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3 years ago

What does Hooke's law say about the relationship of a spring to the force applied to it?

Physics

1 answer:

hoa [83]3 years ago

6 0

Answer:

Explanation:Hooke's law states that " if the elastic limit of an elastic material is not exceeded, the extension,e is directly proportional to the applied force or load.

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What can make something move or cause it to stop

arsen [322]

A push or a pull (force) can make something move or cause it to stop. Hope this helps! :)

5 0

3 years ago

Read 2 more answers

Professor Whitney has a sample of lead and a sample of iron. The samples have equal mass. When Whitney heats the samples, the le

xenn [34]

The temperature increase of a substance is T=Q/m*c, where m is the mass, Q is the energy absorbed and c is the specific heat. So you can conclude that if the lead gets to a higher temperature, it must have a lower specific heat

8 0

3 years ago

An airplane flew from San Francisco to Washington, D.C. Approximately

Arada [10]

Answer:

The solution is given below:

Explanation:

The computation of the speed is shown below

As we know that

Speed = distance ÷ time

where

distance is 2000 km

And, the time is 2.5 hours

SO, the speed is

= 2,000 ÷ 2.5

= 800 km/h

Now the distance would be the same i.e. 2,000 km

but the time is 2 hours

So, the speed is

= 2,000 km ÷ 2 hours

= 1,000 km/hr

The direction should be opposite to the first airplane

5 0

3 years ago

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

Ifa truck starts from rest and it has acceleration of 4 m/s for 5 second

yan [13]

Answer:

Distance 50m

final velocity 20ms^-1

$s = ut + \frac{1}{2} a {t}^{2} \\ = 0 \times t + \frac{1}{2} \times 4 \times {5}^{2} = 50m$

$v = u + at \\ v = 0 + 4 \times 5 \\ v = 20ms^{ - 1}$

6 0

3 years ago