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2 years ago

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Two boys are at the top of a waterslide at Seven Peaks Water Park. One boy (boy A) slips off the top of the tower and falls unob

structed into the pool below. At the same moment the other boy (boy B) starts sliding down the slide and lands in the same pool. Ignoring air resistance and friction, which boy will splash into the water first and which will have the greater speed when he hits the water?

1 answer:

Illusion [34]2 years ago

4 0

Answer: boy B, same

Explanation:

Given

One boy trips off the waterslide while the other starts sliding down

As there is no horizontal velocity, both boys have to travel the same vertical distance.

Their starting vertical velocity is zero and they need to travel the same vertical distance. Therefore, both boys splash water with the same velocity.

The time taken by boy B is more than boy A as boy B will travel some horizontal distance due to slide which will increase its time to reach the bottom.

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7. When will an object's displacement and distance traveled be different?

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If an object changes direction while travelling will an object's displacement and distance travelled be different.

Some people believe that distance and displacement are simply different names for the same quantity. However, distance and displacement are not the same thing. If an object changes direction while travelling, the total distance travelled is greater than the displacement between those two points.

The magnitude of the displacement is always less than or equal to the distance because it is measured along the shortest path between two points.

When the direction of displacement does not change, the magnitude of the displacement and distance are the same. When a body travels in a straight line, for example, its displacement and distance are the same.

Learn more about displacement and distance brainly.com/question/3243551

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2 years ago

Briefly answer the following questions. a) A clock is mounted on the wall. As you look at it, what is the direction of the angul

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Answer:

Explanation:

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b ) If the angular acceleration of a rigid body is zero, the angular velocity will remain constant.

c ) If another planet the same size as Earth were put into orbit around the Sun along with Earth the moment of inertia of the system will increase because the mass of the system increases. Moment of inertia depends upon mass and its distribution around the axis.

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e ) It is not possible that a body has the same moment of inertia for all possible axes because a body can not remain symmetrical about all axes possible. Sphere has same moment of inertia about all axes passing through its centre.

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at the same place.

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3 years ago

If the sun is a medium sized star, why does it look bigger than others?

nata0808 [166]

Answer:

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3 years ago

Read 2 more answers

A fairground ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follow

ale4655 [162]

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2 years ago

A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci

ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity ( $v$ ), measured in meters per second, by using this kinematic equation:

$v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s}$ (1)

Where:

$a$ - Acceleration, measured in meters per square second.

$\Delta s$ - Travelled distance, measured in meters.

$v_{o}$ - Initial velocity, measured in meters per second.

If we know that $v_{o} = 0\,\frac{m}{s}$ , $a = 2.35\,\frac{m}{s^{2}}$ and $\Delta s = 595\,m$ , the final velocity of the rocket is:

$v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}$

$v\approx 52.882\,\frac{m}{s}$

The time associated with this launch ( $t$ ), measured in seconds, is:

$t = \frac{v-v_{o}}{a}$

$t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }$

$t = 22.503\,s$

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

$v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o})$ (2)

Where:

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

If we know that $v_{o} = 52.882\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ , $a = -9.807\,\frac{m}{s^{2}}$ and $s_{o} = 595\,m$ , then the maximum height reached by the rocket is:

$v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})$

$s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}$

$s = 737.577\,m$

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

$s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}$ (2)

Where:

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

$v_{o}$ - Initial speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

If we know that $s_{o} = 595\,m$ , $v_{o} = 52.882\,\frac{m}{s}$ , $s = 0\,m$ and $a = -9.807\,\frac{m}{s^{2}}$ , then the time needed by the rocket is:

$0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}$

$-4.904\cdot t^{2}+52.882\cdot t +595 = 0$

Then, we solve this polynomial by Quadratic Formula:

$t_{1}\approx 17.655\,s$ , $t_{2} \approx -6.872\,s$

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

7 0

2 years ago