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2 years ago

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Physics

1 answer:

Korvikt [17]2 years ago

4 0

Answer:

Second choice

Explanation:

I suppose ME is the same as POTENTIAL energy

ME = mgh = 70 * 10 * 12 = 8400j

Half way down, 4200 j is converted to Kinetic Energy

1/2 m v^2 = 4200

1/2 (70) v^2 = 4200 v = 10.95 m/s

Just before impact ALL of the 8400 j is now KE
1/2 (70)v^2 = 8400 v = 15.49 m/s

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Equation for inclined spring

zalisa [80]

Answer:

dbdbdheh eewr h eahehwGFTT5Q3JFX

Explanation:

FJGXJDGTFJSRRXAGFEWFWDdQDE

6 0

3 years ago

How much elastic potential energy is stored in a bungee cord with a spring constant of 10.0 N/m when the cord is stretched 2.00

Aleonysh [2.5K]

Answer:

 The elastic potential energy stored in the bungee cord = 20 J

Explanation:

potential energy: This is the energy possessed by a body due to its position. The S.I unit of energy is Joules. The mathematical expression for elastic potential energy is given below

E = 1/2ke²................ Equation 1

Where E = elastic potential energy of the spring, k = force constant of the spring, e = extension

Given: K = 10 N/m, e = 2.00 m

Substituting these values into Equation 1

E = 1/2(10)(2)²

E = 5×4

E = 20 Joules.

Therefore the elastic potential energy stored in the bungee cord = 20 J

6 0

3 years ago

The platform height for Olympic divers is 10 m. A 60 kg diver steps off the platform to begin his dive.

azamat

Answer:

a) Ep = 5886[J]; b) v = 14[m/s]; c) W = 5886[J]; d) F = 1763.4[N]

Explanation:

The potential energy can be found using the following expression, we will take the ground level as the reference point where the potential energy is equal to zero.

$E_{p} =m*g*h\\where:\\m = mass = 60[kg]\\g = gravity = 9.81[m/s^2]\\h = elevation = 10 [m]\\E_{p}=60*9.81*10\\E_{p}=5886[J]$

Since energy is conserved, that is, potential energy is transformed into kinetic energy, the moment the harpsichord touches water, all potential energy is transformed into kinetic energy.

$E_{p} = E_{k} \\5886 =0.5*m*v^{2} \\v = \sqrt{\frac{5886}{0.5*60} }\\v = 14[m/s]$

The work is equal to

W = 5886 [J]

We need to use the following equation and find the deceleration of the diver at the moment when he stops his velocity is zero.

$v_{f} ^{2}= v_{o} ^{2}-2*a*d\\where:\\d = 2.5[m]\\v_{f}=0\\v_{o} =14[m/s]\\Therefore\\a = \frac{14^{2} }{2*2.5} \\a = 39.2[m/s^2]$

By performing a sum of forces equal to the product of mass by acceleration (newton's second law), we can find the force that acts to reduce the speed of the diver to zero.

m*g - F = m*a

F = m*a - m*g

F = (60*39.2) - (60*9.81)

F = 1763.4 [N]

3 0

3 years ago

THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THANKS!!! THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THAN

timurjin [86]

Answer:

$f_{o}$ = 391.67 Hz