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damaskus [11]
3 years ago
5

Some giant ocean waves have a wavelength of 25 m and a frequency of

Physics
1 answer:
kolezko [41]3 years ago
3 0

Answer:

Explanation:

The frequency equation for waves is

f=\frac{v}{\lambda} where f is the frequency, v is the velocity, and lambda is the wavelength. Filling in:

.26=\frac{v}{25} so

v = .26(25) and

v = 6.5 meters/second

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If there are 50 people per square kilometer in a city, and the area of the city is 1.5 × 10 square kilometers. What is the total
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Answer:

750 people

Explanation:

From the question,

Number of people in the city = population density×Area of the city

N = D×A.......................... Equagtion 1

Where N = Number of people in the city, D = population density, A = Area of the city.

Given: D = 50 people per square kilometer, A = 1.5×10 square kilometer.

Substitute into equation 1

N = 50(1.5×10)

N = 750 people.

Hence the total number of people in the city is 750 people.

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A man stands still on a moving walkway that is going at a speed of 0.2 m/s to
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Answer:

0.2 m/s east

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Suppose you lift a 20 kg box by a height of 1.0 m.
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Why does the sample on a microscope slide need to be very thin?
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B. so light can shine through it from below.

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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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