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yulyashka [42]
2 years ago
5

A crate with a mass of 175.5 kg is suspended from the end of a uniform boom with a mass of 94.7 kg. The upper end of the boom is

supported by a cable attached to the wall and the lower end by a pivot (marked X) on the same wall.
Calculate the tension in the cable.

Physics
1 answer:
vlabodo [156]2 years ago
6 0

The tension in the cable is mathematically given as

T =266.659 N

<h3>What is the tension in the cable.?</h3>

Generally, the equation for the angle of the boom with horizontal is  mathematically given as

A = tan-1(5 /10)

A= 26.57 \textdegree

The angle of cable with horizontal

B = tan-1(4 / 10)

B= 21.80 degrees

Hence

175.5 * cos26.57 + 94.7 *cos26.57 * 1/2 = T (sin(26.57+21.80)) * 1

T =266.659 N

In conclusion, the tension in the cable.

T =266.659 N

Read more about tension

brainly.com/question/12111847

#SPJ1

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Alternating Current In Europe, the voltage of the alternating current coming through an electrical outlet can be modeled by the
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Answer:

\frac{50}{\pi }Hz

Explanation:

In alternating current (AC) circuits, voltage (V) oscillates in a sine wave pattern and has a general equation as a function of time (t) as follows;

V(t) = V sin (ωt + Ф)            -----------------(i)

Where;

V = amplitude value of the voltage

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Ф = phase difference between voltage and current.

<u><em>Now,</em></u>

From the question,

V(t) = 230 sin (100t)              ---------------(ii)

<em><u>By comparing equations (i) and (ii) the following holds;</u></em>

V = 230

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Ф = 0

<em><u>But;</u></em>

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7 0
3 years ago
A pickup truck and a hatchback car start at the same position. If the truck is moving at a constant 33.2m/s and the hatchback ca
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Answer:

The pickup truck and hatchback will meet again at 440.896 m

Explanation:

Let us assume that both vehicles are at origin at the start means initial position is zero i.e. s_{o} = 0. Both the vehicles will cross each other at same time so we will make equations for both and will solve for time.

Truck:

v_{i} = 33.2 m/s, a = 0 (since the velocity is constant), s_{o} = 0

Using s =s_{o}+v_{i}t+1/2at^{2}

s = 33.2t .......... eq (1)

Hatchback:

a=5m/s^{2}, v_{i} = 0 m/s (since initial velocity is zero), s_{o} = 0

Using s =s_{o}+v_{i}t+1/2at^{2}

putting in the data we will get

s=(1/2)(5)t^{2}

now putting 's' value from eq (1)

2.5t^{2}-33.2t=0

which will give,

t = 13.28 s

so both vehicles will meet up gain after 13.28 sec.

putting t = 13.28 in eq (1) will give

s = 440.896 m

So, both vehicles will meet up again at 440.896 m.

7 0
3 years ago
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