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Luda [366]
3 years ago
10

The box now rests on a frictionless ramp angled at 15◦ . The mover pulls up on a rope attached to the box to move it up the incl

ine. 33 kg F 29 ◦ 15◦ If the rope makes an angle of 29 ◦ with the horizontal, what is the smallest force the mover would have to exert to move the box up the ramp? The acceleration due to gravity is 9.81 m/s 2 .

Physics
1 answer:
Nataly [62]3 years ago
3 0

Answer:

F = 84.61 N

Explanation:

As in the figure, since there is no friction so if component of Force applied along the incline is greater than the component of weight along the incline, then the object will move up the incline.

component of Force along the incline = F cos(23° - 15°) = F cos(8°)

component of weight along the incline = 33*g*sin(15°) = 33*9.81*sin(15°)

Equating the above two components of forces will give the minimum Force required.

F cos(8°) = 33*9.81*sin(15°)

F = 33*9.81*sin(15°) / cos(8°)                  (calculate the value using a scientific calculator)

<u>F = 84.61 N</u>

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True

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A passenger train left station A at 6:00 p.m. Moving with the average speed 45 mph, it arrived at station B at 10:00 p.m. A tran
marishachu [46]
<h2>Average speed of transit train is 60 mph</h2>

Explanation:

Average speed of passenger train = 45 mph

Time taken from station A to station B for passenger train  = 10:00 - 6:00 = 4 hours

Distance between station A to station B = 45 x 4 = 180 miles.

Time taken from station A to station B for transit train  =  4 - 1 = 3 hours

Distance between station A to station B = Average speed of transit train x Time taken from station A to station B for transit train

180 = Average speed of transit train x 3

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8 0
3 years ago
If an object force of 50 N is used to move an object a distance of 20 m, what distance must the object be moved if the input for
steposvetlana [31]

Answer:

\ d_{out} = 100 \ m.

Explanation:

Given data:

F_{in} = 50 \ \rm N

F_{out} = 10 \ \rm N

d_{in} = 20 \ m

Let the distance traveled by the object in the second case be d_{out}.

In the given problem, work done by the forces are same in both the cases.

Thus,

W_{in} = W_{out}

F_{in}.d_{in} = F_{out}.d_{out}

\Rightarrow \ d_{out} = \frac{F_{in}.d_{in}}{F_{out}}

\ d_{out} = \frac{50 \times 20}{10}

\ d_{out} = 100 \ m.

5 0
3 years ago
You just calibrated a constant volume gas thermometer. The pressure of the gas inside the thermometer is 286.0 kPa when the ther
diamong [38]

Answer:

T_{2} = 606.69 K

Explanation:

In that the gas thermometer is a constant volume, it is satisfied that:

\frac{P_{1} }{T_{1} } = \frac{P_{2} }{T_{2} }  

How the boiling water is under regular atmospheric pressure, then

T_{1} = 373 .15 K

Thus

\frac{286000}{373.15} = \frac{465000}{T_{2} }

T_{2} = 606.69 K

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2 years ago
a uniform rule balance on a knife edge at the 55cm mark when a mass of 40g is hung from the 95cm mark. find the weight of the ru
BigorU [14]

Answer:

W = 3.1 N

Explanation:

moments about any convenient point will sum to zero.

I choose summing about the knife edge mark and will assume the ruler of weight W is of uniform construction.

I will assume the ruler weight makes a positive moment

W[55 - 50) - 0.040(9.8)[ 95 - 55] = 0

                                                5W = 15.68

                                                   W = 3.136

7 0
2 years ago
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