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Anastasy
13 days ago
10

A force of only 150 N can lift a 600 N sack of flour to a height of 0.50 m when using a lever as shown in the diagram below. a.

Find the work done on the sack of flour (in J). b. Find the distance you must push with the 150 N force-on the left side (in m). c. Briefly explain the benefit of using a lever to lift a heavy object.

Physics
1 answer:
Softa [3K]13 days ago
8 0
The question lacks clarity. The advantage of lifting with a lever is that it allows you to apply force in a more manageable direction and necessitates far less force to lift an object by balancing the torque exerted by it. For instance, if you aim to lift an object weighing 4N with a force of 2N, utilizing a class 2 lever while maintaining a distance ratio between the body and the force application point from the fulcrum of 1:2 is adequate. In any scenario, one should balance the torque to achieve the desired force.
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(a) Calculate the absolute pressure at the bottom of a fresh- water lake at a depth of 27.5 m. Assume the density of the water i
kicyunya [3294]

Response:

a) P = 370.993\,kPa, b) F = 25.948\,kN

Clarification:

a) The absolute pressure at a depth of 27.5 meters is:

P = P_{atm} + P_{man}

P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}}\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (27.5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)

P = 370.993\,kPa

b) The force applied by the water is:

F = (P - P_{atm})\cdot A

F = (370.993\,kPa-101.3\,kPa)\cdot \left(\frac{\pi}{4} \right)\cdot (0.35\,m)^{2}

F = 25.948\,kN

5 0
1 month ago
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A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then
serg [3582]

Response:

A) W_{ff} =-744.12J

B) F_f=-W_{ff}*sin\theta /hy = 112.75N

C) F_{f2}=207.58N

Clarification:

The question is not fully provided. The complete question was:

A skateboarder with a mass of ms = 54 kg is at the top of a ramp with a height of hy = 3.3 m. He then jumps on his skateboard and goes down the ramp. His speed at the base is vf = 6.2 m/s.  

Part (a) Formulate an expression for the work, Wf, done by the frictional force on the skateboarder in terms of the variables listed in the problem.

Part (b) The ramp is at an angle θ with the ground, where θ = 30°. Formulate an expression for the frictional force's magnitude, fr, between the skateboarder and the ramp.

Part (c) Upon reaching the bottom, the skateboarder continues with speed vf onto a grass-covered flat surface. The friction between the grass and the skateboarder brings him to a halt after 5.00 m. Determine the frictional force, Fgrass in newtons, between the skateboarder and the grass.

For part A), we assess the energy balance to determine the work done by the friction:

W_{ff}=\Delta E

W_{ff}=1/2*m*vf^2-m*g*hy

W_{ff}=-744.12J

For part B), we utilize the previously calculated work:

W_{ff}=-F_f*(hy/sin\theta)   Rearranging for friction force:

F_f=-W_{ff}*sin\theta /hy

F_f=112.75N

For part C), we first ascertain the acceleration through kinematic equations and subsequently find the frictional force using dynamic methods:

Vf^2=Vo^2+2*a*d

Rearranging for 'a':

a=-3.844m/s^2

Now, using dynamics:

|F_f|=|m*a|

|F_f|=207.58N

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Utiliza Scoratic, funciona con cualquier tema.
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A propagating wave in space with electric and magnetic components. These components oscillate at right angles to each other. It
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electromagnetic

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