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hichkok12 [17]
3 years ago
11

Please solve this worksheet for me It’s very important please don’t write unnecessary things or else I’ll report please help me

i request

Physics
1 answer:
Sonja [21]3 years ago
3 0

Answer:

1. 231.25 mL

2. 218.75 kPa

3. 6.21 L

4. 427.5 mL

5. 83.01 mL

6. 307.6 mL

7. 1.15 L

8. 440.4 mL

Explanation:

This questions describes Boyle's law. Using Boyle's law equation to solve the questions as follows:

P1V1 = P2V2

Where;

P1 = initial pressure

P2 = final pressure

V1 = initial volume

V2 = final volume

Please find the answers to each question as an attachment.

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Answer:

Explanation:

given,

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angle of ramp = 22°

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a) for friction less  

  \dfrac{1}{2}mv^2 = \dfrac{1}{2}mu^2 - mgh

             v^2 = u^2 - 2gh

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b) considering the friction

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What creates the changes in step 1,2,3 of the nitrogen cycle
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A box is being moved with a velocity (v) by a force P (parallel to v) along a level horizontal floor. The normal force is (Fn),
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Answer:

Force (P) : Positive

Normal Force (Fn) : Zero

Weight (mg) : Zero

Kinetic Frictional Force (fk) : Negative

Explanation:

The work done by a force on an object is given by the following formula:

W = F.d

W = F d Cosθ

where,

W = Work Done

f = Force Applied

d = displacement

θ = Angle between force and displacement

<u>FOR FORCE (P)</u>:

Since, force P is parallel to the motion of the box. Therefore, θ = 0°

Hence,

W = P d Cos 0°

W = P d(1)

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<u>Therefore, work done by force (P) is Positive.</u>

<u></u>

<u>FOR NORMAL FORCE (Fn) AND WEIGHT (W)</u>:

Since, normal force and weight are perpendicular to the motion of the box. Therefore, θ = 90°

Hence,

W = Fn d Cos 90°= mg d Cos 90°

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<u></u>

<u>FOR KINETIC FRICTIONAL FORCE (fk)</u>:

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Hence,

W = fk d Cos 180°

W = fk d(-1)

W = -fk d

<u>Therefore, work done by Kinetic Frictional Force (fk) is Negative.</u>

<u></u>

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LekaFEV [45]

Answer:

77J

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Hope I helped! ☺

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