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Anna007 [38]
3 years ago
8

Can anyone explain how to do this to me? It is due tomorrow at 9:30am. Thanks.

Physics
1 answer:
Georgia [21]3 years ago
3 0
The 102N acting on the ropes being pulled by eric and kim have some of that force acting horizontally, and some of it vertically. By visualizing it as a right angled triangle, with the hypotenuse the length of the diagonal force, and each side the length of the horizontal and vertical forces, you can use trigonometry to calculate the length of the vertical force. You are told that it is at an angle of 30 with the vertical rope, therefore you know the length of the hypotenuse, and the angle between it and the vertical force, so using trig: (vertical force=x)
x/102=cos(30)
x=102*cos(30)
x=88.33
Therefore the diagonal ropes give a vertical force of 88.33N, and the centre rope, as it acts vertically, gives a vertical force of all 102N. The total:
88.33*2+102=278.66N

I don't know if this is very clear, I hope its good enough to help. If you don't understand, just ask, and I can answer any questions!!! :)
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An object is thrown straight up into the air with an initial speed of 8 m/s, and reaches a greatest height of 15 m before it fal
Ugo [173]

Answer:

The value is T_t =  2.5659 \  s    

Explanation:

From the we are told that  

         The initial speed of the object is  u =  8 \  m/s

         The greatest height it reached is  h  =  15 \  m

Generally from kinematic equation we have that

      v^2 =  u^2 + 2gH

At maximum height v  =  0 m/s

So

      0^2 =  8^2 + 2 *  - 9.8 *  H

=>    H  =  3.27 \  m

Here H is the height from the initial height to the maximum height

So the initial height is mathematically represented as  

      s =  h - H

=>    s =  15 - 3.27

=>    s =  11.73 \  m

Generally the time taken for the object to reach maximum height is mathematically evaluated using kinematic equation as follows

            v  =  u + (-g) t

At maximum height v  =  0 m/s

           0 = 8 - 9.8t

=>         t = 0.8163 \  s

Generally the time taken for the object to move from the maximum height to the ground is mathematically using kinematic equation as follows

       h  =  ut_1 + \frac{1}{2}  gt_1^2

Here the initial velocity is  0 m/s given that its the velocity at maximum height

Also  g is positive because we are moving in the direction of gravity  

So

       15  =  0* t  +  4.9 t^2

=>      t_1  =  1.7496

Generally the total time taken is mathematically represented as

          T_t =  t + t_1

=>        T_t =  0.8163  +   1.7496

=>        T_t =  2.5659 \  s            

 

6 0
3 years ago
A 1.50-mm-diameter glass sphere has a charge of + 1.60 nC. What speed does an electron need to orbit the sphere 1.60 mm above th
saveliy_v [14]

Answer :

Velocity will be 3.28\times 10^{-11}m/sec

Explanation:

We have given glass surface has a diameter of 1.5 mm

And charge q = 1.60 nC

Radius of electrons orbit r = height of electron above surface + radius of sphere  = =1.6+\frac{1.5}{2}=2.35mm = 0.00235m

Force on electron is given by F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}, here q is charge on sphere and e is charge on electron

F=\frac{1}{4\pi \epsilon _0}\frac{qe}{r^2}=\frac{kqe}{r^2}=\frac{9\times 10^9\times 1.6\times 10^{-9}\times 1.6\times 10^{-19}}{0.00235^2}=4.172\times 10^{-13}N

This force work as centripetal force

So F=\frac{mv^2}{r}

4.172\times 10^{-13}=\frac{9.11\times 10^{-31}v^2}{0.00235}

v = =0.0328\times 10^{-9}=3.28\times 10^{-11}m/sec

   

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Answer:

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Second Choice

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