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Harman [31]
3 years ago
8

A 4.40-m-long, 500 kg steel uniform beam extends horizontally from the point where it has been bolted to the framework of a new

building under construction. A 70 kg construction worker stands at the far end of the beam.
What is the magnitude of the torque about the bolt due to the worker and the weight of the beam?
Physics
1 answer:
icang [17]3 years ago
4 0

Answer:

Torque=13798.4 N.m

Explanation:

Given data

Mass of beam m₁=500 kg

Mass of the person m₂=70 kg

length of steel r₁=4.40m

center of gravity of the beam is at r₂=r₁/2 =4.40/2 = 2.20m

To find

Torque

Solution

Torque due to beam own weight

T_{1}=m_{1}gr_{1}\\ T_{1}=500*2.2*9.8\\T_{1}=10780N.m

Torque due to person

T_{2}=m_{2}r_{2}g\\T_{2}=70*(4.40)*(9.8)\\T_{2}=3018.4 N.m

Now for total torque

T_{total}=T_{1}+T_{2}\\T_{total}=10780+3018.4\\T_{total}=13798.4N.m

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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

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 = \frac{50}{\pi }Hz

Therefore, the frequency of the voltage is \frac{50}{\pi }Hz

7 0
3 years ago
A ball is thrown straight up from ground level. It passes a 2-m-high window. The bottom of the window is 7.5 m off the ground. T
slamgirl [31]

Answer:

u=14.48m/s

Explanation:

From the question we are told that:

Height of window h=2m

Height of window off the ground h_g=7.5m

Time to fall and drop t=1.3s

 

Generally the Newton's equation motion  is mathematically given by

 s=ut+\frac{1}{2}at^2

Where

h=ut+\frac{1}{2}at^2

2=u1.3-\frac{1}{2}*9.8*1.3^2

2=u1.3-8.281

u=7.91m/s^2  

Generally the Newton's equation motion  is mathematically given by

2as=v^2-u^2

Where

-2gh_g=v^2-u^2

-2*9.8*7.5=(7.91)^2-u^2

-147=62.5681-u^2

u=\sqrt{209.5681}

u=14.48m/s

Therefore the  ball’s initial speed

u=14.48m/s

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A speeding Thunderbird left skid marks on the road that were 76.7 m long when it came to a stop. If the acceleration of the car
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Answer: 39.2 m/s

Explanation:

You can use the kinematic equation:

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We know the final velocity because it says it came to a stop. So now all we gotta do is plug in.

0^2=v_i^2+2(-10)(76.7)\\v_i^2=1,534\\v_i=\sqrt{1534} \\=39.166 m/s

4 0
3 years ago
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