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3 years ago

7

A marching band is practicing. The band director sees the drum major hit a drum. How far away is the director if he hears the dr

umbeat in 0. 40 seconds? the speed of sound is 330 m/s.

1 answer:

ivanzaharov [21]3 years ago

7 0

Answer:

825 m

Explanation:

330 m/s

1 / 0.40 = 2.5

330 x 2.5 = 825

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PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

AlexFokin [52]

sorry I'm late but the answer is the 2nd option

8 0

3 years ago

Suppose that a nascar race car is moving to the right with a constant velocity of +86 m/s. What is the average acceleration of t

Stels [109]

(a) As the car is moving with constant velocity, it means the rate change of velocity does not change, therefore the average acceleration of the car is zero.

Thus, there is no acceleration, when velocity is constant.

(b) Average acceleration,

$a =\frac{ v-u}{\Delta t}$

Here, v is final velocity and u is the initial velocity and $\Delta t$ is the time interval.

As twelve seconds later, the car is halfway around the track and traveling in the opposite direction with the same speed, therefore

$a =\frac{ -86 \ m/s-86 \ m/s }{12 \ s} = - 14 .3 \ m/s^2$

Thus, the average acceleration of the car is $14 .3 \ m/s^2$ in the direction to the left.

6 0

3 years ago

The cheetah, the fastest of land animals, can run a distance of 274 m in 8.65 s at its top speed. What is the cheetahs top speed

OverLord2011 [107]

Answer:

31.68 meters per second.

Explanation:

Speed is equals to Distance divided by Time, so, this means that speed is inversely proportional to time and directly proportional to distance.

Here, Distance is given which is 274 meters.

Time taken to cover the given distance is 8.65 seconds.

So, by putting the values of distance and time in the formula of speed we will get the top speed of Cheetah which is 31.68 meters per second.

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3 years ago

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,00

Nutka1998 [239]

Answer:

$6.99535\times 10^{-6}\ V/m$

Explanation:

P = Power Output = 1000 W

r = Radius = 35000000 m

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

c = Speed of light = $3\times 10^8\ m/s$

Intensity of Electric radiation is given by

$I=\dfrac{P}{A}\\\Rightarrow I=\dfrac{P}{4\pi r^2}\\\Rightarrow I=\dfrac{1000}{4\pi\times 35000000^2}\ W/m^2$

Intensity of Electric radiation is given by

$I=\dfrac{1}{2}c\epsilon_0E_0\\\Rightarrow E_0=\sqrt{\dfrac{2I}{c\epsilon_0}}\\\Rightarrow E_0=\sqrt{\dfrac{2\times \dfrac{1000}{4\pi\times 35000000^2}}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E_0=6.99535\times 10^{-6}\ V/m$

The amplitude of the electric field vector is $6.99535\times 10^{-6}\ V/m$

6 0

4 years ago

If the capacity of the lungs of an adult were 3.8 L, how many moles of air would the lungs contain at body temperature and atmos

Ahat [919]

Answer:

0.14955 mol

Explanation:

We can solve this problem using the ideal gas law

$P \ V = \ n \ R \ T$

where P is the pressure, V the volume, n the number of moles, R the ideal gas constant and T the temperature.

We can use the atmospheric pressure as 1 atm, and the body temperature as 36.5 °C, in Kelvin this is:

$T_{body} = 36.5 \° C = (36.5 + 273.15) K = 309.65 \ K$

The ideal gas constant is:

$R = 0.082057 \frac{L \ atm}{ K \ mol}$

taking all this in consideration, the number of moles will be:

$n = \frac{P \ V}{ R \ T }$

$n = \frac{1 \ atm * 3.8 \ L }{ 0.082057 \frac{L \ atm}{ K \ mol}$ * 309.65 \ K } [/tex]

$n = 0.14955 \ mol$

3 0

3 years ago