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Pani-rosa [81]
3 years ago
12

How much greater is the light-collecting area of a 6-meter telescope than a 3-meter telescope?.

Physics
1 answer:
Fittoniya [83]3 years ago
8 0

Answer:

2x greater

Explanation:

3x2=6

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Middle C also known as "do" on the fixed do-solfege scale, is a note used when playing or singing music. The frequency of middle
saveliy_v [14]

Answer:

198.2m/s

Explanation:

Speed of a wave(v) is the product of the frequency of the wave (f) and its wavelength(¶).

Mathematically, v = f/¶

Given frequency of the middle C = 261.63Hz

Wavelength = 131.87cm

Converting this to meters we have;

131.87/100 = 1.32m

Speed of the sound = 261.63/1.32

Speed of the sound = 198.20m/s

Therefore the speed of sound for middle C is 198.2m/s

8 0
4 years ago
Two cars are moving. The first car has twice the mass of the second car but only half as much kinetic energy. When both cars inc
Zanzabum

Answer:

20 mi

Exp3tanation:

I did the same question in a quizz

8 0
3 years ago
O prevent tank rupture during deep-space travel, an engineering team is studying the effect of temperature on gases confined to
Ira Lisetskai [31]

To calculate for the pressure of the system, we need an equation that would  relate the number of moles (n), pressure (P), and temperature (T) with volume (V). There are a number of equations that would relate these values however most are very complex equations. For simplification, we assume the gas is an ideal gas. So, we use PV = nRT.<span>

PV = nRT  where R is the universal gas constant
P = nRT / V</span>

<span>P = 3.40 mol ( 0.08205 L-atm / mol-K ) (251 + 273.15 K) / 1.75 L </span>

<span>P = 83.56 atm</span>

<span>
</span>

<span>Therefore, the pressure of the gas at the given conditions of volume and temperature would be 83.56.</span>

7 0
3 years ago
When the drivers pass each other, the driver of the red car is to toss a package of contraband to the other driver. To catch the
Brrunno [24]

Answer:

D=387.28m

Explanation:

At the moment where the toss is made X_R = X_G, so we need both equations:

For the red car:

X_R=\frac{a_R*t^2}{2}   With initial speed of 0 and acceleration of 6.12m/s^2.

For the green car:

X_G=Xo + V_G*t   With V_G = 60km/h*\frac{1000m}{1km} * \frac{1h}{3600s} = 16.66m/s   and Xo = 200m

Since both positions will be the same:

\frac{a_R*t^2}{2}=Xo+V_G*t   Solving for t:

t1 = -5.8s  and   t1 =11.25s

Replacing t = 11.25 on either equation to find the displacement:

D = X_R = \frac{a_R*t^2}{2} = 387.28m

3 0
4 years ago
Motion of a dog is shown in this velocity vs time graph.
Tcecarenko [31]

Answer:

8,3

Explanation:

3 0
3 years ago
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