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3 years ago

3. A 92 kg Tarzan is holding on to a level 22m vine. He swings on the vine. What will his speed at the bottom of the swing be?

1 answer:

6 0

To solve this problem, we must always remember that energy is conserved. In this case, since he is falling down, he has highest potential energy at the top and zero at bottom. While his kinetic energy is zero at the top since he started from rest and highest at the bottom. We can also say that Potential Energy lost is Kinetic Energy gained thus,

- ΔPE = ΔKE ---> one is negative since PE is losing energy

- m g (h2 – h1) = 0.5 m (v2^2 – v1^2)

Where,

m = mass of tarzan (cancel that out)

g = gravitational acceleration

h2 = height at the bottom= 0

h1 = height at top = 22 m

v2 = velocity at the bottom

v1 = velocity at top = 0 (started from rest)

Therefore substituting all values:

- 9.8 (- 22) = 0.5 (v2^2)

v2 = 20.77 m / s (ANSWER)

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What is the reason eclipses do not happen every full moon and new moon

ioda

The answer is D, hope this helped!

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2 years ago

Verify that the linear speed of an ultracentrifuge is about 0.50 km's, and Earth in its orbit is about 30 km/s by calculating:

FrozenT [24]

Answer:

a) Indeed, the linear speed of the ultracentrifuge is 0.524 kilometers per second.

b) Indeed, the linear speed of the Earth in its orbits about the Sun is approximately 30 kilometers per second.

Explanation:

The linear speed of the particle ( $v$ ), measured in kilometers per second, rotating in a circular pattern is calculated by the following formula:

$v = R\cdot \omega$ (1)

Where:

$R$ - Radius, measured in kilometers.

$\omega$ - Angular speed, measured in radians per second.

Now we proceed to calculate the linear speed of each element:

a) Ultracentrifuge

If we know that $\omega \approx 5235.988\,\frac{rad}{s}$ and $R = 1\times 10^{-4}\,km$ , then the linear velocity is:

$v = (1\times 10^{-4}\,km)\cdot \left(5235.988\,\frac{rad}{s} \right)$

$v = 0.524\,\frac{km}{s}$

Indeed, the linear speed of the ultracentrifuge is 0.524 kilometers per second.

b) Earth

The Earth is 150 million kilometers away from the Sun and takes 365 days to complete one revolution around the Sun. First, we calculate angular speed of the planet:

$\omega = \frac{2\pi}{T}$ (2)

Where $T$ is the period, measured in seconds.

If we know that $T = 31536000\,s$ , then the angular speed of the Earth is:

$\omega = \frac{2\pi}{31536000\,s}$

$\omega = 1.992\times 10^{-7}\,\frac{rad}{s}$

Now, we determine the linear speed:

$v = (1.5\times 10^{8}\,km)\cdot \left(1.992\times 10^{-7}\,\frac{rad}{s} \right)$

$v = 29.88\,\frac{km}{s}$

Indeed, the linear speed of the Earth in its orbits about the Sun is approximately 30 kilometers per second.

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2 years ago

What is the currency for performing work and what is its unit?

ExtremeBDS [4]

The currency for performing work is called : Energy

And Energy's unit of measurement is : Joule

The amount of energy you've inputed will always equal the output.

For example, in order to do a 500 joules of work, you need 500 joules of energy

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2 years ago

At the base of a frictionless icy hill that rises at 25.0∘ above the horizontal, a toboggan has a speed of 11.9 m/s toward the h

forsale [732]

Answer:

17.10 m

Explanation:

When the toboggan stops, we have:

$N-w_y=0\\F-w_x=0$

The x-component of weight is the product of the weight and the sine of the angle above the horizontal, so the y-component of W is the product of the weight and the cosine of the angle above the horizontal.

$N-mgcos(25^\circ)=0\\F-mgsin(25^\circ)=0\\F=mgsin(25^\circ)(1)$

The work-energy principle states that the change in the kinetic energy of an object is equal to the net work done on the object.

$\Delta K=W\\K_f-K_i=F\cdot h\\\frac{m(v_f)^2}{2}-\frac{m(v_i)^2}{2}=Fhcos(180^\circ)$

Recall that $v_f=0$ , replacing (1):

$-\frac{m(v_i)^2}{2}=mgsin(25^\circ)h(-1)\\h=\frac{(v_i)^2}{2gsin(25^\circ)}\\h=\frac{(11.9\frac{m}{s})^2}{2(9,8\frac{m}{s^2})sin(25^\circ)}\\h=17.10 m$

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3 years ago

Dash is standing on his frictionless skateboard with three balls, each with a mass of 155 g, in his hands. The combined mass of

Tomtit [17]

Answer:

1.0326 m/s

Explanation:

mass of each ball, m = 155 g = 0.155 kg

mass of Dash and his skateboard, M = 65 kg

Speed of skateboard moving backward, v = 1.04 m/s

Let the speed of the balls in forward direction is u.

By using the conservation of momentum

Momentum in backward direction = Momentum in forward direction

(M + 3m) x u = M x v

(65 + 3 x 0.155) x u = 65 x 1.04

65.465 u = 67.6

u = 1.0326 m/s

Thus, the speed of balls in forward direction is 1.0326 m/s.

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3 years ago