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77julia77 [94]
3 years ago
15

Real world trampolines lose energy since they are damped springs with much internal friction. How much energy does the sumo wres

tler lose on each bounce in this situation? b) How can a gymnast keep a constant bounce height in a real world trampoline?
Physics
1 answer:
mote1985 [20]3 years ago
6 0

The answer to the first question is that, the sumo wrestler loses energy equal to the absorbed energy of the trampoline plus the air friction drag loss. With the second question, the gymnast should maintain energy equal to the Potential energy gymnast has at the peak of bounce height by exerting little force to compensate for the energy spent on the internal friction of the trampolines action.

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A charge of 5.4 C experiences a force of 25.0 in an electric field. What is the strength of electric field at that point ? If th
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The strength of the electric field at that point and the force would this charge experiences at that point will be 4.587 N/C and  12.38 N.

<h3></h3><h3>What is the electric field strength?</h3>

The electric field strength is defined as the ratio of electric force to charge.

Given data;

q₁ = 5.4 C

F₁ is the electric force in case1

E is the electric field =?

F₂ is the electric force in case 2

q₂ is the charge 2

The strength of the electric field at that point is;

F₁=Eq₁

E₁=F/q₁

E₁=25.0 N / 5.4 C

E₁=4.587 N/C

The force would this charge experience at that point when the charge is 2.7 C;

F₂=Eq₂

F₂=4.587 N/C × 2.7 C

F₂ = 12.38 N

Hence the strength of the electric field at that point and the force would this charge experiences at that point will be 4.587 N/C and  12.38 N.

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8 0
2 years ago
. A spring has a length of 0.200 m when a 0.300-kg mass hangs from it, and a length of 0.750 m when a 1.95-kg mass hangs from it
kap26 [50]

Answer:

29.4 N/m

0.1  

Explanation:

a) From the restoring Force we know that :  

F_r = —k*x  

the gravitational force :  

F_g=mg  

Where:

F_r is the restoring force .

F_g is the gravitational force

g is the acceleration of gravity

k is the constant force  

xi , x2 are the displacement made by the two masses.

Givens:

<em>m1 = 1.29 kg</em>

<em>m2 = 0.3 kg  </em>

<em>x1   = -0.75 m  </em>

<em>x2 = -0.2 m </em>

<em>g   = 9.8 m/s^2  </em>

Plugging known information to get :

F_r =F_g

-k*x1 + k*x2=m1*g-m2*g

k=29.4 N/m

b) To get the unloaded length 1:  

l=x1-(F_1/k)

Givens:

m1 = 1.95kg , x1 = —0.75m  

Plugging known infromation to get :

l= x1 — (F_1/k)  

= 0.1  

 

3 0
3 years ago
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