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Luba_88 [7]
3 years ago
14

A 75.0-kg painter climbs a ladder that is 2.75 m long leaning against a vertical wall. The ladder makes a 30.0° angle with the w

all. (a) How much work does is done by gravity on the painter? (b) Does the answer to part (a) depend on whether the painter climbs at constant speed or accelerates up the ladder?
Physics
2 answers:
Schach [20]3 years ago
6 0
<h3><u>Answer;</u></h3>

a) 178.6125 Joules

b) Work done does not depend on constant speed or acceleration up the ladder

<h3><u>Explanation</u>;</h3>

Work done =Fd times the cosine of angle between;

where F is force d is distance m is the mass a is acceleration

.F=ma. In this case a=g=9.81 m/s^2 and cos 30=0.866

Therefore;  

W= 75 × 9.81 × 2.75 × .866  

   = 178.6125 Joules

b. The work done by the painter does not depend on whether the painter climbs at constant speed or accelerates up the ladder.

W=3376.945 Joules

Mazyrski [523]3 years ago
4 0

Answer:

Work is defined as the movement caused by a force. In the case of someone climbing something, the person is doing force against gravity, so the work needed to lift an object of mass M by a height H is equal to W = MgH where h is the gravity acceleration.

In this case, M = 75kg and we can obtain the height by the Pythagorean theorem, where the ladder is the hypotenuse of a triangle.

If we want to only know the displacement in the y-axis, then we need to compute the adjacent cathetus, this is:

H = 2.75m*cos(30°) = 2.38m

and we know that g= 9.8m/s^2

so we can compute the work and obtain:

W = 2.38m*9.8m/s^2*75kg = 1739,3J

b) work does not depend on the velocity, the fact that he is ascending means that there is a force applied on him that fights against the gravity. Non the less, if there was a force bigger applied to him in the y-direction (which may accelerate him, the previous calculation is thinking that the force done is equal to the gravitational force) the work can be bigger. The work calculated before is the minimal work needed to make him ascend the height H.

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The circuit you should use to find the open-circuit voltage is
fiasKO [112]

Answer:

Incomplete questions check attachment for circuit diagram.

Explanation:

We are going to use superposition

So, we will first open circuit the current source and find the voltage Voc.

So, check attachment for open circuit diagram.

From the diagram

We notice that R3 is in series with R4, so its equivalent is given below

Req(3-4) = R3 + R4

R(34) = 20+40 = 60 kΩ

Notice that R2 is parallel to the equivalent of R3 and R4, then, the equivalent of all this three resistor is

Req(2-3-4) = R2•R(34)/(R2+R(34))

R(234) = (100×60)/(100+60)

R(234) = 37.5 kΩ

We notice that R1 and R(234) are in series, then, we can apply voltage divider rule to find voltage in R(234)

Therefore

V(234) = R(234) / [R1 + R(234)] × V

V(234) = 37.5/(25+37.5) × 100

V(234) = 37.5/62.5 × 100

V(234) = 60V.

Note, this is the voltage in resistor R2, R3 and R4.

Note that, R2 is parallel to R3 and R4. Parallel resistor have the same voltage, then voltage across R2 equals voltage across R34

V(34) = 60V.

Now, we also know that R3 and R4 are in series,

So we can know the voltage across R4 which is the Voc we are looking for.

Using voltage divider

V4 = Voc = R4/(R4 + R(34)) × V(34)

Voc = 40/(40+60) × 60

Voc = 24V

This is the open circuit Voltage

Now, finding the short circuit voltage when we short circuit the voltage source

Check attachment for circuit diagram.

From the circuit we notice that R1 and R2 are in parallel, so it's equivalent becomes

Req(1-2) = R1•R2/(R1+R2)

R(12) = 25×100/(25+100)

R(12) = 20 kΩ

We also notice that the equivalent of Resistor R1 and R2 is in series to R3. Then, the equivalent resistance of the three resistor is

Req(1-2-3) = R(12) + R(3)

R(123) = 20 + 20

R(123) = 40 kΩ

We notice that, the equivalent resistance of the resistor R1, R2, and R3 is in series to resistor R4.

So using current divider rule to find the current in resistor R4.

I(4) = R(123) / [R4+R(123)] × I

I(4) = 40/(40+40) × 8

I(4) = 4mA

Then, using ohms law, we can find the voltage across the resistor 4 and the voltage is the required Voc

V = IR

V4 = Voc = I4 × R4

Voc = 4×10^-3 × 40×10^3

Voc = 160V

Then, the sum of the short circuit voltage and the open circuit voltage will give the required Voc

Voc = Voc(open circuit) + Voc(short circuit)

Voc = 24 + 160

Voc = 184V.

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<span>No, because the truck applies more pressure than the bridge can support.</span>
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MrMuchimi
F=mg=Gm1m2/r^2
g=Gm2/r^2
g=2Gm2/(2r)^2=2Gm2/4r^2=Gm2/2r^2
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Answer:

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Explanation:

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