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timama [110]
3 years ago
14

Jed drops a 18 kg box off of the Eiffel tower. After 6.1 seconds, how fast is the box moving (neglect air resistance.)

Physics
2 answers:
boyakko [2]3 years ago
4 0

Answer:

Speed, v = 60 m/s

Explanation:

Given that,

Mass of the box, m = 18 kg

Time, t = 6.1 seconds

Initially the box is at rest, u = 0

Let v is the final speed of the box i.e. after 6.1 seconds. We can calculate it using first equation of motion as :

v=u+at

Here, a = g

v=gt

v=9.8\times 6.1

v = 59.78 m/s

or

v = 60 m/s

So, after 6.1 seconds, the box is moving with a speed of 60 m/s. Hence, this is the required solution.

GuDViN [60]3 years ago
3 0
The box fall with acceleration 10m/s2 due to gravity

velocity/ 6.1 = 180
velocity = 1098 m/s

the answer is A.
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The strength of the electric field at a certain distance from a point charge is represented by E. What is the strength of the el
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Answer:

E

Explanation:

Using Coulomb's law equation

Force of the charge = k qQ /d²

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E = ( K Qq/ d² ) / q

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E = KQ / d²

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A 10 g particle undergoes SHM with an amplitude of 2.0 mm and a maximum acceleration of magnitude 8.0 multiplied by 103 m/s2, an
Nat2105 [25]

Answer:

a)T=0.0031416s

b)v_{max}=6.283\frac{m}{s}

c) E=0.1974J

d)F=80N

e)F=40N

Explanation:

1) Important concepts

Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".

2) Part a

The equation that describes the simple armonic motion is given by X=Acos(\omega t +\phi)    (1)

And taking the first and second derivate of the equation (1) we obtain the velocity and acceleration function respectively.

For the velocity:

\frac{dX}{dt}=v(t)=-A\omega sin(\omega t +\phi)   (2)

For the acceleration

\frac{d^2 X}{dt}=a(t)=-A\omega^2 cos(\omega t+\phi)   (3)

As we can see in equation (3) the acceleration would be maximum when the cosine term would be -1 and on this case:

A\omega^2=8x10^{3}\frac{m}{s^2}

Since we know the amplitude A=0.002m  we can solve for \omega like this:

\omega =\sqrt{\frac{8000\frac{m}{s^2}}{0.002m}}=2000\frac{rad}{s}

And we with this value we can find the period with the following formula

T=\frac{2\pi}{\omega}=\frac{2 \pi}{2000\frac{rad}{s}}=0.0031416s

3) Part b

From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:

v_{max}=A\omega =0.002mx2000\frac{rad}{s}=4\frac{m rad}{s}=4\frac{m}{s}

4) Part c

In order to find the total mechanical energy of the oscillator we can use this formula:

E=\frac{1}{2}mv^2_{max}=\frac{1}{2}(0.01kg)(6.283\frac{m}{s})^2=0.1974J

5) Part d

When we want to find the force from the 2nd Law of Newton we know that F=ma.

At the maximum displacement we know that X=A, and in order to that happens cos(\omega t +\phi)=1, and we also know that the maximum acceleration is given by::

|\frac{d^2X}{dt^2}|=A\omega^2

So then we have that:

F=ma=mA\omega^2

And since we have everything we can find the force

F=ma=0.01Kg(0.002m)(2000\frac{rad}{s})^2 =80N

6) Part e

When the mass it's at the half of it's maximum displacement the term cos(\omega t +\phi)=1/2 and on this case the acceleration would be given by;

|\frac{d^2X}{dt^2}|=A\omega^2 cos(\omega t +\phi)=A\omega^2 \frac{1}{2}

And the force would be given by:

F=ma=\frac{1}{2}mA\omega^2

And replacing we have:

F=\frac{1}{2}(0.01Kg)(0.002m)(2000\frac{rad}{s})^2 =40N

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