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Eddi Din [679]
3 years ago
13

17. The four points built into the camera to ensure that each exposure can be oriented

Physics
1 answer:
3241004551 [841]3 years ago
7 0
The diameter of the circle is 18 m. Eugene incorrectly says that the circumference of the circle is about 113.04 m. What mistake did Eugene make? Use 3.14 for pi.
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The equation for the chemical reaction shown is not balanced. What number should replace the question mark to balance this equat
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The equation is balanced as it is.
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The transfer is perpendicular

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FCFrdfAQefgEwdfgedfsaewfdedasfewreagargferdfares
Ivan

Answer: AAAAAAAAGGGGGHHHHJJJGSSSUUUUUUUUYCCFVGBHNJM

Explanation: YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET YEET

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3 years ago
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Jill does twice as much work as Jack does and in half the time. Jill's power output is Group of answer choices one-fourth as muc
Musya8 [376]

Answer:

Second Choice.

Explanation:

Jack's Power = W/t

Jill's Power = 2W/(0.5)*t

2/0.5 = 4

Jill's Power = 4*W/t

Jill's Power is 4 times greater than Jack's

Second Choice

3 0
2 years ago
For a caffeinated drink with a caffeine mass percent of 0.65% and a density of 1.00 g/mL, how many mL of the drink would be requ
slava [35]

Explanation:

First we will convert the given mass from lb to kg as follows.

        157 lb = 157 lb \times \frac{1 kg}{2.2046 lb}

                   = 71.215 kg

Now, mass of caffeine required for a person of that mass at the LD50 is as follows.

         180 \frac{mg}{kg} \times 71.215 kg

         = 12818.7 mg

Convert the % of (w/w) into % (w/v) as follows.

      0.65% (w/w) = \frac{0.65 g}{100 g}

                           = \frac{0.65 g}{(\frac{100 g}{1.0 g/ml})}

                           = \frac{0.65 g}{100 ml}

Therefore, calculate the volume which contains the amount of caffeine as follows.

   12818.7 mg = 12.8187 g = \frac{12.8187 g}{\frac{0.65 g}{100 ml}}

                       = 1972 ml

Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.

5 0
3 years ago
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