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Lera25 [3.4K]
3 years ago
8

What tension must a 50.0 cm length of string support in order to whirl an attached 1,000.0 g stone in a circular path at 5.00 m/

s?
Physics
1 answer:
galina1969 [7]3 years ago
4 0

Explanation:

Assuming the circular path is horizontal, the sum of forces in the centripetal direction is:

∑F = ma

T = mv²/r

T = (1.0000 kg) (5.00 m/s)² / (0.500 m)

T = 50.0 N

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What could be the possible answer to the question ?<br><br>thankyou ~​
Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

  • The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

T_1 = \mathbf{\dfrac{M \cdot g}{2}}

Which gives;

M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}

T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})}  = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}

F₀ = T₂·sin(37°)

Which gives;

F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2}  \approx  \mathbf{0.377  \cdot M \cdot g}

  • F₀ ≈ <u>0.377·M·g</u>

<u />

Learn more about equilibrium of forces here:

brainly.com/question/6995192

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2 years ago
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One problem with weight training as a way to improve overall health is that the results of a weight-training program are not mea
Ede4ka [16]

One problem with weight training as a way to improve overall health is that the results of a weight-training program are not measurable.

B.False

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A pulley lifts a 72-N load with a force of 24-N. The input distance is 2m and the output distance is 0.5m. What is the efficienc
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Answer:

Explanation:

Work done on the lever ( input energy ) = force applied x input distance

= 24 N x 2m = 48 J

Work done by the lever ( output energy ) = load x output distance

= 72 N x 0.5m = 36 J

efficiency = output energy / input energy

= 36 J  / 48 J

= 3 / 4 = .75

In percentage terms efficiency = 75 % .

5 0
3 years ago
A block of mass m is attached to a rope wound around the outer rim of a disk of radius R and moment of inertia I, which is free
Hoochie [10]

Answer:

Explanation:

I is the moment of inertia of the pulley, α is the angular acceleration of the pulley and T is the tension in the rope. Let a is the linear acceleration.

The relation between the linear acceleration and the angular acceleration is

a = R α   .... (1)

According to the diagram,

T x R = I x α

T x R = I x a / R      from equation (1)

T = I x a / R²      .... (2)

mg - T = ma    .... (3)

Substitute the value of T from equation (2) in equation (3)

mg - \frac{Ia}{R^{2}}=ma

a=\frac{mg}{m+\frac{I}{R^{2}}}

T is the acceleration in the system

Substitute the value of a in equation (2)

T = \frac{I}{R^{2}}\times \frac{mg}{m+\frac{I}{R^{2}}}

T=\frac{I\times mg}{I+mR^{2}}

This is the tension in the string.

4 0
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Which of the following factors most affect the rate at which waves erode land features along the shore?
OLEGan [10]

Answer:

The biggest factor affecting coastal erosion is the strength of the waves breaking along the coastline. A wave's strength is controlled by its fetch and the wind speed. Longer fetches & stronger winds create bigger, more powerful waves that have more erosive power.

Explanation:

hope it helps !

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