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vaieri [72.5K]
2 years ago
5

Un acróbata de 60.0 kg está unido a un cordón de bungee con un resorte de 10.0 m de longitud . Salta de un puente que abarca un

río desde una altura de 50.0 m. Cuando finalmente se detiene, el cordón tiene una longitud estirada de 30.0 m. Trata el acrobata como una masa puntual, y no tener en cuenta el peso de la cuerda de bungee. Asumiendo que la constante de resorte del bungee es de 80.3 N/m, cual es el total de energía potencial en relación con el agua cuando el hombre deja de caer? *
Physics
1 answer:
Ede4ka [16]2 years ago
8 0

Answer:

What kind of languege is this please?

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On what factor does speed of sound depend​
velikii [3]

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For sound waves to travel, there is a requirement of medium and density of the medium is considered to be one of the factors on which the speed of sound depends. When the medium is dense, the molecules in the medium are closely packed which means that the sound travels faster.

Explanation:

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Swimmers at the beach are tanning on towels. Which method of heat transfer is responsible for their tan? *
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ratiation

Explanation:

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2 years ago
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A psychologist studies if internet uses causes isolation
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A) experimental because he isn’t sure and is testing out
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2 years ago
A satellite with a mass of 2,000 ㎏ is inserted into an orbit that is twice the Earth' s radius. W hat is the force of gravity on
dolphi86 [110]

Answer:

option (b) 4900 N

Explanation:

m = 2000 kg, R = 6380 km = 6380 x 10^3 m, Me = 5.98 x 10^24 kg, h = R

F = G Me x m / (R + h)^2

F = G Me x m / 2R^2

F = 6.67 x 10^-11 x 5.98 x 10^24 x 2000 / (2 x 6380 x 10^3)^2

F = 4900 N

4 0
3 years ago
Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma
AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

m_1 = Mass of first car = 120 kg

m_2 = Mass of second car

u_1 = Initial Velocity of first car = 14 m/s

u_2 = Initial Velocity of second car = 0 m/s

v_1 = Final Velocity of first car = -2 m/s

v_2 = Final Velocity of second car

For perfectly elastic collision

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}

Applying in the next equation

v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s

m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg

Mass of second car = 160 kg

Velocity of second car = 12 m/s

5 0
3 years ago
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