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3 years ago

Why is the restoring force in Hooke's law a negative value?

2 answers:

Rudiy273 years ago

8 0

It's a negative value because the negative sign on the spring's force means the force exerted opposes the spring's displacement

Colt1911 [192]3 years ago

4 0

Explanation:

Hooke's law :

According to the Hooke's law, The force is the directly proportional to the displacement of the string.

On the other hand,

In a simple harmonic motion, The restoring force is directly proportional to the displacement and acts in the opposite direction to that of displacement.

The restoring force is defined as,

$F\propto -x$

$F = -kx$

Where,

F = restoring force

k = restoring constant

x = displacement

The negative sign shows the force opposes the string's displacement

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Interactive Solution 9.1 presents a model for solving this problem. The wheel of a car has a radius of 0.300 m. The engine of th

Murrr4er [49]

Answer:

740 N

Explanation:

We are given that

Radius,r=0.3 m

Torque, $\tau=222 Nm$

We have to find the magnitude of the static frictional force.

According to question

Torque by engine=Torque by static friction

$222=f\times r$

$f=\frac{222}{r}$

$f=\frac{222}{0.3}$

$f=740 N$

Hence, the magnitude of static frictional force=740 N

8 0

3 years ago

1. A flywheel begins rotating from rest, with an angular acceleration of 0.40 rad/s. a) What will its angular velocity be 3 seco

Sidana [21]

Answer:

(a) 1.2 rad/s

(b) 1.8 rad

Explanation:

Applying,

(a) α = (ω-ω')/t................ Equation 1

Where α = angular acceleration, ω = final angular velocity, ω' = initial angular velocity, t = time.

From the question,

Given: α = 0.40 rad/s², t = 3 seconds, ω' = 0 rad/s (from rest)

Substitute these values into equation 1

0.40 = (ω-0)/3

ω = 0.4×3

ω = 1.2 rad/s

(b) Using,

∅ = ω't+αt²/2.................. Equation 2

Where ∅ = angle turned.

Substitutting the values above into equation 2

∅ = (0×3)+(0.4×3²)/2

∅ = 1.8 rad.

5 0

3 years ago

Find the charge on the capacitor in an LRC-series circuit when L = 1 2 h, R = 10 Ω, C = 0.01 f, E(t) = 50 V, q(0) = 1 C, and i(0

alex41 [277]

Answer:

Explanation:

Given that,

In an LRC circuit

L = 1/2h

R = 10 Ω,

C = 0.01 f

E(t) = 50 V,

q(0) = 1 C, and

i(0) = 0 A.

q(t) = C

We can to fine the charge after a long time, let say t→∞

The Kirchoff second law for the system is

L•dq²/dt + R•dq/dt + q/C = E(t)

Divide through by L

dq²/dt + R/L •dq/dt + q/LC = E(t)/L

Now inserting the values of R, L, C and E

dq²/dt+10/½ •dq/dt +q/½×0.01=50/½

dq²/dt + 20•dq/dt + 200q = 100

Let solve the differential equation

First : homogenous solution

Using D operator

D² + 20D +200 = 0

Solving the quadratic equation using formula method

D = (-b±√b²-4ac)/2a

D = (-20±√20²-4×1×200) /2

D = (-20±√400-800)/2

D = (-20±20•i)/2

D = -10±10•i

So we have a complex solution

Then, the complementary solution is

q(t) = e(-10t)[ Acos10t + BSin10t]

A and B are constant

Let find the particular solution using the method of undetermine coefficient

Let assume particular solution of

q(t) = C, I.e q(t) Is a constant

So, inserting this into the equation below

dq²/dt + 20•dq/dt + 200q = 100

200q = 100

q = 100/200

q = ½

Then, the particular solution is ½

So, the total solution is the sum of particular solution and complementary solution

q = e(-10t)[ Acos10t + BSin10t] + ½

Using the initial conditions

q(0) = 1

1 = e(0) [ACos0 + BSin0] +½

1 = A+½

A = ½

Also i(0) = 0

I(t) = q'(t)

Then,

q'(t) = -10•e(-10t)[ Acos10t + BSin10t] + e(-10t)[ -10Asin10t + 10BCos10t]

0 = -10e(0) [ ACos0 + BSin0] + e(0)[-10ASin0 +10BCos0]

0 = -10(A) + 10B

A=B=½

So the general equation becomes

q(t) = e(-10t)[ ½cos10t + ½Sin10t] + ½

So, as t→∞, the aspect of e(-10t) become zero

So the charge stabilizes at q = ½C after a long time

q = ½C as t→∞

6 0

3 years ago

Read 2 more answers

10 points

Nana76 [90]

Answer:

Explanation:

You didn't fill in the proper masses which is why you never got an answer to this. But that's ok...I got you. I happen to know what they are! We will use the universal law of gravitation and the gravitational constant to solve this.

$F_g=\frac{Gm_1m_2}{r^2}$ and filling in:

$F_g=\frac{(6.67*10^{-11})(5.98*10^{24})(7.36*10^{22})}{(3.84*10^8)^2}$ The denominator is the radius of the earth plus the radius of the moon plus the distance between their surfaces, just FYI.

That gives us that

$F_g=1.99*10^{20}N$ Not sure what your choices entail, but I'd have to say, taking into consideration that maybe your problem didn't figure in the distance between the surfaces, you'd be at choice B.

5 0

2 years ago

A dragster going at 5 m/s north increases its velocity to 23 m/s north in 4 seconds. What is its acceleration during this time i

m_a_m_a [10]

Acceleration = (change in speed) / (time for the change)

Acceleration = (23 - 5) m/s / 4s = 18/4 m/s² = <em>4.5 m/s²</em>

6 0

3 years ago

Read 2 more answers