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GenaCL600 [577]
3 years ago
8

Problem: A barbell consists of two small balls, each with mass m at the ends of a very low mass rod of length d. The barbell is

mounted on the end of a low-mass rigid rod of length b. This apparatus is started in such a way that while the rod rotates clockwise with angular speed ω1, the barbell rotates clockwise about its center with an angular speed ω2. What is the total angular momentum of this system about point B?
Physics
1 answer:
zepelin [54]3 years ago
7 0

Answer:

mass of ball 1=m1

mass of ball 2=m2

velocity of ball=r1w1

velocity of ball 2=r2w2

Total angular momentum=m1*v1+m2*v2

but

v1=r1*w1

v2=r2*w2

Substitute values in above equation

Total angular momentum of the system=m1*r1*w1+m2*r2*w2

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Melanie completes a long distance run at an average speed of 6 mph. If it takes her 3 hours, how far did she run?
Rama09 [41]

Answer:

18 miles

Explanation:

The average speed is 6 mph

Melanie ran for 3 hours

Speed × Time = Distance

So, 6 mph × 3 h = 18 miles

6 0
3 years ago
I stretch a rubber band and "plunk" it to make it vibrate in its fundamental frequency. I then stretch it to twice its length an
Nikitich [7]

Answer:

The new frequency (F₂ ) will be related to the old frequency by a factor of one (1)

Explanation:

Fundamental frequency = wave velocity/2L

where;

L is the length of the stretched rubber

Wave velocity = \sqrt{\frac{T}{\frac{M}{L}}}

Frequency (F₁) = \frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}

To obtain the new frequency with respect to the old frequency, we consider the conditions stated in the question.

Given:

L₂ =2L₁ = 2L

T₂ = 2T₁ = 2T

(M/L)₂ = 0.5(M/L)₁ = 0.5(M/L)

F₂ = \frac{\sqrt{\frac{2T}{0.5(\frac{M}{L})}}}{4*L} = \frac{\sqrt{4(\frac{T}{\frac{M}{L}}})}{4*L} = \frac{2}{2} [\frac{\sqrt{\frac{T}{\frac{M}{L}}}}{2*L}] = F_1

Therefore, the new frequency (F₂ ) will be related to the old frequency by a factor of one (1).

7 0
3 years ago
A circular swimming pool has a diameter of 12 m. The circular side of the pool is 3 m high, and the depth of the water is 2.5 m.
Anestetic [448]

Answer:

Explanation:

Diameter of pool = 12 m

radius of pool, r = 6 m

Total height raised, h = 3 + 2.5 = 5.5 m

density of water, d = 1000 kg/m³

Mass of water, m = Volume of water x density

m = πr²h x d

m = 3.14 x 6 x 6 x 5.5 x 1000

m = 113040 kg

Work = m x g x h

W = 113040 x 9.8 x 5.5

W = 6092856 J

7 0
3 years ago
Imagine you’re an engineer making a string of battery powered holiday lights. If a bulb burns out current cannot flow through th
Anit [1.1K]

Answer:

The 2 light bulbs can be connected in parallel to each other to avoid disconnection when one bulb burns out.

Explanation:

The parallel connection is required not series. A parallel connection is the connection of electronic components (e.g bulbs, LED, resistors, capacitors etc) in such a way that the same voltage is supplied across the ends of the components.  While in a series connection, the components are connected to each other end-to-end.

As regard the question, parallel connection ensures that the brightness any of the bulbs is not affected with respect to the other bulbs. And other bulbs continue to function when any burns out. The 2 light bulbs should be connected in parallel to the baterry to avoid disconnection of all the bulbs.

4 0
3 years ago
A motorcycle traveling at a speed of 44.0 mi/h needs a minimum of 44.0 ft to stop. If the same motorcycle is traveling 79.0 mi/h
Tasya [4]

Answer:

141.78 ft

Explanation:

When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.

Calculating the acceleration using one of Newton's equations of motion:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 44 mi/h\\\\s = 0.00833 mi\\\\=> 0^2 = 44^2 + 2 * a * 0.00833\\\\=> 1936 = -0.01666a\\\\a = -116206.48 mi/h^2 or -14.43 m/s^2

Note: The negative sign denotes deceleration.

When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2

Hence, we can find the minimum stopping distance using:

v^2 = u^2 + 2as\\\\v = 0 mi/h\\\\u = 79 mi/h\\\\a = -116206.48 mi/h\\\\=> 0^2 = 79^2 + (2 * -116206.48 * s)\\\\6241 = 232412.96s\\\\s = \frac{6241}{232412.96} \\\\s = 0.0268531 mi = 141.78 ft

The minimum stopping distance is 141.78 ft.

4 0
3 years ago
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