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dybincka [34]
3 years ago
13

As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff ideal sprin

gs arranged side by side so that they are parallel to each other. When you push the platform, you compress the springs. You do 80.0 J of work when you compress the springs 0.180 m from their uncompressed length.
What magnitude of force must you apply to hold the platform in this position?
Physics
1 answer:
shepuryov [24]3 years ago
6 0

Answer:

The magnitude of force must you apply to hold the platform in this position = 888.89 N

Explanation:

Given that :

Workdone (W) = 80.0 J

length x = 0.180 m

The equation  for this  work done by the spring is expressed as:

W = \frac{1}{2}k_{eq}x^2

Making the spring constant k_{eq} the subject of the formula; we have:

k_{eq} = \frac{2W}{x^2}

Substituting our given values, we have:

k_{eq} = \frac{2*80}{0.180^2}

k_{eq} = 4938.27 \ N.m^{-1}

The magnitude of the force that must be apply  to the hold platform in this position is given by the formula :

F = k_{eq}x

F = 4938.27*0.180

F = 888.89 N

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Answer:

Answer:

Explanation:

Given that

K=8.98755×10^9Nm²/C²

Q=0.00011C

Radius of the sphere = 5.2m

g=9.8m/s²

1. The electric field inside a conductor is zero

εΦ=qenc

εEA=qenc

net charge qenc is the algebraic sum of all the enclosed positive and negative charges, and it can be positive, negative, or zero

This surface encloses no charge, and thus qenc=0. Gauss’ law.

Since it is inside the conductor

E=0N/C

2. Since the entire charge us inside the surface, then the electric field at a distance r (5.2m) away form the surface is given as

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F=8.98755E9×0.00011/5.2²

F=36561.78N/C

The electric field at the surface of the conductor is 36561N/C

Since the charge is positive the it is outward field

3. Given that a test charge is at 12.6m away,

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3 years ago
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Reread the introduction section of this activity. Write a brief conclusion for this activity
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Explanation:

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Answer:

The thrown rock strike 2.42 seconds earlier.

Explanation:

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So now we have an equation and unkown value.

for the thrown rock

\frac{1}{2}(9.8)*t^2+29*t-300=0

for the dropped rock

\frac{1}{2}(9.8)*t^2+0*t-300=0

solving both equation with the quadratic formula:

\frac{-b\±\sqrt{b^2-4*a*c} }{2*a}

we have:

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A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo
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Solution :

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