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shepuryov [24]
2 years ago
15

A skateboarder, starting from the top of a ramp 4,5 m above the ground, skates down the ramp. The mass of the skateboarder and h

is board is 65 kg
Use the principle of conservation of mechanical energy to calculate the magnitude of the velocity of the skateboarder when he reaches the ground
Physics
2 answers:
valkas [14]2 years ago
8 0

Answer:

Explanation:

The potential energy (PE) of the skateborder (and board) is given by mgh (mass, gravity acceleration, and height).  Assuming no friction of any kind, the magnitude of the skateboarder's velocity will be the result of converting the potential energy to kenetic energy (KE), (1/2)mv².  

[We only know the magnitude of the velocity since we aren't told in which direction the skateboarder is headed.]

1.  Calculate PE:

mgh

(65kg)*(9.61m/s²)*(4.5m) = 2811 kgm²/s² (Joules)

2.  Calculate Speed from KE

Since we are assuming no friction nor other forces, all of the PE will be converted to KE, so we can write:

PE = (1/2)mv²

2811 kgm²/s² = (1/2)*(65kg)v²

v² = (2811 kgm²/s²)/((1/2)*(65kg))

v² = 86.49 m²/s²

v = 9.3 m/s  :  The magnitude of the velocity.  

===================

I vote pointing the skateboarder west, where the gold is.  His velocity would be 9.3 m/s West.

lozanna [386]2 years ago
6 0

Answer:

Kinetic Energy at bottom=2886.5J (when g=9.8m/s²)

<em><u>OR</u></em>

2925J (if g=10m/s²)

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answer : short period oscillations frequency  = 0.063 rad / sec

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Explanation:

first we have to state the general form of the equation

= ( S^2 + 2\alpha _{p} w_{np} S + w^{2} _{np} ) (S^{2} + 2\alpha _{s} w_{ns}S + w^{2} _{ns}  ) = 0

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w_{np}  = Natural frequency of plugiod oscillation

\alpha _{p} = damping ratio of plugiod  oscilations

comparing the general form with the given equation

w^{2} _{np}  = 18.2329

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(20 points) You are at the center of a boat and have been rowing the boat for a long time. You weigh only 80 kg and your 120 kg
valina [46]

Answer:

Explanation:

From the given information:

Let the first weight be m_ 1 = 80 kg

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The weight of  Bubba be m_3 = 60 kg

Also, since you and Budda are a distance of 4m to each other, then the length to which both meet buddy will be:

x_1 = x_3 = \dfrac{4}{2} \\ \\ = 2

The length of the boat be x_2 = 4 m

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We can find the center of mass of the system by using the formula:

X_{CM} = \dfrac{m_1x_1+m_2x_2+m_3x_3}{m_1+m_2+m_3} \\ \\ X_{CM} = \dfrac{(80 \times 2)+(120\times4)+(60\times2)}{80+120+60} \\ \\ X_{CM} = \dfrac{160+480+120}{260} \\ \\ \mathbf{X_{CM} = 2.923}

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One of your summer lunar space camp activities is to launch a 1130 kg1130 kg rocket from the surface of the Moon. You are a seri
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Answer:

∆U = 2.296×10^10Joules

Explanation:

Gravitational potential energy is defined as the energy possessed by an object under the influence of gravity due to its virtue of position.

Potential energy U = Fr where;

F is the force of attraction between the masses of the moon and the rocket.

r is the radius or height of the object.

From Newton's law of universal gravitation, F = GMm/r²

Potential energy U = (-GMm/r²)×r

Potential energy U = -GMm/r

The force is negative because the objects act upward.

M is the mass of the rocket

m is the mass of the moon

Gravitational potential energy possessed by the rocket

U1 = -GMm/r1

r1 is the altitude covered by the rocket

Gravitational potential energy possessed by the Moon

U2 = -GMm/(r2+r1)

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Change in gravitational potential energy ∆U = U2-U1

∆U = -GMm/(r2+r1)-(-GMm/r1)

∆U = -GMm/(r2+r1) + GMm/r1

∆U = -GMm{1/(r2+r1)-1/r1}

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G = 6.67×10^-11m³/kgs²

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