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2 years ago

Quinn accelerates her skateboard along a straight path from 0 m/s to 4.0 m/s

Physics

1 answer:

umka2103 [35]2 years ago

8 0

initial velocity=u=0m/s
Final velocity=v=4m/s
Time=t=2.5s

$\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf\longmapsto Acceleration=\dfrac{4-0}{2.5}$

$\\ \sf\longmapsto Acceleration=\dfrac{4}{2.5}$

$\\ \sf\longmapsto Acceleration=1.6m/s^2$

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There are two tuning forks struck at the same time. One tuning fork is tuned to a frequency note of 500 Hz. The other is tuned t

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Answer: 26 beats in 2 seconds

Explanation:

The number of beats per second = frequency of tuning fork 1 - frequency of tuning fork 2.

Given :

- frequency of tuning fork 1 = 500Hz

- frequency of tuning fork 2 = 487Hz

Thus,

Beats per second = 500 - 487

= 13 beats per second

Therefore in two(2) seconds, you will have 2 x 13 = 26 seconds.

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2 years ago

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2 years ago

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

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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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2 years ago

As the captain of the scientific team sent to Planet Physics, one of your tasks is to measure g. You have a long, thin wire labe

spin [16.1K]

Answer:

1.19 m/s²

Explanation:

The frequency of the wave generated in the string in the first experiment is f = n/2l√T/μ were T = tension in string = mg were m = 1.30 kg weight = 1300 g , μ = mass per unit length of string = 1.01 g/m. l = length of string to pulley = l₀/2 were l₀ = lent of string. Since f is the second harmonic, n = 2, so

f = 2/2(l₀/2)√mg/μ = 2(√mg/μ)/l₀ (1)

Also, for the second experiment, the period of the wave in the string is T = 2π√l₀/g. From (1) l₀ = 2(√mg/μ)/f and from (2) l₀ = T²g/4π²

Equating (1) and (2) we ave

2(√mg/μ)/f = T²g/4π²

Making g subject of the formula

g = 2π√(2√(m/μ)/f)/T

The period T = 316 s/100 = 3.16 s

Substituting the other values into , we have

g = 2π√(2√(1300 g/1.01 g/m)/200 Hz)/3.16

g = 2π√(2 × 35.877/200 Hz)/3.16

g = 2π√(71.753/200 Hz)/3.16

g = 2π√(0.358)/3.16

g = 2π × 0.599/3.16

g = 1.19 m/s²

6 0

2 years ago

Point charges of 28.0 µC and 42.0 µC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el

olga_2 [115]

Answer: a) electric field will be zero at zero meters apart

b) for smaller charge q, E = 6.048X10^6N/m towards away from the charge,

for bigger charge Q, E = 4.032X10^6N/m

Explanation:

Detailed explanation and calculation is shown in the image below.

6 0

2 years ago