms

Aleonysh [2.5K]

Explanation:

It is given that,

Power of EM waves, P = 1800 W

We need to find the intensity at a distance of 5 m. Also, the rms value of the electric field.

Intensity,

$I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{1800}{4\pi\times (5)^2}\\\\I=5.72\ W/m^2$

The formula that is used to find the rms value of the electric field is as follows :

$I=\epsilon_o cE^2_{rms}$

c is speed of light and $\epsilon_o$ is permittivity of free space

So,

$E_{rms}=\sqrt{\dfrac{I}{\epsilon_o c}}\\\\E_{rms}=\sqrt{\dfrac{5.72}{8.85\times 10^{-12}\times 3\times 10^8}}\\\\E_{rms}=46.41\ V/m$

Hence, this is the required solution.

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3 years ago

*WILL MARK BRAINLIEST FOR RIGHT ANSWER* How much current must be applied across a 60 Ω light bulb filament in order for it to co

sp2606 [1]

Answer: The current must be equal to $\frac{\sqrt{33} }{6}$ amps, or ~0.9574 amps.

Explanation:

You can find the current in amperes using ohms and watts from this formula:

$I = \sqrt{\frac{P}{R} }$

Where P represents power in watts, R represents resistance in ohms, and I represents current in amperes.

You can then substitute 60 and 55 into the equation to find I:

$I = \sqrt{\frac{55}{60} } \\I = \frac{\sqrt{55} }{\sqrt{60} }$

Then, simplify the denominator:

$I = \frac{\sqrt{55} }{2\sqrt{15} }$

Rationalize the denominator:

$I = \frac{\sqrt{55} }{2\sqrt{15} } * \frac{\sqrt{15} }{\sqrt{15} } = \frac{\sqrt{825} }{30}$

Simplify the numerator by finding its factors:

$I = \frac{5\sqrt{33} }{30} = \frac{\sqrt{33} }{6}$

The current must be equal to $\frac{\sqrt{33} }{6}$ amps, or ~0.9574 amps.

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3 years ago

A heat pump operates on a Carnot heat pump cycle with a COP of 8.7. It keeps a space at 248C by consum-ing 2.15 kW of power. Det

Marizza181 [45]

Answer:

Heat of the reservoir is 461.38 K or 188.1 °C

The heating load is 18.705 kW

Explanation:

COP = 8.7

working temperature $T_{h}$ = 248 °C = 248 + 273.3 = 521.3 K

work power W = 2.15 kW

reservoir temperature $T_{c}$ = ?

heating load Q = ?

We know that

COP = Q/W

Q = COP x W = 8.7 x 2.15 = 18.705 kW

Also,

COP = $\frac{T_{h} }{T_{h}- T_{c} }$ = $\frac{521.3}{521.3- T_{c} }$

8.7 = $\frac{521.3}{521.3- T_{c} }$

4535.31 - 8.7 $T_{c}$ = 521.3

4535.31 - 521.3 = 8.7 $T_{c}$

4014.01 = 8.7 $T_{c}$

$T_{c}$ = 4014.01/8.7 = 461.38 K

or 461.38 -273.3 = 188.1 °C

3 0

3 years ago