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Dennis_Churaev [7]
3 years ago
5

What are the mechanics of a dunk tank?

Physics
1 answer:
mr_godi [17]3 years ago
3 0
Step 1

Get a big plastic tank filled with water. You can use an industrial sized barrel, storage tank or some sort of above ground pool.

Step 2

Take a large board, wide enough to fit across the tank, and cut out a rectangular piece from the middle for the seat.

Step 3

Attach the seat back into the support board with heavy duty hinges at the back so that it can swing down freely.

Step 4

Attach a heavy duty door latch of the type used in public restrooms to the underside of the seat so that, when the latch is closed, the seat can't swing down. Use screws and epoxy to attach and reinforce it.

Step 5

Horizontally mount a bicycle wheel on the far side of the tank. It should be mounted on the same side that the latch is going to be on. Mount it in such a way that it will spin freely.

Step 6

Create a target. You can use a simple piece of wood, a caricature, or anything else. The point is to make something for people to throw at.

Step 7

Nail the target to the end of the lever arm.

Step 8

Attach a large lever arm to the bicycle wheel. This is the board that the target is going to be mounted on, so it should reach out far to the side of the wheel.

Step 9

Cover the seat board with something comfortable to sit on. This is going to be the seat, so you want to use something padded. Foam attached with contact cement would be an excellent choice, but even a thin layer of cloth or vinyl would work.

Step 10

Attach the longer board across the top of the tank so that the other board swings downward when released. Depending on what the tank is made of, you could use epoxy, nails, brackets or any combination.

Step 11

Attach a string from the bicycle wheel to the latch so that, when the lever arm swings back, the latch pulls out and releases the seat.

Step 12

Fill the pool with water...

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Captain John Stapp is often referred to as the "fastest man on Earth." In the late 1940s and early 1950s, Stapp ran the U.S. Air
valentina_108 [34]

Answer:

The sled needed a distance of 92.22 m and a time of 1.40 s to stop.

Explanation:

The relationship between velocities and time is described by this equation: v_f=v_0+a*t, where v_f is the final velocity, v_0 is the initial velocity, a the acceleration, and t is the time during such acceleration is applied.

Solving the equation for the time, and applying to the case: t=\frac{v_f-v_0}{a}=\frac{0\frac{m}{s}-282\frac{m}{s}  }{-201\frac{m}{s^2} }=1.40s, where v_f=0\frac{m}{s} because the sled is totally stopped, v_0=282\frac{m}{s} is the velocity of the sled before braking and, a=-201\frac{m}{s^2} is negative because the deceleration applied by the brakes.

In the other hand, the equation that describes the distance in term of velocities and acceleration:x_f-x_0=v_0*t+\frac{1}{2}*a*t^2, where x_f-x_0 is the distance traveled, v_0 is the initial velocity, t the time of the process and, a is the acceleration of the process.

Then for this case the relationship becomes: x_f-x_0=282\frac{m}{s} *1.40s+\frac{1}{2}(-201\frac{m}{s})*(1.40s)^2=94.22m.

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4 0
3 years ago
Read 2 more answers
A thundercloud has an electric charge of 48.8 C near the top of the cloud and –41.7 C near the bottom of the cloud. The magnitud
IceJOKER [234]

Answer: 1.51 km

Explanation:

<u>Coulomb's Law:</u> The electrostatic force between two charge particles Q: and Q2 is directly proportional to product of magnitude of charges and inversely proportional to square of separation distance between them.

Or,   \vec{F}=k \frac{Q_{1} Q_{2}}{r^{2}}

Where Q1 and Q2 are magnitude of two charges and r is distance between them:

<u>Given:</u>

Q1 = Charge near top of cloud = 48.8 C

Q2 = Charge near the bottom of cloud = -41.7 C

Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N

k = 8.99 x 109Nm^2/C^2

<u>So,</u>

\begin{aligned}&7.98 \times 10^{6}=\left(8.99 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right) \frac{48.8 \mathrm{C} \times 41.7 \mathrm{C}}{\mathrm{r}^{2}} \\&r=\sqrt{\frac{1.8294 \times 10^{13}}{7.98 \times 10^{6}}}=1.514  \times 10^{3} \mathrm{~m}=1.51 \mathrm{~km}\end{aligned}

Therefore, the separation between the two charges (r) = 1.51 km

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A). Scale.
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Answer: its a

Explanation: hope this help im sorry if its wrong

 

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