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Karo-lina-s [1.5K]
2 years ago
10

A crane is used to lower weights into a lake for an underwater construction project. Determine the tension in the cable of the c

rane due to a concrete sphere with a radius of 2 m when the sphere is lowered into the water. The concrete has a specific gravity of 2.8.
Physics
1 answer:
atroni [7]2 years ago
5 0

Answer:

Tension in the cable = 591292.8 N or 591.3 kN

Explanation:

Tension in the cable = Weight of object in air - upthrust/bouyancy due to the liquid

Weight of object in air = mass * acceleration due to gravity, g

Since mass is not given, it is obtained from the formula, mass = density * volume

Volume of the sphere = 4/3πr³ where r, radius = 2 m; π = 22/7

volume = 4/3 * 22/7 * (2m)³

volume of sphere = 33.52 m³

Density of concrete sphere = specific gravity * density of water

where, specific gravity = 2.8, density of water = 1000 kg/m³

density of concrete =  2.8 * 1000 kg/m³

density of concrete = 2800 kg/m³

acceleration due to gravity, g = 9.8 m/s²

Thus, weight of concrete = 2800 kg/m³ * 33.52 m³ * 9.8 m/s² = 919788.8 N

Upthrust = density of water * volume of sphere * g

Upthrust = 1000 kg/m³ * 33.52 m³ * 9.8 m/s² = 328496 N

The tension in the cable is the calculated as below;

Tension in the cable = Weight of object in air - upthrust/bouyancy due to the liquid

Tension in cable = 919788.8 N - 328.496 N = 591292.8 N

Therefore, tension in the cable is 591292.8 N or 591.3 kN

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Marcia pokes fun at her husband Brian because he loves to sing in the shower every morning, but
Masteriza [31]

Answer:

<em>Since I can see no choices, I answered it in my own understanding.</em>

Brian - amplitude and frequency

Marcia - amplitude and longitudinal wave

Explanation:

"Sound" and "sound waves" are essential part of a person's life. They can be used for<u> communicating</u> and <u>detecting some object</u>s.

Brian loves singing in the shower which means that he is using a greater amplitude. Amplitude refers to the<em> intensity of the sound </em>or the amount of energy that a sound carries. When one sings in the shower, the sound cannot travel very far. It bounces immediately back to the person singing thus, making the sound bigger. Brian is also using a <em>different range of </em><em>frequency</em><em> compared to his normal way of talking.</em> The frequency of a normal male voice is normally 85 to 180 Hz. A person singing may have a frequency as high as 1,500 Hz.

Marcia talks loudly on the phone. This means that she is also using a greater amplitude because the intensity of her voice is big. Since she is using the telephone, this means that her voice travels in a longitudinal wave through the telephone. This allows her voice to reach to the person on the other end of the line.

7 0
3 years ago
Please help on this one?
Sati [7]
I’m pretty sure it’s A
4 0
3 years ago
A flutist assembles her flute in a room where the speed of sound is 342m/s . When she plays the note A, it is in perfect tune wi
USPshnik [31]

Answer:

a.3Hz

b.0.0034m

Explanation:

First, we know the flute is an open pipe, because open pipe as both end open and a close organ pipe as only one end close.

The formula relating the length and he frequency is giving as

f=\frac{nv}{2l}\\.

a.we first determine the length of the flute at the fundamental frequency i.e when <em>n</em>=1 and when the speed is in the 342m/s

Hence from

f=\frac{nv}{2l}\\\\l=\frac{342}{2*440}\\ l=0.389m\\.

since the value of the length will remain constant, we now use the value to determine the frequency when the air becomes hotter and the speed becomes 345m/s.

f=\frac{nv}{2l} \\f=\frac{345}{2*0.389}\\f=443.4Hz

Hence the require beat is

B=/f_{1}-f_{2}/\\B=/440-443/\\B=3Hz.

b. since the length is dependent also on the speed and frequency, we determine the new length when she plays with a fundamental frequency when the speed of sound is 345m/s

using the formula

L_{new}=\frac{v}{2f}\\\\L_{new}=\frac{345}{2*440}\\\\L_{new}=0.39204

Now to determine the extension,

L_{extend}=L_{new}-L_{old}\\L_{extend}=0.39204- 0.38864\\L_{extend}=0.0034m\\

4 0
3 years ago
a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it wi
Lelu [443]

a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s .The amplitude of the subsequent oscillations 48.13 cm/s

a 1.25 kilogram block is fastened to a spring with a 17.0 newtons per meter spring constant. Given that K is equal to 14 Newtons per meter and mass equals 10.5 kg. The block is then struck with a hammer by a student while it is at rest, giving it a speedo of 46.0 cm for a brief period of time. The required energy provided by the hammer, which is half mv squared, is transformed into potential energy as a result of the succeeding oscillations. This is because we know that energy is still available for consultation. So access the amplitude here from here. He will therefore be equal to and by. Consequently, the Newton's spring constant is 14 and the value is 10.5. The velocity multiplied by 0.49

Speed at X equals 0.35 into amplitude, or vice versa. At this point, the spirit will equal half of K X 1 squared plus half. Due to the fact that this is the overall energy, square is equivalent to half of a K square or an angry square. amplitude is 13 and half case 14 x one is 0.35. calculate that is equal to initial velocities of 49 squares and masses of 10.5. This will be divided in half and start at about 10.5 into the 49-square-minus-14. 13.42 into the entire square in 20.35. dividing by 10.5 and taking the square as a result. 231 6.9 Six centimeters per square second. 10.5 into 49 sq. 14. 2 into a 13.42 square entire. then subtract 10.5 from the result to get the square. So that is 48.13cm/s.

To learn more about oscillations Please click on the given link:

brainly.com/question/26146375

#SPJ4

This is incomplete question Complete Question is:

a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s . what are The amplitude of the subsequent oscillations?

4 0
1 year ago
44.2 cm + 0.123 cm = cm
trapecia [35]
The answer is 44.323 cm
3 0
3 years ago
Read 2 more answers
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