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kicyunya [14]
2 years ago
7

A heavy ball with a weight of 150 N is hung from the ceiling of a lecture hall on a 4.0-m-long rope. The ball is pulled to one s

ide and released to swing as a pendulum, reaching a speed of 5.6 m/s as it passes through the lowest point.
What is the tension in the rope at that point?
Engineering
1 answer:
shusha [124]2 years ago
6 0

Answer:

The tension in the rope at the lowest point is 270 N

Explanation:

Given;

weight of the ball, W = 150 N

length of the rope, r = 4 m

velocity of the ball, v = 5.6 m/s

When the ball passes through the lowest point, the tension on the rope is the sum of weight of the ball and centripetal force.

T = W + F

Centripetal force, F = mv²/r

where;

m is the mass of the ball

m = W/g

m = 150 / 9.8 = 15.306 kg

Centripetal force, F = mv²/r

F = (15.306 x 5.6²)/4

F = 120 N

T = W + F

T = 150 + 120

T = 270 N

Therefore, the tension in the rope at the lowest point is 270 N

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The smallest area of each cable if the stress is not to exceed 90MPa in bronze is 43.6 mm² and 120MPa in steel is 32.7 mm².

<h3>What is normal stress?</h3>

If the direction of deformation force is perpendicular to the cross-sectional area of ​​the body, the stress is called normal stress. Changes in wire length and body volume will be normal.

σ = P/A

Where, σ = Normal stress

P = Pressure

A = Area

1 Kg = 9.81 N

800 kg = 7848 N

Since the rod is half bronze and half steel

800 kg = 7848/2

= 3924 N

Pₙ = Fₙ = 3924 N                       [n = Bronze]

Pₓ =  3924 N                             [x = steel]

Given,

σₙ = 90MPa

σₓ = 120MPa

Aₙ = ?

Aₓ = ?

Aₙ = Pₙ/σₙ

Aₙ = 3924/90

Aₙ = 43.6 mm²

Aₓ = Pₓ/σₓ

Aₓ = 3924/120

Aₓ = 32.7 mm²

To know more about normal stress, visit:

brainly.com/question/28012990

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4 0
1 year ago
Match the descriptions to the respective stage of team
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Answer:

development

forming stage - The team transitions into this stage after completion of its first successful project.

storming stage - In this stage, comfort and trust between the members is high.

norming stage - In this stage, the team members come together for the first time.

performing stage- In this stage, the members have a better idea of what the team expects of them.

5 0
2 years ago
14. You’re setting up a home network for your neighbor, who is a music teacher. She has students visiting her home regularly for
nata0808 [166]

Answer:

The feature to configure is Isolation Option which can either be Guest network or Wireless Isolation

Create a guest ssid separate from the Internal Wi-Fi used at home. The guest ssid ensures that the  parents have a separate point to the internet from the teacher.

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Explanation:

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3 years ago
State two main consumers of electricity transmitted at low frequency Calculate the primary voltage of a transformer if the prima
bagirrra123 [75]

Answer:

1) a. Customers requiring AC electric power transmission for powering remote devices which may include a subsea transmission system where power is distributed to subsea devices

b. Customers that utilize commutator-type motors

2) The primary voltage is 150 volts

Explanation:

The parameters given are;

Number of primary winding, N_p = 50 turns

Number of secondary winding, N_s = 150 turns

Voltage in secondary winding, V_s = 450 volts

Voltage in primary winding = V_p

The relation between N_p, N_s, V_s and V_p is as follows;

\dfrac{V_p}{V_s}  = \dfrac{N_p}{N_s}

Which gives;

V_p  = \dfrac{N_p \times V_s}{N_s}

From which we have;

V_p  = \dfrac{50\times 450}{150} = 150 \ volts

The primary voltage = 150 volts.

7 0
3 years ago
Read 2 more answers
Seawater has a specific density of 1.025. What is its specific volume in m^3/kg (to 3 significant figures of accuracy, tolerance
Svetradugi [14.3K]

Answer:

specific\ volume=0.00097\ m^3/kg

Explanation:

Given that

Specific gravity of sea water = 1.025

So density of sea water = 1.025 x 1000 kg/m^3

Density of sea water = 1025  kg/m^3

We know that

Density=\dfrac{mass}{Volume}   ---1

Specific volume

specific\ volume=\dfrac{Volume}{mass}    ---2

From equation 1 and 2

We can say that

specific\ volume=\dfrac{1}{density}\ m^3/kg

specific\ volume=\dfrac{1}{1025}\ m^3/kg

specific\ volume=0.00097\ m^3/kg

6 0
3 years ago
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