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3 years ago

According to Hooke's law, the force necessary to stretch or compress a spring is proportional to what value?

Physics

1 answer:

ankoles [38]3 years ago

8 0

Answer:

displacement

Explanation:

Hookes Law describes the elastic properties of materials only in the range in which the force and displacement are proportional (displacement = change in length).

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A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net = 4.444 N

Dmitriy789 [7]

Answer:

The value of third charge is 0.8μC.

Explanation:

Given that.

Magnitude of net force=4.444 N

According to figure,

Suppose, First charge = 2.4 μC

Second charge = 6.2 μC

Distance r₁ = 9.8 cm

Distance r₂ = 2.1 cm

We need to calculate the value of r

Using Pythagorean theorem

$r=\sqrt{(r_{1})^2+(r_{2})^2}$

Put the value into the formula

$r=\sqrt{(9.8)^2+(2.1)^2}$

$r=10.02\ cm$

We need to calculate the force

Using formula of force

$F_{12}=\dfrac{kq_{1}q_{2}}{(r)^2}$

Force F₁₂,

$F_{12}=\dfrac{9\times10^{9}\times2.4\times10^{-6}\times6.2\times10^{-6}}{(10.02\times10^{-2})^2}$

$F_{12}=13.33\ N$

$F_{21}=-13.33\ N$

Force F₂₃,

$F_{23}=\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(10.02)^2}$

We need to calculate the value of third charge

$F_{net}=F_{21}+F_{23}$

$4.444=-13.33+\dfrac{9\times10^{9}\times6.2\times10^{-6}\times q_{3}}{(5.01)^2}$

$q_{3}=\dfrac{(4.444+13.33)\times(5.01\times10^{-2})^2}{9\times10^{9}\times6.2\times10^{-6}}$

$q_{3}=7.99\times10^{-7}\ C$

$q_{3}=0.8\times10^{-6}\ C$

Hence, The value of third charge is 0.8μC.

4 0

3 years ago

A helicopter travels from point C to point A to perform a medical supply drop. The helicopter then needs to land at point B. How

Ainat [17]

Answer:

32 degrees

Explanation:

8 0

3 years ago

A block slides down an incline plane that makes a 30 degree angle with the

Alona [7]

Hi there!

We know that:

Force due to gravity = Mgsinθ

Force due to friction = μMgcosθ

Let the positive direction be directed in the direction of the block's acceleration, which is downward.

Thus:

ΣF = Mgsinθ - μMgcosθ

Solving for acceleration requires diving all terms by the mass, so:

a = gsinθ - μgcosθ

Substitute in given values. (g = 9.8 m/s²)

a = 9.8sin(30) - 0.3(9.8)cos(30) = 2.354 m/s²

7 0

3 years ago

Lila is a track and field athlete.She has to complete four laps around the track, which is 400 meters. The race took her 6 minut

hoa [83]

V=d/t
V=?
d=400m(4)
=1600m
t=6 min.
=360 s

V=1600m/360s
V=4.4m/s

3 0

3 years ago

Two blocks of masses 20 kg and 8.0 kg are connected togetherby

OleMash [197]

Answer:

14 N

Explanation:

The tension in the second string is puling both the masses of 20 kg and 8 kg with acceleration of 0.5 m s⁻²

So tension in the second string = total mass x acceleration

= 28 x .5 = 14 N . Ans..

4 0

3 years ago