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3 years ago

What statement is true about a spring?

2 answers:

8 0

I think B is the most correct, because logically it's harder to bend a stiffer spring than it is to bend a softer one. Also, I don't think length comes into play. So B.

brilliants [131]3 years ago

4 0

Answer:

B. The stiffer the spring, the greater the force required to restore its position.

Explanation:

The 'restoring force' (the force required to restore the position of a spring) is given by:

$F=-kx$

where

F is the restoring force

k is the spring constant, which determines its stiffness (the higher the value of k, the stiffer the spring)

x is the displacement of the spring relative to its natural length

From the formula, we see that the higher the value of k (so, the stiffer the spring), the larger the restoring force, so the correct option is

B. The stiffer the spring, the greater the force required to restore its position

You might be interested in

For a damped simple harmonic oscillator, the block has a mass of 1.2 kg and the spring constant is 9.8 N/m. The damping force is

ArbitrLikvidat [17]

Answer:

a) t=24s

b) number of oscillations= 11

Explanation:

In case of a damped simple harmonic oscillator the equation of motion is

m(d²x/dt²)+b(dx/dt)+kx=0

Therefore on solving the above differential equation we get,

x(t)=A₀ $e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)$

where A(t)=A₀ $e^{\frac{-bt}{2m}}$

A₀ is the amplitude at t=0 and

$w'$ is the angular frequency of damped SHM, which is given by,

$w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

Now coming to the problem,

Given: m=1.2 kg

k=9.8 N/m

b=210 g/s= 0.21 kg/s

A₀=13 cm

a) A(t)=A₀/8

⇒A₀ $e^{\frac{-bt}{2m}}$ =A₀/8

⇒ $e^{\frac{bt}{2m}}=8$

applying logarithm on both sides

⇒ $\frac{bt}{2m}=ln(8)$

⇒ $t=\frac{2m*ln(8)}{b}$

substituting the values

$t=\frac{2*1.2*ln(8)}{0.21}=24s(approx)$

b) $w'=\sqrt{\frac{k}{m}-\frac{b^{2}}{4m^{2}} }$

$w'=\sqrt{\frac{9.8}{1.2}-\frac{0.21^{2}}{4*1.2^{2}}}=2.86s^{-1}$

$T'=\frac{2\pi}{w'}$ , where $T'$ is time period of damped SHM

⇒ $T'=\frac{2\pi}{2.86}=2.2s$

let $n$ be number of oscillations made

then, $nT'=t$

⇒ $n=\frac{24}{2.2}=11(approx)$

8 0

3 years ago

Need help solving this

Nostrana [21]

Answer:

cant see picture

Explanation:

5 0

3 years ago

Difference between freefall and weightlessness.

Margaret [11]

Answer:

Differences between freefall and weightlessness are as follows:

<h3><u>Freefall</u></h3>

When a body falls only under the influence of gravity, it is called free fall.
Freefall is not possible in absence of gravity.
A body falling in a vacuum is an example of free fall.

<h3><u>Weightlessness</u></h3>

Weightlessness is a condition at which the apparent weight of body becomes zero.
Weightlessness is possible in absence of gravity.
A man in a free falling lift is an example of weightlessness.

Hope this helps....

Good luck on your assignment....

6 0

3 years ago

W=90 kg×4n/kg<br>Please halp me i need it

kiruha [24]

I'm guessing that this is a problem to find the weight of a 90kg mass on a planet where the acceleration of gravity is 4 m/s^2. (Much less gravity than Earth, a little more than Mars.)

Just do the multiplication, and you get

360 Newtons.

5 0

3 years ago

The sound level measured in a room by a person watching a movie on a home theater system varies from 40 dB during a quiet part t

UNO [17]

Answer:

$\alpha=-3.01dB$