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Eduardwww [97]
3 years ago
5

Rony fills a bucket with water and whirls it in a vertical circle to demonstrate that the water will not spill from the bucket a

t the top of the loop. If the length of the rope from his hand to the centre of the bucket is 1.24 m, what is the minimum tension in the rope (at the top of the swing)? How slow can he swing the bucket? Explain your answer.
Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
8 0

Answer:

0 N, 3.49 m/s

Explanation:

Draw a free body diagram for the bucket at the top of the swing.  There are two forces acting on the bucket: weight and tension, both downwards.

If we take the sum of the forces in the radial direction, where towards the center is positive:

∑F = ma

W + T = m v² / r

The higher the velocity that Rony swings the bucket, the more tension there will be.  The slowest he can swing it is when the tension is 0.

W = m v² / r

mg = m v² / r

g = v² / r

v = √(gr)

Given that r = 1.24 m:

v = √(9.8 m/s² × 1.24 m)

v = 3.49 m/s

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ololo11 [35]

Answer:

Depending on the locomotive, it can use steam, batteries or a diesel engine.  

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7 0
3 years ago
Transverse thrusters are used to make large ships fully maneuverable at low speeds without tugboat assistance. A transverse thru
suter [353]

Answer:

Thrust developed = 212.3373 kN

Explanation:

Assuming the ship is stationary

<u>Determine the Thrust developed</u>

power supplied to the propeller ( Punit ) = 1900 KW

Duct distance ( diameter ; D  ) = 2.6 m

first step : <em>calculate the area of the duct </em>

A = π/4 * D^2

   =  π/4 * ( 2.6)^2  = 5.3092 m^2

<em>next : calculate the velocity of propeller</em>

Punit = (A*v*β ) / 2  * V^2     ( assuming β = 999 kg/m^3 ) also given V1 = 0

∴V^3 = Punit * 2 / A*β

         = ( 1900 * 10^3 * 2 ) / ( 5.3092 * 999 )

hence V2 = 8.9480 m/s

<em>Finally determine the thrust developed </em>

F = Punit / V2

  = (1900 * 10^3) / ( 8.9480)

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8 0
3 years ago
In a local bar, a customer slides an empty beer mug down the counter for a refill. the height of the counter is 1.15 m. the mug
yarga [219]

<span>A.    </span>Let’s say the horizontal component of the velocity is vx and the vertical is vy. <span>
Initially at t=0 (as the mug leaves the counter) the components are v0x and v0y. 
<span>v0y = 0 since the customer slides it horizontally so applied force is in the x component only.

<span>The equations for horizontal and vertical projectile motion are:
x = x0 + v0x t 
y = y0 + v0y t - 1/2 g t^2 = y0 - 1/2 g t^2 </span></span></span>

Setting the origin to be the end corner of the counter so that x0=0 and y0=0, hence:

x = v0x t

y = - 1/2 g t^2 

Given value are: x=1.50m and y=-1.15m (y is negative since mug is going down)

<span>1.50m = v0x t    ---->  v0x= 1.50/t</span>

<span>-1.15m = -(1/2) (9.81) t^2    -----> t =0.4842 s</span>

Calculating for v0x:

v0x = 3.10 m/s

<span>B.    </span>v0x is constant since there are no other horizontal forces so, v0x=vx=3.10m/s

vy can be calculated from the formula:

<span>vy = v0y + at         where a=-g (negative since going down)</span>

vy = -gt = -9.81 (0.4842)

vy = -4.75 m/s

 

Now to get the angle below the horizontal, tan(90-Ø) = -vx/vy

tan(90-Ø )= 3.1/4.75

Ø = 56.87˚<span> below the horizontal</span>

4 0
3 years ago
Read 2 more answers
Cuántas veces es mayor la masa del protón que la del electrón?
viktelen [127]

Answer:

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Explanation:

6 0
3 years ago
A student is watching waves come in from the ocean. He noticed that the first wave he saw (Wave A) had twice the amplitude of th
Alika [10]

Answer:

Wave A

<em>I</em><em> </em><em>hope this</em><em> </em><em>helps</em><em> </em>

8 0
2 years ago
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