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.
Calculate the tension in the cable.
1 answer:
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)
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
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Answer:
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
ω = angular frequency = 2 π f [f = cyclic frequency or simply, frequency]
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
ω = 100
Ф = 0
<em><u>But;</u></em>
ω = 2 π f = 100
2 π f = 100 [divide both sides by 2]
π f = 50
f = Hz
Therefore, the frequency of the voltage is Hz
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. = 0. Both the vehicles will cross each other at same time so we will make equations for both and will solve for time.
Truck:
= 33.2 m/s, a = 0 (since the velocity is constant),
= 0
Using
s = 33.2t .......... eq (1)
Hatchback:
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= 0 m/s (since initial velocity is zero),
= 0
Using
putting in the data we will get
now putting 's' value from eq (1)
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.