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Ratling [72]
3 years ago
7

A 30.0 kg object rests on a flat, frictionless surface. A rope lifts up on the object with a force of 309 N. What is the acceler

ation of the object?
Physics
1 answer:
saveliy_v [14]3 years ago
4 0
Total force = Frope - mg = 309-300
total force = ma
9 = 30a
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What is the mechanical advantage of a nail puller where you exert a force 45 cm from the pivot and the nail is 1.8 cm on the oth
Mandarinka [93]

Answer:

Explanation:

Mechanical Advantage is the ratio of the distance of the input load (Li)from the pivot to the output load applied to the pivot(Lo)

MA = Li/Le

Given;

Li = 45cm

Lo = 1.8cm

MA = 45/1.8

MA = 25

Hence the mechanical advantage is 25

Also MA is expressed in terms of the force ratio which is the ratio of the Load to the effort applied.

MA = Load/Effort

Given

Load = 1250N

MA = 25

Effort = ?

Substitute

25 = 1250/Effort

Effort = 1250/25

Effort = 50N

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A cylinder with a moving piston expands from an initial volume of 0.250 L against an external pressure of 2.00 atm. The expansio
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The final volume of the gas is 144.25 L

Explanation:

For an ideal gas kept at constant pressure, the work done by the gas on the surroundings is given by

W=p\Delta V = p(V_f - V_i)

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p is the pressure of the gas

V_i is the initial volume

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For the gas in the cylinder in this problem,

p = 2.00 atm

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And we also know the work done,

W = 288 J

So we can solve the equation for V_f, the final volume:

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Answer with Explanation:

a. Option d is true.

a negatively charged plane parallel to the end faces of the cylinder

b. Radius of cylinder, r=0.66m

Magnitude of electric field, E=300 N/C

We have to find the net flux through the closed surface.

Net electric flux,\phi=-2 EA=-2E(\pi r^2)

\phi=-2\times 300\times (3.14\times (0.66)^2)

\phi=-820.67 Nm^2/C

c.

Net charge,Q=\epsilon_0\times \phi

Where

\epsilon_0=8.85\times 10^{-12}

Q=-820.67\times 8.85\times 10^{-12}

Q=-7.26\times 10^{-9} C

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We are sending a 30 Mbit file from source host A to destination host B. All links in the path between source and destination hav
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Answer:

t=1.5\times 10^{-4}\ s

Explanation:

Given:

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<u>Now the delay in data transfer from source to destination for each 10 Mb:</u>

t'=\frac{s}{v}

t'=\frac{10^4}{2\times 10^8}

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<u>Now this time is taken for each 10 Mb of data transfer and we have 30 Mb to transfer:</u>

So,

t=3\times t'

t=3\times 5\times 10^{-5}

t=1.5\times 10^{-4}\ s

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3 years ago
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