PHYSICS COLLEGE

A 2.07-kg fish is attached to the lower end of an unstretched vertical spring and released. The fish drops 0.131 m before momentarily coming to rest. (a) What is the spring constant of the spring? (b) What is the period of the oscillations of the fish? ?

Answers

Answer 1

Answer:

Answer:

part a)

k = 310 N/m

part b)

T = 0.51 s

Explanation:

Part A)

As per work energy theorem we have

Work done by gravity + work done by spring = change in kinetic energy

$mgx - (1)/(2)kx^2 = 0$

$(2.07)(9.8)(0.131) - (1)/(2)k(0.131)^2 = 0$

now we will have

$k = 310 N/m$

Part B)

Time period of oscillation is given as

$T = 2\pi\sqrt{(m)/(k)}$

$T = 2\pi\sqrt{(2.07)/(310)}$

$T = 0.51 s$

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A car travels 13 km in a southeast direction and then 16 km 40 degrees north of east. What is the car's resultant direction?

Answers

Answer:

21.48 km 2.92° north of east

Explanation:

To find the resultant direction, we need to calculate a sum of vectors.

The first vector has module = 13 and angle = 315° (south = 270° and east = 360°, so southeast = (360+270)/2 = 315°)

The second vector has module 16 and angle = 40°

Now we need to decompose both vectors in their horizontal and vertical component:

horizontal component of first vector: 13 * cos(315) = 9.1924

vertical component of first vector: 13 * sin(315) = -9.1924

horizontal component of second vector: 16 * cos(40) = 12.2567

vertical component of second vector: 16 * sin(40) = 10.2846

Now we need to sum the horizontal components and the vertical components:

horizontal component of resultant vector: 9.1924 + 12.2567 = 21.4491

vertical component of resultant vector: -9.1924 + 10.2846 = 1.0922

Going back to the polar form, we have:

$module = √(horizontal^2 + vertical^2)$

$module = √(460.0639 + 1.1929)$

$module = 21.4769$

$angle = arc\ tangent(vertical/horizontal)$

$angle = arc\ tangent(1.0922/21.4491)$

$angle = 2.915\°$

So the resultant direction is 21.48 km 2.92° north of east.

In the study of sound, one version of the law of tensions is:f1= f2 √ (F1/F2)

If f1= 300, F2= 60, and f2=260, find f1 to the nearest unit.

Answers

Answer:

F1 = 80

Explanation:

f1= f2 √ (F1/F2)

Where f1 = 300, f2 = 260 and F2 = 60

Putting in the above formula

300 = 260√(F1/60)

Dividing both sides by 260

=> 1.15 = √(F1/60)

Squaring both sides

=> 1.33 = F1/60

Multiplying both sides by 60

=> F1 = 80

A remote controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by v= [5.00 m/s – (0.0180 m/s3)t^2 ]i+[2.00 m/s + (0.550 m/s2)t ]j .a) What are ax(t) and ay(t), the x- and y- components of cars acceleration as a function of time?
b) What are the magnitude and direction of the velocity of the car at t= 8 sec?
c) What is the magnitude and direction of cars acceleration at t=8 sec

Answers

Maybe try A for your answer

A uniform, 4.5 kg, square, solid wooden gate 2.0 m on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.2 kg raven flying horizontally at 4.5 m/s flies into this door at its center and bounces back at 1.5 m/s in the opposite direction. What is the angular speed of the gate just after it is struck by the unfortunate raven?

Answers

Answer:

Explanation:

Mass of the gate, $m_1 = 4.5 kg$

Mass of the raven, $m_2 = 1.2 kg$

Initial speed of raven, $v_1 = 4.5 m/s$

Final speed of raven, $v_2 = - 1.5 m/s$

Moment of Inertia of the gate about the axis passing through one end:

$I = (1)/(3) m_1 a^2\nI = (1)/(3) *4.5 * 2^2\nI = 6 kg m^2$

Angular momentum of the gate, $L = I \omega$

$L = 5.33 \omega$

Using the law of conservation of angular momentum:

$m_2 v_f (a/2) + I\omega = m_2v_i (a/2)\nI\omega = m_2 (a/2)(v_i - v_f)\n$

The interatomic spring stiffness for tungsten is determined from Young's modulus measurements to be 90 N/m. The mass of one mole of tungsten is 0.185 kg. If we model a block of tungsten as a collection of atomic "oscillators" (masses on springs), what is one quantum of energy for one of these atomic oscillators? Note that since each oscillator is attached to two "springs", and each "spring" is half the length of the interatomic bond, the effective interatomic spring stiffness for one of these oscillators is 4 times the calculated value given above. Use these precise values for the constants: ℏ = 1.0546 10-34 J · s (Planck's constant divided by 2π) Avogadro's number = 6.0221 1023 molecules/mole kB = 1.3807 10-23 J/K (the Boltzmann constant)

Answers

Answer:

Explanation:

solution below

Final answer:

The quantum of energy for one atomic oscillator in tungsten, given the effective interatomic spring stiffness of 360 N/m, the mass of one tungsten atom as 3.074 x 10^-25 kg, and the reduced Planck's constant of 1.0546 x 10^-34 J · s, can be calculated to be approximately 1.33 x 10^-21 J.

Explanation:

To calculate the quantum of energy for one atomic oscillator in tungsten, we will consider the model of an atom being connected to two springs, both having an effective interatomic spring stiffness of four times the given value (90 N/m). This value thus becomes 360 N/m.

One mole of tungsten has a mass of 0.185 kg, thus the mass of one atom can be determined by dividing this value by Avogadro's number (6.0221 x 10^23 molecules/mole), which gives approximately 3.074 x 10^-25 kg.

The quantum of energy, or the energy of one quantum (the smallest possible energy increment), is given by the formula E = ħω, where ħ is the reduced Planck's constant (1.0546 x 10^-34 J · s) and ω is the angular frequency, given by sqrt(k/m), where k is the spring constant and m is the mass.

Substituting the known values into these equations gives ω= sqrt((360)/(3.074 x 10^-25)) and E= (1.0546 x 10^-34) x sqrt((360)/(3.074 x 10^-25)), which results in a quantum of energy of approximately 1.33 x 10^-21 J.

Learn more about Quantum Energy here:

brainly.com/question/28175160

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Solenoid A has total number of turns N length L and diameter D. Solenoid B has total number of turns 2N, length 2L and diameter 2D. Inductance of solenoid A is 8 times inductance of solenoid B
1/4 of inductance of solenoid B
same as inductance of solenoid B
1/8 of inductance of solenoid B
four times of inductance of solenoid B

Answers

Answer:

∴Inductance of solenoid A is $\frac18$ of inductance of solenoid B.

Explanation:

Inductance of a solenoid is

$L=N\frac\phi I$

$=N(B.A)/(I)$

$=N(\mu_0NI)/(l.I)A$

$=(\mu_0N^2A)/(l)$

$=(\mu_0N^2)/(l).\pi(\frac d2)^2$

$=\mu_0\pi(N^2d^2)/(4l)$

N= number of turns

$l$ = length of the solenoid

d= diameter of the solenoid

A=cross section area

B=magnetic induction

$\phi$ = magnetic flux

$I$ = Current

Given that, Solenoid A has total number of turns N, length L and diameter D

The inductance of solenoid A is

$=\mu_0\pi(N^2D^2)/(4L)$

Solenoid B has total number of turns 2N, length 2L and diameter 2D

The inductance of solenoid B is

$=\mu_0\pi((2N)^2(2D)^2)/(4.2L)$

$=\mu_0\pi(16 N^2D^2)/(4.2L)$

Therefore,

$\frac {\textrm{Inductance of A}}{\textrm{Inductance of B}}=(\mu_0\pi(N^2D^2)/(4L))/(\mu_0\pi(16 N^2D^2)/(4.2L))$

$\Rightarrow \frac {\textrm{Inductance of A}}{\textrm{Inductance of B}}=\frac18$

$\Rightarrow {\textrm{Inductance of A}}=\frac18* {\textrm{Inductance of B}}$

∴Inductance of solenoid A is $\frac18$ of inductance of solenoid B.

Hi there!

We can begin by calculating the inductance of a solenoid.

Recall:
$L = (\Phi _B)/(i)$

L = Inductance (H)
φ = Magnetic Flux (Wb)

i = Current (A)

We can solve for the inductance of a solenoid. We know that its magnetic field is equivalent to:
$B = \mu _0 (N)/(L)i$

And that the magnetic flux is equivalent to:
$\Phi _B = \int B \cdot dA = B \cdot A$

Thus, the magnetic flux is equivalent to:
$\Phi _B = \mu _0 (N)/(L)iA$

The area for the solenoid is the # of loops multiplied by the cross-section area, so:
$A_(total)= N * A$

$\Phi _B = \mu _0 (N^2)/(L)iA$

Using this equation, we can find how it would change if the given parameters are altered:
$\Phi_B ' = \mu_0 ((2N)^2)/(2L) i * 4A$

**The area will quadruple since a circle's area is 2-D, and you are doubling its diameter.

$\Phi'_B = (4)/(2) * 4(\mu_0 (N)/(L)iA) = 8\mu_0 (N)/(L)iA$

Thus, Solenoid B is 8 times as large as Solenoid A.

Solenoid A is 1/8 of the inductance of solenoid B.

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