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Over [174]
1 year ago
6

A body weighs 1000 lbf when exposed to a standard earth gravity of g=32.174 ft/s^2. a. What is its mass in kg? b. What is its we

ight in N if on moon (gmoon = 1.62 m/s^2)? c. How fast will it accelerate if under a force of 400 lbf on moon and on the earth in m/s.
Physics
1 answer:
allsm [11]1 year ago
4 0

Answer

given,

weight = 1000 lbf

g = 32.174 ft/s²

mass =\dfrac{1000}{32.174}

m = 31 slug

1 slug = 14.593 kg

m = 31 x 14.593

m = 452.383 Kg

b) weight on moon

  W = m g

   W = 452.383 x 1.62

   W= 732.86 N

c) we know,

   F = m  a

  400 lbf = 31 slugs x a

  a = 12.90 ft/s²

1 ft = 0.304 m

  a = 3.92 m/s²

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Answer:

Show attached picture

Explanation:

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We know that the meter's internal resistance is 1000 times higher than the bulb's resistance:

R_M = 1000 R (1)

Both  the meter and the bulb are connected in parallel to the battery, so they both have same potential difference at their terminals:

V_M = V_R

Using Ohm's law, V=RI, we can rewrite the previous equation as:

R_M I_M = R I_R

where

I_M is the current in the meter

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2 years ago
A 450g mass on a spring is oscillating at 1.2Hz. The totalenergy of the oscillation is 0.51J. What is the amplitude.
Volgvan

Answer:

A=0.199

Explanation:

We are given that  

Mass of spring=m=450 g==\frac{450}{1000}=0.45 kg

Where 1 kg=1000 g

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We have to find the amplitude of oscillations.

Energy of oscillator=E=\frac{1}{2}m\omega^2A^2

Where \omega=2\pi\nu=Angular frequency

A=Amplitude

\pi=\frac{22}{7}

Using the formula

0.51=\frac{1}{2}\times 0.45(2\times \frac{22}{7}\times 1.2)^2A^2

A^2=\frac{2\times 0.51}{0.45\times (2\times \frac{22}{7}\times 1.2)^2}=0.0398

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2 years ago
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It will take 4 sec rock to comes its original point

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It is given that the rock comes to its original point

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19.6=4.9t

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3 0
2 years ago
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A shift in one fringe in the Michelson-Morley experiment corresponds to a change in the round-trip travel time along one arm of
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Explanation:

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-3 m/s
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