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sergejj
3 months ago
7

a closed tank is partially filled with glycerin. if the air pressure in the tank is 6 lb/in.2 and the depth of glycerin is 10 ft

, what is the pressure in lb/ft2 at the bottom of the tank

Physics
1 answer:
Ostrovityanka [3.2K]3 months ago
4 0

Answer:

note:

The solution is attached because of a calculation error. Please refer to the attachment.

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A ceiling fan has five blades, each with a mass of 0.34 kg and a length of 0.66 m. The fan is operating in its "low" setting at
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An ideal spring is mounted horizontally, with its left end fixed. The force constant of the spring is 170 N/m. A glider of mass
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4 0
2 months ago
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

8 0
2 months ago
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