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Alex777 [14]
3 years ago
11

Suppose a pendulum bob is swinging back and forth. If the highest point is 12.5 cm above the lowest point, what would be speed o

f the bob in m/s when it passes the lowest point (no friction or air resistance)? Think Conservation of Mechanical Energy.
Physics
1 answer:
Colt1911 [192]3 years ago
7 0

Answer:

The speed of the bob  when it passes the lowest point  V = 1.57 \frac{m}{s}

Explanation:

Given data

H = 12.5 \ cm = 0.125 \ m\\

When a pendulum swinging back & forth then at highest point the velocity is zero and lowest point velocity is maximum.

Velocity at lowest point is given by  

V = \sqrt{2 g H}

V = \sqrt{2 (9.81)(0.125)}

V = 1.57 \frac{m}{s}

Therefore the speed of the bob  when it passes the lowest point V = 1.57 \frac{m}{s}

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tan(\theta)=\frac{h-0.8}{1}\\
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6 0
3 years ago
An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the coast. Laying pipe in the ocean is twi
shusha [124]

Answer:

Rectangular path

Solution:

As per the question:

Length, a = 4 km

Height, h = 2 km

In order to minimize the cost let us denote the side of the square bottom be 'a'

Thus the area of the bottom of the square, A = a^{2}

Let the height of the bin be 'h'

Therefore the total area, A_{t} = 4ah

The cost is:

C = 2sh

Volume of the box, V = a^{2}h = 4^{2}\times 2 = 128            (1)

Total cost, C_{t} = 2a^{2} + 2ah            (2)

From eqn (1):

h = \frac{128}{a^{2}}

Using the above value in eqn (1):

C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}

C(a) = 2a^{2} + \frac{256}{a}

Differentiating the above eqn w.r.t 'a':

C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}

For the required solution equating the above eqn to zero:

\frac{4a^{3} - 256}{a^{2}} = 0

\frac{4a^{3} - 256}{a^{2}} = 0

a = 4

Also

h = \frac{128}{4^{2}} = 8

The path in order to minimize the cost must be a rectangle.

8 0
3 years ago
You are traveling 70 mph (31.3 m/s) and slam on the brakes to avoid hitting another car. How far do you travel if it takes you 8
Arlecino [84]

You traveled a distance of 620.075 meters if it takes you 8.5 seconds to stop.

<u>Given the following data:</u>

  • Initial velocity, U = 31.3 m/s  
  • Time, t = 8.5 seconds.

We know that acceleration due to gravity (a) for an object is equal to 9.8 meter per seconds square.

To find the distance traveled, we would use the second equation of motion:

Mathematically, the second equation of motion is given by the formula;

S = ut + \frac{1}{2} at^2

Where:

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Substituting the parameters into the formula, we have;

S = 31.3(8.5) + \frac{x}{y} (9.8)(8.5^2)\\\\S = 266.05 + 4.9(72.25)\\\\S = 266.05 + 354.025

<em>Distance, S</em><em> = </em><em>620.075 meters.</em>

Therefore, you traveled a distance of 620.075 meters if it takes you 8.5 seconds to stop.

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