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bagirrra123 [75]
3 years ago
9

after a large snowstorm you shovel 2000 kilograms of snow off your side walk in 1 hour. you lift the shovel to an average height

of 1.25 meters while you are piling the snow in your yard. how much work did you do?
Physics
1 answer:
solmaris [256]3 years ago
8 0

Lifting a mass to a height, you give it gravitational potential energy of

       (mass) x (gravity) x (height)  joules.

To give it that much energy, that's how much work you do on it.

If 2,000 kg gets lifted to 1.25 meters off the ground, its potential energy is

         (2,000) x (9.8) x (1.25) = 24,500 joules.

If you do it in 1 hour (3,600 seconds), then the average power is

           (24,500 joules) / (3,600 seconds) = 6.8 watts.

None of these figures depends on whether the load gets lifted all at once,
or one shovel at a time, or one flake at a time.

But this certainly is NOT all the work you do.  When you get a shovelful
of snow 1.25 meters off the ground, you don't drop it and walk away, and
it doesn't just float there. You typically toss it, away from where it was laying
and over onto a pile in a place where you don't care if there's a pile of snow
there. In order to toss it, you give it some kinetic energy, so that it'll continue
to sail over to the pile when it leaves the shovel.  All of that kinetic energy
must also come from work that you do ... nobody else is going to take it
from you and toss it onto the pile.


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A train traveling at 25 m/s is blowing its whistle at 440 Hz as it crosses a level crossing. You are waiting at the crossing and
ohaa [14]

Answer:

b) 472HZ, 408HZ

Explanation:

To find the frequencies perceived when the bus approaches and the train departs, you use the Doppler's effect formula for both cases:

f_o=f\frac{v_s+v_o}{v_s-v}\\\\f_o=f'\frac{v_s-v_o}{v_s+v}\\\\

fo: frequency of the source = 440Hz

vs: speed of sound = 343m/s

vo: speed of the observer = 0m/s (at rest)

v: sped of the train

f: frequency perceived when the train leaves us.

f': frequency when the train is getTing closer.

Thus, by doing f and f' the subjects of the formulas and replacing the values of v, vo, vs and fo you obtain:

f=f_o\frac{v_s-v}{v_s+v_o}=(440Hz)\frac{340m/s-25m/s}{340m/s}=408Hz\\\\f'=f_o\frac{v_s+v}{v_s-v_o}=(440Hz)\frac{340m/s+25m/s}{340m/s}=472Hz

hence, the frequencies for before and after tha train has past are

b) 472HZ, 408HZ

6 0
3 years ago
In which situation is work not being done?
almond37 [142]

AS

work done =W = F.d = F d cosФ     (Ф is angle between force F and displacement d) If a body/object is moving on a smooth surface (friction-less surface ) .There is no force acting on that body.  F=0 so W=FdcosФ= (0)dcosФ ⇒ W=0

Now if a body is facing some amount of force but under the action of force there is no displacement covered. d=0 so W =FdcosФ= F(0)cosФ ⇒W=0

example:  A person is applying a force on rigid wall but wall remains at rest there is no displacement occurs in wall.

The third term upon which work done  dependent is angle between force and displacement i.e Ф. If Ф=90° then W= FdcosФ= Fdcos90⇒ W=0   ( as cos 90°=0)

5 0
3 years ago
21. Calculate the acceleration of the bus from point D to E. Show your work.
Marat540 [252]

21) Acceleration from D to E: 1 m/s^2

22) The acceleration of the bus from D to E is 1 m/s^2

Explanation:

21)

The acceleration of an object is equal to the rate of change of velocity of the object. Mathematically:

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

where

u is the initial velocity

v is the final velocity

t is the time elapsed

In this problem, we want to measure the acceleration of the bus from point D to point E. We have:

- Initial velocity at point D: u = 0

- Final velocity at point E: v = 5 m/s

- Time elapsed from D to E: t = 21 - 16 = 5 s

Therefore, the acceleration between D and E is

a=\frac{5-0}{5}=1 m/s^2

22) This question is the same as 21), so the result is the same.

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

4 0
3 years ago
Scientists are busy in invention of the devices which run from solor energy​
timama [110]

Answer:

because solar enegry is a perpetual source of enegry and by using it non renewable sources of enegry can be preserved and also prevents from enegry crisis.

hope its helpful to uh

3 0
4 years ago
With 51 gallons of fuel in its tank, the airplane has a weight of 2390.7 pounds. What is the weight of the plane with 81 gallons
Shtirlitz [24]

Answer: 2561.7 pounds

Explanation:

If we assume the total weight of an airplane (in pounds units) as a <u>linear function</u> of the amount of fuel in its tank (in gallons) and we make a Weight vs amount of fuel graph, which resulting slope is 5.7, we can use the slope equation of the line:

m=\frac{Y-Y_{1}}{X-X_{1}}  (1)

Where:

m=5.7 is the slope of the line

Y_{1}=2390.7pounds is the airplane weight with  51 gallons of fuel in its tank (assuming we chose the Y axis for the airplane weight in the graph)

X_{1}=51gallons is the fuel in airplane's tank for a total weigth of 2390.7 pounds (assuming we chose the X axis for the a,ount of fuel in the tank in the graph)

This means we already have one point of the graph, which coordinate is:

(X_{1},Y_{1})=(51,2390.7)

Rewritting (1):

Y=m(X-X_{1})+Y_{1}  (2)

As Y is a function of X:

Y=f_{(X)}=m(X-X_{1})+Y_{1}  (3)

Substituting the known values:

f_{(X)}=5.7(X-51)+2390.7  (4)

f_{(X)}=5.7X-290.7+2390.7  (5)

f_{(X)}=5.7X+2100  (6)

Now, evaluating this function when X=81 (talking about the 81 gallons of fuel in the tank):

f_{(81)}=5.7(81)+2100  (7)

f_{(81)}=2561.7  (8)   This means the weight of the plane when it has 81 gallons of fuel in its tank is 2561.7 pounds.

3 0
3 years ago
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