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Aliun [14]
3 years ago
5

A window washer is standing on a scaffold supported by vertical ropes at each end. The scaffold weighs 200. N and is 3.00 m long

. What is the tension in each rope when the 700. N worker stands 1.00 m from one end?

Physics
1 answer:
seropon [69]3 years ago
5 0
<h2>Tension in ropes are 333.33 N and 566.67 N.</h2>

Explanation:

Refer the given figure showing the arrangement.

For having stability we have torque at A and B are zero.

        Sum of Torques at A = 0

        700 x 1 + 200 x 1.5 - Q x 3 = 0

                               3Q = 1000

                                  Q = 333.33 N

        Sum of Torques at B = 0

        200 x 1.5 + 700 x 2 - P x 3 = 0

                               3P = 1700

                                  P = 566.67 N  

Tension in ropes are 333.33 N and 566.67 N.

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15 Points , Physics HW, can someone please show me step by step how to calculate percent difference for this problem. My numbers
timofeeve [1]

The percent difference between two numbers x and y is given by

\dfrac{|y-x|}x \times 100\%

The absolute value is there because we only care about the absolute percent difference, and not taking into account whether we go from x to y or vice versa. If we remove them, we have two possible interpretations of percent difference.

For example, the (absolute) percent difference between 3 and 6 is

\dfrac{|6-3|}3 \times 100\% = 100\%

In other words, we add 100% of 3 to 3 to end up with 6. This is the same as the percent difference going from 3 to 6. On the other hand, the percent difference going from 6 to 3 is

\dfrac{3-6}3\times100\%=-50\%

which is to say, we take away 50% of 6 away from 6 to end up with 3.

"Make comparisons to object measurements" tells us that the differences should be computed relative to "measurements for object". In other words, take x from the left column and y from the right column.

\dfrac{|7.1-7.3|}{7.1} \times 100\% \approx 2.82\%

\dfrac{|4.8-5.0|}{4.8} \times 100\% \approx 4.17\%

\dfrac{|7.2-7.5|}{7.2} \times 100\% \approx 4.17\%

3 0
2 years ago
A driven RLC circuit is being driven by an AC emf source with a maximum current of 2.75 A and maximum voltage of 150 V. The curr
weqwewe [10]

Answer:

(a). Z = 54.54 ohm

(b). R = 36 ohm

(c). The circuit will be Capacitive.

Explanation:

Given data

I = 2.75 A

Voltage = 150 V

\phi = 0.85 rad = 48.72°

(a). Impedance of the circuit is given by

Z = \frac{V}{I}

Z = \frac{150}{2.75}

Z = 54.54 ohm

(b). We know that resistance of the circuit is given by

R = \frac{Z}{\sqrt{1 + \tan^{2}\phi } }

Put the values of Z & \phi in above formula we get

R = \frac{54.54}{\sqrt{1 + \tan^{2} ( \ 48.72) } }

R = 36 ohm

(c). Since the phase angle is negative so the circuit will be Capacitive.

3 0
3 years ago
Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.
DiKsa [7]

Answer:0.061

Explanation:

Given

T_C=300 k

Temperature of soup T_H=340 K

heat capacity of soup c_v=33 J/K

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any  instant

efficiency is given by

\eta =\frac{dW}{Q}=1-\frac{T_C}{T}

dW=Q(1-\frac{T_C}{T})

dW=c_v(1-\frac{T_C}{T})dT

integrating From T_H to T_C

\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT

W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT

W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}

W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]

Now heat lost by soup is given by

Q=c_v(T_C-T_H)

Fraction of the total heat that is lost by the soup can be turned is given by

=\frac{W}{Q}

=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}

=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}

=\frac{300-340-300\ln (\frac{300}{340})}{300-340}

=\frac{-40+37.548}{-40}

=0.061

4 0
3 years ago
Two coconuts fall freely from rest at the same time, one from a tree twice as high as the other. If the coconut from the shorter
valentinak56 [21]

Wee can use here kinematics

as we know that

y = v*t + \frac{1}{2} at^2

for shorter tree we know that

y = 0 + \frac{1}{2}*9.8 * 2^2

y = 19.6 meter

now since we know that other tree is twice high

So height of other tree is y = 39.2 m

now again by above equation

y = v*t + \frac{1}{2} at^2

39.2 = 0 + \frac{1}{2}*9.8 * t^2

t = 2.83 s

so the time taken is 2.83 s

4 0
3 years ago
DON'T ADD LINKS!
Amanda [17]

Two, four, eight and sixteen cells are produced from first, second, third and fourth cell division respectively.

<h3>How many cells are produced? </h3>

When the body cell passes through cell division process, it divides into two daughter cells. In the second cell division, two cells divides into four cells. In the third cell division, the four cells divides into eight cells. In the fourth cell division, the eight cells divides and sixteen cells are produced.

So we can conclude that two, four, eight and sixteen cells are produced from first, second, third and fourth cell division respectively.

Learn more about cell division here: brainly.com/question/796780

#SPJ1

6 0
2 years ago
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