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Andru [333]
3 years ago
9

Meghan and Kailey are attempting to push a piano down the hallway, but as they turn up the 8thgrade hallway they are faced with

a hill. Meghan and Kailey are pushing with a force of 10N each, but the heavy piano is exerting a force of 50N in the opposite direction. Will the girls be able to push the piano? What happens if two other students begin to help by pulling on the piano. Danny pulls with a force of 17N and Liam pulls with a force of 16N. How much force is being exerted on the piano, and will the piano move?
Physics
2 answers:
Alex Ar [27]3 years ago
6 0

Answer: GIVE ME MIN

Explanation:

stich3 [128]3 years ago
5 0
Meghan and Kailey make a pushing force of 20 newtons together, so no they will not push the piano. However with Danny and Liam they make a total force of 53 Newtons so they should be able to push it up the hill.
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an object at rest starts accelerating if it travels 30 meters to end up going 10 m/s what was it’s acceleration
vfiekz [6]

First we have to calculate the time taken to travel the distance 30 m, is

t = \frac{distance}{velocity} = \frac{30 \ m}{10 \ m/s } =  3 \ s.

Now from equation of motion,

v = u + at

Given, v = 10 \ m/s .

As object starts from rest, so  u = 0.

Substituting these values in above equation, we get

 10 \ m/s = 0 + a \times 3 \ s \\\\ a = \frac{10}{3}  = 3.33 \ m/s^2.

Thus, the acceleration is 3.33 \ m/s^2

3 0
4 years ago
When a piano tuner strikes both the A on the piano and a 440 Hz tuning fork, he hears a beat every 2 seconds. The frequency of t
AysviL [449]

Answer:

The  frequency is  f = 439.5 \  Hz        

Explanation:

From the question we are told that

   The frequency of the  tuning fork is  f_t  =  440 \ Hz

   The beat period is  T  =  2 \  s

Generally the beat frequency is mathematically represented as

       f_b  =  \frac{1}{T}

       f_b  =  \frac{1}{2}

      f_b  = 0.5 \  Hz

The  beat frequency is also represented mathematically as

     f_b  =  f_t  -  f

Where  f is the frequency of the piano

 So

       f =  440 -  0.5  

       f = 439.5 \  Hz        

3 0
3 years ago
A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi
slega [8]
We divide the thin rectangular sheet in small parts of height b and length dr. All these sheets are parallel to b. The infinitesimal moment of inertia of one of these small parts is
dI =r^2*dm
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The moment of inertia is
I= \int\limits^{-a/2}_{a/2} {r^2*dm} = M \int\limits^{-a/2}_{a/2} r^2(b*dr)/(ab)=(M/a)(r^3/3) (from (-a/2) toI=(M/3a)(a^3/8 +a^3/8)=(Ma^2)/12 (a/2))



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