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Gnesinka [82]
3 years ago
15

If your front lawn is 18.0 feet wide and 20.0 feet long, and each square foot of lawn accumulates 1050 new snow flakes every min

ute, how much snow (in kilograms) accumulates on your lawn per hour? assume an average snow flake has a mass of 2.00 mg
Physics
1 answer:
Ivanshal [37]3 years ago
8 0
<span>First, we need to determine the entire area of your front line by multiplying its length times its width.
18.0*20.0 = 360.0 square feet
We can use the rate of accumulation of snow, combined with this figure, to determine how much snow accumulates on your lawn per minute.
360.0 sq ft * 1050 flakes/min/sq ft = 378,000 flakes/min
We can then use the mass of a snowflake to calculate total snow accumulation per minute.
378,000 flakes/min * 2.00 mg/flake = 756,000 mg/min
Finally, we can use this number to determine accumulation per hour.
756,000 mg/min * 60 min/hr =
45,360,000 mg/hr</span>
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Electrical energy is used to turn the blades of a fan. The amount of energy transformed is seen here : 750 J electrical energy i
Charra [1.4K]

Electrical energy is used to run the fan

Here as per given condition 750 J of electrical energy is used to run the fan which is converted into Kinetic energy as 400 J

So here we can see that 350 J of energy is lost against many other type of frictional and resistive loses.

So here we can say that out of 750 J of energy only 400 J is used to run the fan and rest amount of energy is lost against friction.

also we can say that efficiency of this fan will be

\eta = \frac{Useful}{Total}

\eta = \frac{400}{750}

\eta = \frac{8}{25}


3 0
2 years ago
Occupants in a single space shuttle in orbit feel weightless. describe a scheme whereby occupants in a pair of shuttles (or even
agasfer [191]
The   scheme  whereby occupants  in  a  pair  of  shuttles is  as  follows
use  a  strong  cable  with  large  weight  on  the  end
Then  use  the  orbital  naneuvering   system(OMS)  to   set  the   whole  work  as  spinning   about   their  common  center of  gravity.
8 0
3 years ago
A skateboarder rolls off a 2.5 m high bridge into the river. If the skateboarder was originally moving at 7.0 m/s, how much time
saul85 [17]

Answer:

  t = 0.714 s and  x = 5.0 m

Explanation:

This is a projectile throwing exercise, in this case when the skater leaves the bridge he goes with horizontal speed

         vₓ = 7.0 m / s

Let's find the time it takes to get to the river

         y = y₀ + v_{oy} t - ½ g t²

the initial vertical speed is zero and when it reaches the river its height is zero

        0 = y₀ + 0 - ½ g t²

        t = \sqrt{\frac{2y_o}{g}  }

        t = ra 2 2.5 / 9.8

        t = 0.714 s

the distance traveled is

       x = vₓ t

       x = 7.0 0.714

       x = 5.0 m

3 0
3 years ago
A 40kg load is raised to a height of 25m. If the operation requires 1 min, find the power required​
OlgaM077 [116]

Answer:

163.33 Watts

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 40 Kg

Height (h) = 25 m

Time (t) = 1 min

Power (P) =..?

Next, we shall determine the energy. This can be obtained as follow:

Mass (m) = 40 Kg

Height (h) = 25 m

Acceleration due to gravity (g) = 9.8 m/s²

Energy (E) =?

E = mgh

E = 40 × 9.8 × 255

E = 9800 J

Finally, we shall determine the power. This can be obtained as illustrated below:

Time (t) = 1 min = 60 s

Energy (E) = 9800 J

Power (P) =?

P = E/t

P = 9800 / 60

P = 163.33 Watts

Thus, the power required is 163.33 Watts

8 0
2 years ago
A stone is thrown horizontally at a speed of 10.0 m/s from the top of a cliff 139.4 m high.
Lesechka [4]

Answer:

a) 14.2sec

b) 1394m away if horizontal speed never changes

c) 9.8m/s

Explanation:

5 0
2 years ago
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