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3 years ago

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The pirate ship tie at the amusement park is a giant pendulum that riders sit in. It swings back and forth, with a maximum veloc

ity of 20 m/s. If the combined mass of the ride and the passengers is 8000 kg, what is the maximum height of the boat?

2 answers:

kondor19780726 [428]3 years ago

8 0

Law of conservation of energy states that KE=PE, so

1/2*mv^2=mgh,

1/2*v^2=gh,

h=v^2/2g=20^2/(2*9.8)=20.41 m

kkurt [141]3 years ago

4 0

Here as we know that there is no loss of energy

so we can say that maximum kinetic energy will become gravitational potential energy at its maximum height

So here we have

$\frac{1}{2}mv^2 = mgh$

here we have

v = 20 m/s

m = 8000 kg

now from above equation we have

$\frac{1}{2}(8000)(20^2) = (8000)(9.8)h$

$h = \frac{200}{9.8)$

$h = 20.4 m$

so maximum height is 20.4 m

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In 1977 off the coast of Australia, the fastest speed by a vessel on the water

fenix001 [56]

Answer: 154.08 m/s

Explanation:

Average acceleration $a_{ave}$ is the variation of velocity $\Delta V$ over a specified period of time $\Delta t$ :

$a_{ave}=\frac{\Delta V}{\Delta t}}$

Where:

$a_{ave}=1.80 m/s^{2}$

$\Delta V=V_{f}-V_{o}$ being $V_{o}=0$ the initial velocity and $V_{f}$ the final velocity

$\Delta t=85.6 s$

Then:

$a_{ave}=\frac{V_{f}-V_{o}}{\Delta t}}$

Since $V_{o}=0$ :

$a_{ave}=\frac{V_{f}}{\Delta t}}$

Finding $V_{f}$ :

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3 years ago

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the

alex41 [277]

Answer:

(a) $v_1 = a_1t_1 = 1.76 t_1$

(b) It won't hit

(c) 110 m

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

$v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1$

(b) The velocity of the car before the driver begins braking is

$v_1 = 1.76*20 = 35.2m/s$

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

$a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2$

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$s_2 = v_1t_2 + a_2t_2^2/2$

$s_2 = 35.2*5 - 7.04*5^2/2 = 88 m$

Also the distance from start to where the driver starts braking is

$s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352$

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

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enyata [817]

Answer:

$W = 19.8 J$

Explanation:

14 lb force is required to stretch the spring by 4 inch distance

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