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anzhelika [568]
3 years ago
10

Your friend provides a solution to the following problem. Evaluate her solution. The problem: Jim (mass 50 kg) steps off a ledge

that is 2.0 m above a platform that sits on top of a relaxed spring of force constant 8000 N/m. How far will the spring compress while stopping Jim? Your friend's solution: (50kg)(9.8m/s2)(2.0m)=(1/2)(8000N/m)x x=0.25m4.
Physics
1 answer:
Mekhanik [1.2K]3 years ago
5 0

Answer:

No its wrong

Correct compression is 0.41

Explanation:

After jumping from 2m height on spring attached platform platform is compressed a distance x(let).

So work done by gravity on Jim is converted into spring potential energy.

k=8000 N/m

mass of Jim =50 kg

\frac{1}{2}k(x)^{2} =m*g(2+x)

\frac{1}{2}8000(x)^{2} =50*9.8(2+x)

8.16x^{2} =2 +x

Solve this quadratic one solution is positive and other is negative.

positive one is our answer = 0.41 m

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Two particles are fixed to an x axis: particle 1 of charge q1 = 2.78 × 10-8 c at x = 15.0 cm and particle 2 of charge q2 = -3.24
Oksi-84 [34.3K]
Refer to the attached figure. Xp may not be between the particles but the reasoning is the same nonetheless.
At xp the electric field is the sum of both electric fields, remember that at a coordinate x for a particle placed at x' we have the electric field of a point charge (all of this on the x-axis of course):
E=\frac{1}{4\pi\varepsilon_0}\frac{q}{(x-x')^2}
Now At xp we have:
\frac{1}{4\pi\varepsilon_0}\frac{q_1}{(x_p-x_1)^2}-\frac{1}{4\pi\varepsilon_0}\frac{3.29q_1}{(x_p-x_2)^2}=0
\implies (x_p-x_1)^2=\frac{(x_p-x_2)^2}{3.29}\\
\implies(1-\frac{1}{3.29})x_p^2+2(\frac{x_2}{3.29}-x_1)x_p+x_1^2-\frac{x_2^2}{3.29}=0
Which is a second order equation, using the quadratic formula to solve for xp would give us:
xp=\frac{-(\frac{x_2}{3.29}-x_1)-\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}
or
xp=\frac{-(\frac{x_2}{3.29}-x_1)+\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}
Plug the relevant values to get both answers.
Now, let's comment on which of those answers is the right answer. It happens that BOTH are correct. This is simply explained by considring the following.

Let's place a possitive test charge on the system This charge feels a repulsive force due to q1 but an attractive force due to q2, if we place the charge somewhere to the left of q2 the attractive force of q2 will cancel the repulsive force of q1, this translates to a zero electric field at this x coordinate. The same could happen if we place the test charge at some point to the right of q1, hence we can have two possible locations in which the electric field is zero. The second image shows two possible locations for xp.

6 0
3 years ago
A bucket of water is being raised from a well using a rope. If the bucket of water has a mass of 6.2 kg, how much force (in N) m
Sholpan [36]

Answer:

6.2N force

Explanation:

According to Newton's second law of motion, force is equal to the product of the mass of a body and its acceleration. Mathematically,

Force = mass × acceleration

Given mass of bucket of water = 6.2kg

acceleration of the bucket = 1m/s²

Force exerted on the rope = 6.2 × 1

= 6.2N

5 0
3 years ago
The pointer of an analog meter is connected to a
dangina [55]
C. coil suspended by bearings. 
<span>but im not 100% sure</span>
8 0
3 years ago
Read 2 more answers
When air resistance is ignored, _____ of the projectile affect(s) the range and maximum height of the projectile.
FromTheMoon [43]

When air resistance is ignored, initial velocity of the projectile affect the range and maximum height of the projectile.

Projectile is a missile designed to be fired from a rocket or gun.

A projectile is the object that is propelled by the application of an external force and then moves freely under the influence of gravity and air resistance.

The range is defined as the distance between the launch point and the point where the projectile hits the ground.

The height from the ground at the top most position of projectile is referred to as maximum height.

When air resistance is ignored, initial velocity of the projectile affect the range and maximum height of the projectile.

Learn more about maximum height click here brainly.com/question/6261898

#SPJ4

8 0
1 year ago
6. Aaron will run from home at y mph and walk back at x mph. how much distance does he need to travel to spend a total of t hour
Kisachek [45]

Answer:

Total distance, d=\dfrac{xyt}{(x+y)}

Explanation:

It is given that,

Speed of Aaron from home is y mph and walk back at x mph. Let t is the total time he spend in walking and jogging. Let d is the distance covered.

We he moves from home to destination, time is equal to, \dfrac{d}{x}

Similarly, when he move back to home, time taken is equal to \dfrac{d}{y}

Total time taken is equal to :

\dfrac{d}{x}+\dfrac{d}{y}=t

d(\dfrac{1}{x}+\dfrac{1}{y})=t

d=\dfrac{t}{(\dfrac{1}{x}+\dfrac{1}{y})}

d=\dfrac{xyt}{(x+y)}

So, the distance he speed in walking and jogging is \dfrac{xyt}{(x+y)}. Hence, this is the required solution.

8 0
3 years ago
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