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Brut [27]
3 years ago
10

If your front lawn is 16.016.0 feet wide and 20.020.0 feet long, and each square foot of lawn accumulates 13501350 new snowflake

s every minute, how much snow, in kilograms, accumulates on your lawn per hour? Assume an average snowflake has a mass of 2.20 mg.
Physics
1 answer:
asambeis [7]3 years ago
4 0

Answer:

The mass of snow accumulated in the lawn in 1 hour equals 57.024 kilograms

Explanation:

Given

Width of lawn = 16 feet

Length of lawn = 20 feet

Thus total area of lawn equals 16\times 20=320sq.feet\

Now it is given that 1 square foot accumulates 1350 new snowflakes each minute thus number of snowflakes accumulated by 320 square feet in 1 minute equals

320\times 1350=432000

Now it is given that average mass of each snowflake is 2.20mg=2.20\times 10^{-3}g=2.20\times 10^{-6}kg

Hence the mass accumulated per minute equals 432000\times 2.20\times 10^{-6}=0.9504kg

Now since there are 60 minutes in 1 hour thus the mass accumulated in 1 hour equals 0.9504kg\times 60=57.024kg

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Two sacks contain the same number of identical apples and are separated by a distance r. The two
andrezito [222]

Answer:

The appropriate response will be "F_2=\frac{3}{4}F". A further explanation is given below.

Explanation:

According to the question,

⇒ F_1=\frac{G(m_1 m_2)}{r^2} ....(equation 1)

and,

⇒ F_2=\frac{G(m_1 m_2)}{r^2} ....(equation 2)

Now,

On dividing both the equations, we get

⇒ \frac{F_1}{F_2}=\frac{\frac{G(2)(2)}{(1)^2}}{\frac{G(1)(3)}{(1)^2}}

⇒ \frac{F_1}{F_2}= \frac{4}{3}

⇒ F_2=\frac{3}{4}F

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3 years ago
1. Who helped Mendeleev ride 1200 miles to a university in Moscow only to be rejected?
RUDIKE [14]

Answer: that guy

Explanation:

8 0
3 years ago
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What two major uses does this (H-R)diagram have for astronomers?
EastWind [94]
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7 0
3 years ago
A projectile is fired straight upward at 173 m/s. how fast is it moving at the instant it reaches the top of its trajectory?
Fittoniya [83]
The top of the trajectory is the point where it changes from rising to falling. At that exact instant, its vertical speed is zero.
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A student throws a baseball at a large gong 52 m away and hears the sound of the gong 1.73333 s later. The speed of sound in air
Arte-miy333 [17]

Speed = 52/1.57576= 33 m/s

Explanation:

Distance = speed * time

Given , the distance traveled by the baseball = 52 m

Speed of sound in air = 330 m/s

Total time  = 1.73333 s .

Total time for the student to hear the sound of the gong is the sum of the time take for the baseball to reach the gong and the time taken by the sound to travel back.

Distance traveled by the sound is 52 m and the speed is 330 m/s

So time taken by the sound to travel back = distance traveled / the speed

=> time = 52/330 s = 0.15757 s

Time taken by the base ball to reach the gong is the total time - the time taken by the sound

=> time taken by base ball = 1.73333 - 0.15757 = 1.57576s

Speed of the base ball to reach the gong = distance / time

Speed = 52/1.57576= 33 m/s

3 0
3 years ago
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