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3 years ago

13

Microphones and loudspeakers are used in an auditorium because the sound waves at the stage compared to the sound waves at the b

ack of the auditorium will have different…

1 answer:

MaRussiya [10]3 years ago

6 0

Acoustics (sounds)
Hope the this helps!

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When a very cold air mass covers half of the united states, a very warm air mass often covers the other half. explain how this h

Rudiy27

<span>Maritime tropical air masses develop over warm waters present in the tropics and Gulf of Mexico, where heat and moisture are carried to to the overlying air from the water below.
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</span><span> Tropical air masses having northward movement carry warm moist air into the United States, thus increasing the potential for condensation. Generally the southern states experience tropical air masses. But, in winter season, southerly winds ahead of migrating cyclones <span>sometimes transport tropical air mass towards north.
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4 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

Which quantity is a scalar quantity?

Vsevolod [243]

The answer is Area ,area

5 0

2 years ago

Using a horizontal force of 200 n, we intend to move a wooden cabinet across a floor at a constant velocity. What is the frictio

grin007 [14]

By using the second law of Newton, the frictional force is 200N.

We need to know about the second law of Newton (force) to solve this problem. The total force applied an object is proportional to the mass of object and acceleration. It can be defined as

∑F = m . a

where F is force, m is mass and a is acceleration.

From the question above, we know that

F1 = 200N

v = constant therefore (a = 0 m/s²)

By using second law of Newton, we get

∑F = m . a

F1 - Ffriction = m . 0

200 - Ffriction = 0

Ffriction = 200 N

Hence, the frictional force is 200N.

Find more on force at: brainly.com/question/25239010

#SPJ4

5 0

1 year ago

A rectangular dam is 101 ft long and 54 ft high. If the water is 35 ft deep, find the force of the water on the dam (the density

blsea [12.9K]

To solve this problem we will begin by finding the pressure through density and average depth. Later we will find the Force, by means of the relation of the pressure and the area.

$P = \rho h$

Here,

h = Depth average

$\rho$ = Density

Moreover,

$\text{Density of water}= \rho = 62.4lb/ft^3$

Replacing,

$P = (62.4lb/ft^3)(\frac{35}{2}ft)$

$P = 1092 lb/ft^2$

Finally the force

$\text{Force} = \text{Pressure}\times \text{Area of dam with water acting on it}$

$F = 1092lb/ft^2(101ft*52ft)$

$F = 5.735*10^6lbf$

6 0

2 years ago