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ArbitrLikvidat [17]

3 years ago

9

A body oscillates with simple harmonic motion along the x axis. Its displacement varies with time according to the equation x =

5 sin (πt + π/3). What is the phase of the motion (in rad) at t = 2 s?

1 answer:

frutty [35]3 years ago

7 0

Answer:

θ = (7π / 3) rad

Explanation:

given,

displacement of simple harmonic motion along x-axis

equation is given as

x = 5 sin (π t + π/3 )

general equation of simple harmonic motion

x = A sin θ

θ is the phase angle

θ = π t + π/3

at t = 2 s

$\theta =\pi \times 2 +\dfrac{\pi}{3}$

$\theta =\dfrac{7\pi}{3}\ rad$

Phase of the motion at t =2 s is θ = (7π / 3) rad

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What must be the distance between point charge q1 = 28.0 μC and point charge q2 = −57.0 μC for the electrostatic force between t

vlabodo [156]

Answer:

1.686 m

Explanation:

From coulomb's law,

F = kq1q2/r² ...................................... Equation 1

Where F = electrostatic force between the two charges, q1 = first charge, q2 = second charge, r = distance between the charges.

making r the subject of the equation,

r = √(kq1q2/F).......................... Equation 2

Given: F = 5.05 N, q1 = 28.0 μC = 28×10⁻⁶ C, q2 = 57.0 μC = 57.0×10⁻⁶ C

Constant: k = 9.0×10⁹ Nm²/C².

Substituting into equation 2

r = √(9.0×10⁹×28×10⁻⁶×57.0×10⁻⁶/5.05)

r = √(14364×10⁻³/5.05)

r = √(14.364/5.05)

r = √2.844

r = 1.686 m

r = 1.686 m.

Thus the distance must be 1.686 m

6 0

3 years ago

A person is standing on a level floor. His head, upper

BabaBlast [244]

Answer:

$y_{cg} = 1.03 m$

Explanation:

Given data:

weigh (head+arms + head) w_1 = 438 N

centre of gravity y_1= 1.28 m

weigh (upper leg) w_2 = 144 N

Center of gravity y_2 = 0.760 m

weigh ( lower leg + feet) = 87 N

centre of gravity = y_3 = 0.250 m

location of center of gravity $= \frac{w_1 y_1 + w_2 y_2 + w_3 y_3}{w_1 +W_2 +w_3}$

$y_{cg} = \frac{438 \times 1.28 + 144\times 0.760 + 87 \times 0.250}{438+144+87}$

$y_{cg} = 1.03 m$

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3 years ago

A student (m = 68 kg) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a

Gnesinka [82]

Answer:

5.7141 m

Explanation:

Here the potential and kinetic energy will balance each other

$mgh=\frac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}$

This is the initial velocity of the system and the final velocity is 0

t = Time taken = 0.04 seconds

F = Force = 18000 N

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}$

From Newton's second law

$F=ma\\\Rightarrow F=m\frac{v-u}{t}\\\Rightarrow 18000=68\frac{0-\sqrt{2gh}}{0.04}\\\Rightarrow \frac{18000}{68}\times 0.04=-\sqrt{2\times 9.81\times h}\\\Rightarrow 10.58823=-\sqrt{2\times 9.81\times h}$

Squarring both sides

$112.11061=2\times 9.81\times h\\\Rightarrow h=\frac{112.11061}{2\times 9.81}\\\Rightarrow h=5.7141\ m$

The height from which the student fell is 5.7141 m

5 0

3 years ago

As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arr

Nitella [24]

Answer:

<u><em>1. Part A: 648N</em></u>
<u><em></em></u>
<u><em>2. Part B: 324J</em></u>
<u><em></em></u>
<u><em>3. Part C: 1,296N</em></u>

Explanation:

<em><u>1. Part A:</u></em>

The magnitude of the force is calculated using the Hook's law:

$|F|=k\Delta x$

You know $\Delta x=0.250m$ but you do not have $k$ .

You can calculate it using the equation for the work-energy for a spring.

The work done to compress the springs a distance $\Delta x$ is:

$Work=\Delta PE=(1/2)k(\Delta x)^2$

Where $\Delta PE$ is the change in the elastic potential energy of the "spring".

Here you have two springs, but you can work as if they were one spring.

You know the work (81.0J) and the length the "spring" was compressed (0.250m). Thus, just substitute and solve for k:

$81.0J=(1/2)k(0.250m)^2\\\\k=2,592N/m$

In reallity, the constant of each spring is half of that, but it is not relevant for the calculations and you are safe by assuming that it is just one spring with that constant.

Now calculate the magnitude of the force:

$|F|=k\Delta x=2,592N/m\times 0.250m=648N$

<u><em></em></u>

<u><em>2. Part B. How much additional work must you do to move the platform a distance 0.250 m farther?</em></u>

<u><em></em></u>

The additional work will be the extra elastic potential energy that the springs earn.

You already know the elastic potential energy when Δx = 0.250m; now you must calculate the elastic potential energy when Δx = 0.250m + 0.250m = 0.500m.

$\Delta E=(1/2)2,592n/m\times(0.500m)^2=324J$

Therefore, you must do 324J of additional work to move the plattarform a distance 0.250 m farther.

<em><u></u></em>

<em><u>3. Part C</u></em>

<u><em></em></u>

<u><em>What maximum force must you apply to move the platform to the position in Part B?</em></u>

The maximum force is when the springs are compressed the maximum and that is 0.500m

Therefore, use Hook's law again, but now the compression length is Δx = 0.500m

$|F|=k\Delta x=2,592N/m\times 0.500m =1,296N$

8 0

3 years ago

The maximum lift-to-drag ratio of the World War I Sopwith Camel was 7.7. If the aircraft is in flight at 5000 ft when the engine

andre [41]

The related concepts to solve this problem is the Glide Ratio. This can be defined as the product between the height of fall and the lift-to-drag ratio. Mathematically, this expression can be written as,

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$R = 5000ft (7.7)$

$R = 38500ft$

Converting this units to miles.

$R = 38500ft (\frac{1mile}{5280ft})$

$R = 7.2916miles$

Therefore the glide in terms of distance measured along the ground is 7.2916miles

3 0

4 years ago