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3 years ago

On a nice winter day at the South Pole, the temperature rises to −54°F. What is the approximate temperature in degrees Celsius?

2 answers:

5 0

<h2>Answer:</h2><h3>-47.7778</h3><h2>Explanation:</h2><h3>To translate degrees Fahrenheit to degrees Celsius, deduct 30 from the temperature in degrees Fahrenheit and divide by 2 to get the temperature in degrees Celsius. = 27 °C (In fact, 84 °F is equal to 28.89 °C—again, a close approximation!).</h3>

Evgesh-ka [11]3 years ago

3 0

Answer:

Its going to be about -47.78°C

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An apple with a mass of 0.95 kilograms hangs from 3.0 meters above the ground. What is the potential energy of the apple

Vesnalui [34]

As Potential energy =mgh

m= 0.95kg

h=3 meter

g = 9.8 m/sec^2. ( acceleration due to gravity)

So P.E =(0.95)(9.8)(3)kgm^2/s^2

P.E =27.93 joules

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3 years ago

Although all sports require a person to be in shape to some degree, some

Alexandra [31]

Answer:

Chess

Explanation:

Chess is considered a sport

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3 years ago

A 60-W light bulb radiates electromagnetic waves uniformly in all directions. At a distance of 1.0 m from the bulb, the light in

krek1111 [17]

1. If we increase the distance to twice it's original value, the light intensity is reduced by one-fourth, the light intensity would be:
I0/4

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B0/2

3. average energy density is inversely proportional to the square of the distance, so the new average energy density is:
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4 years ago

A motorist inflates the tires of her car to a pressure of 180 kPa on a day when the temperature is -8.0° C. When she arrives at

GuDViN [60]

Apply Gay-Lussac's law:

P/T = const.

P = pressure, T = temperature, the quotient of P/T must stay constant.

Initial P and T values:

P = 180kPa, T = -8.0°C = 265.15K

Final P and T values:

P = 245kPa, T = ?

Set the initial and final P/T values equal to each other and solve for the final T:

180/265.15 = 245/T

T = 361K

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3 years ago

A steel beam that is 5.50 m long weighs 332 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending

ElenaW [278]

Explanation:

The given data is as follows.

Length of beam, (L) = 5.50 m

Weight of the beam, ( $W_{b}$ ) = 332 N

Weight of the Suki, ( $W_{s}$ ) = 505 N

After crossing the left support of the beam by the suki then at some overhang distance the beam starts o tip. And, this is the maximum distance we need to calculate. Therefore, at the left support we will set up the moment and equate it to zero.

$\sum M_{o} = 0$

$-W_{s} \times x + W_{b} \times 1.5$ = 0

x = $\frac{W_{b} \times 1.5}{W_{s}}$

= $\frac{332 N \times 1.5}{505 N}$

= 0.986 m

Hence, the suki can come (2 - 0.986) m = 1.014 from the end before the beam begins to tip.

Thus, we can conclude that suki can come 1.014 m close to the end before the beam begins to tip.

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4 years ago