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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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ikadub [295]

Answer:

the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.  

Explanation:

This problem asks that we determine whether or not a critical flaw in a wide plate is subject to detection  given the limit of the flaw detection apparatus (3.0 mm), the value of KIc (98.9 MPa m), the design stress (sy/2 in which s y = 860 MPa), and Y = 1.0.

ac=1/\pi (\frac{Klc}{Ys} )^{2}\\ ac=1/\pi(\frac{98.9}{(1)(860/2)} )^{2}\\  ac=0.0168m\\ac=16.8mm

Therefore, the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.  

5 0
4 years ago
A water skier is pulled behind a boat. When the skis push down on the water, what is the reaction force?
IrinaVladis [17]
The answer is C) The water pushed up on the skis

The water reacts to the downward force of the skis by pushing back up against the skis. 
8 0
4 years ago
Read 2 more answers
The old rubber boot has two leaks. The top of the boot is 0.3 m higher than the leaks. What is the velocity of the water coming
Otrada [13]

Answer:

Leak 1 = 3.43 m/s

Leak 2 = 2.42 m/s

Explanation:

Given that the top of the boot is 0.3 m higher than the leaks. 

Let height H = 0.3m and the acceleration due to gravity g = 9.8 m/s^2

From the figure, the angle of the leak 1 will be approximately equal to 45 degrees. While the leak two can be at 90 degrees.

Using the third equation of motion under gravity, we can calculate the velocity of leak 1 and 2

Find the attached files for the solution and figure

7 0
3 years ago
In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 1850 J of work is done on the gas
hammer [34]

Answer:

The value of change in internal energy of the gas = + 1850 J

Explanation:

Work done on the gas (W) =  - 1850 J

Negative sign is due to work done on the system.

From the first law  we know that Q = Δ U + W ------------- (1)

Where Q = Heat transfer to the gas

Δ U = Change in internal energy of the gas

W = work done on the gas

Since it is adiabatic compression of the gas so heat transfer to the gas is zero.

⇒ Q = 0

So from equation (1)

⇒ Δ U = - W ----------------- (2)

⇒ W = - 1850 J (Given)

⇒ Δ U = - (- 1850)

⇒ Δ U = + 1850 J

This is the value of change in internal energy of the gas.

7 0
3 years ago
a car initially at rest move with the constant accerates along straght line read after it's spread increase and finally related
nasty-shy [4]

Answer:

32km per hour

Explanation:

Explanation:

In first case v = a t

==> a t = 40 km p h

Now distance covered S1 + S2 + S3

S1 = 1/2 a t^2 and S3 = 1/2 a t^2

But S2 = 3t * 40 = 120 t km

Hence total distance = at^2 + 120 t

Time taken (total) = t + 3t + t = 5 t

Hence average speed = at^2 + 120 t / 5 t

Cancelling t we have at + 120 / 5 = 40 + 120 / 5 = 160/5 = 32 km per hour

8 0
3 years ago
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