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3 years ago

11

Find the frequency of a wave with a period of 0.5 seconds

1 answer:

strojnjashka [21]3 years ago

8 0

Frequency of the wave is 2 per second

Explanation:

Frequency is the number of times waves pass at a particular point of time. Here, time period = 0.5 s

Frequency is given by the formula

f = 1/T, where f is the frequency and T is the time period

⇒ f = 1/0.5 = 2 per second

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Rhythms that occur faster and slower than the beat are. Select one:. a. incorrect. b. not synchronized with the time signature.

zlopas [31]

Rhythms that occur faster and slower than the beat are b.<span>not synchronized with the time signature. The synchronization follows the same beat or rhythm. If the time signature say is lower than the original, then the rhythm should be faster. Otherwise, the rhythm is slower than the original one.</span>

4 0

3 years ago

If the radius of an atom is 60 pm and the radius of the Earth is 6000 km, by how many orders of

kherson [118]

Answer:

1 x 10¹⁷

Explanation:

Given data:

Radius of the earth = 6000km

Radius of an atom = 60pm

Now, how many orders is the radius of the earth larger than an atom

Solution:

To solve this problem, let us express both quantity as the same unit;

1000m = 1km

6000km = 6000 x 10³m = 6 x 10⁶m

60pm;

1 x 10⁻¹²m = 1pm

60pm = 60 x 1 x 10⁻¹²m = 6 x 10⁻¹¹m

Now;

The order: $\frac{6 x 10^{6} }{6 x 10^{-11} }$ = 1 x 10¹⁷

6 0

2 years ago

Suppose that you are headed toward a plateau 55 meters high. If the angle of elevation to the top of the plateau is 40degrees,

Iteru [2.4K]

Answer:

$x=65.55m$

Explanation:

Let x be the distance to the shore

From trigonometry properties:

$tan(40^{o} )=\frac{55m}{x} \\x=\frac{55m}{tan(40^{o} )} \\x=65.55m$

3 0

3 years ago

What are examples of non mechanical energy

ExtremeBDS [4]

Atoms, molecules, electrons, photons, protons etc.

3 0

3 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