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gizmo_the_mogwai

1 month ago

15

Ocean waves are observed to travel to the right along the water surface during a developing storm. A Coast Guard weather station

observes that there is a vertical distance from high point to low point of 4.6 meters and a horizontal distance of 8.6 meters between adjacent crests. The waves splash into the station once every 6.2 seconds. Determine the amplitude, wavelength, frequency and period of these waves. Write out the equation that governs this behavior.

1 answer:

inna [3.1K]1 month ago

6 0

The amplitude is 2.3 m. The wavelength is 8.6 m. The frequency equals 0.16 Hz. The period lasts for 6.25 seconds. The governing equation for this behavior is. The details are illustrated in the initial uploaded image.

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A closed system of mass 10 kg undergoes a process during which there is energy transfer by work from the system of 0.147 kJ per

Softa [3030]

Result: -50.005 kJ

Details:

Provided Data

mass of the system = 10 kg

work done = 0.147 kJ/kg

Elevation change $(\Delta h)=-50 m$

initial speed $(v_1)=15 m/s$

Final Speed $(v_2)=30 m/s$

Specific internal Energy $(\Delta U)=-5 kJ/kg$

according to the first Law of thermodynamics

$Q-W=\Delta H$

$\Delta H=\Delta KE+ \Delta PE +\Delta U$

where KE represents kinetic energy

PE indicates potential energy

U denotes internal Energy

$\Delta H=\frac{m(v_2^2-v_1^2)}{2}+mg(\Delta h)+\Delta U$

$Q=W+\Delta KE+ \Delta PE +\Delta U$

$Q=0.147\times 10+\frac{10\cdot (30^2-15^2)}{2}+10\cdot 9.81\cdot (-5)-5\times 10$

Q = 1.47 + 3.375 - 4.850 - 50

Q = -50.005 kJ

5 0

2 months ago

A ball rolls up a slope. At the end of three seconds its velocity is 20 cm/s; at the end of eight seconds its velocity is 0. Wha

serg [3582]

Answer:

$a_{acceleriation}=-4cm/s^{2}\\ or\\ a_{acceleriation}=-0.04m/s^{2}$

Explanation:

Data provided

initial velocity v₀=20 cm/s at time t=3s

final velocity vf=0 at time t=8 s

Required

Average Acceleration for the interval from 3s to 8s

Solution

Acceleration can be defined as the first derivative of velocity concerning time

$a_{acceleriation} =\frac{dv_{velocity}}{dt_{time}}\\a_{acceleriation} =\frac{v_{f}-v_{o} }{dt}\\ a_{acceleriation} =\frac{0-20cm/s }{8s-3s}\\ a_{acceleriation}=-4cm/s^{2}\\ or\\ a_{acceleriation}=-0.04m/s^{2}$

8 0

1 month ago

A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in 1.90 s. You may ignore air

Sav [3153]

Answer:

Height (h) = 17 m

Velocity (v) = 18.6 m/s

Explanation: This problem can be solved using kinematic motion equations.

Given Data

Initial velocity (u) = 0

Acceleration (a) = g

Time (t) = 1.9 seconds

First, we calculate the height.

$s=ut+0.5at^2\\h=ut+0.5at^2\\h=0*1.9+0.5*9.81*1.9^2\\h=17.707 m$

Then, we find the final velocity

$v=u+at\\v=0+9.81*1.9\\v=18.639$

The acceleration graph is a linear representation described by y=9.8, as it remains constant:

The velocity graph can be represented by y=9.8x (where y signifies velocity and x indicates time):

The displacement graph can be described as y=4.9x^2 (with x as time and y as displacement):

These graphs apply exclusively from x=0 to x=1.9, so disregard other sections of the graphs.

5 0

2 months ago

A 1.97-pF capacitor with a plate area of 5.86 cm2 and separation between the plates of 2.63 mm is connected to a 9.0-V battery a

inna [3103]

If the plate separation is modified after the battery is disconnected, the updated distance between plates is 9.21 mm

If changes are made while the battery remains connected, the new separation becomes 0.11 mm

The capacitance for an air-filled parallel plate capacitor can be expressed as:

$C=\frac{\epsilon_0A}{d}$

In this equation, $\epsilon_0$ refers to the permittivity of free space, A stands for the plate area, and D represents the separation distance.

Thus,

$C \alpha \frac{1}{d}$ .......(1)

Therefore, should the distance between the plates shift from d₁ to d₂, the capacitance ratio in both scenarios can be represented as:

$\frac{C_1}{C_2} =\frac{d_2}{d_1}$ ......(2)

Scenario (i)

When the capacitor is fully charged and then disconnected from the battery before adjusting the plate distance, the charge will remain steady while the capacitance varies.

The initial energy E₁ stored in the capacitor can be expressed as:

$E_1=\frac{Q^2}{2C_1}$ ......(3)

Once the separation changes to d₂, capacitance becomes C₂, but the charge Q remains unchanged.

Thus,

$E_2=\frac{Q^2}{2C_2}$ ......(4)

By dividing equation (4) by (3),

$\frac{E_2}{E_1} =\frac{C_1}{C_2}$

According to equation (2),

$\frac{E_2}{E_1} =\frac{C_1}{C_2}=\frac{d_2}{d_1}$

This results in a 3.5 fold increase in energy.

$\frac{E_2}{E_1} =\frac{d_2}{d_1}=3.5\\ d_2=3.5*2.63 mm\\ =9.205 mm=9.21 mm$

Scenario (2)

If the capacitor is kept connected to the power source, the voltage V across the plates will remain unchanged.

The initial energy is described as

$E_1=\frac{1}{2} C_1V^2$ ......(5)

The final energy when the plate separation transitions to d₂ can be written as:

$E_2=\frac{1}{2} C_2V^2$ .....(6)

Referencing equations (5) and (6)

$\frac{E_2}{E_1} =\frac{C_2}{C_1}$

From equation (2),

$\frac{E_2}{E_1} =\frac{C_2}{C_1}=\frac{d_1}{d_2}$

Thus, in this particular scenario,

$\frac{E_2}{E_1} =\frac{d_1}{d_2}\\d_2=\frac{d_1}{3.5} \\ =\frac{2.63 mm}{3.5} \\ =0.109 mm=0.11 mm$

Therefore,

Adjusting plate separation after battery disconnection yields 9.21 mm

If modified while connected, the new separation measures 0.11 mm

6 0

1 month ago

A 5 kg object near Earth's surface is released from rest such that it falls a distance of 10 m. After the object falls 10 m, it

Ostrovityanka [3204]

Response:D

Clarification:

Provided

mass of object $m=5 kg$

Distance traveled $h=10 m$

resulting velocity $v=12 m/s$

energy conservation occurs starting when the object begins its descent and reaches a speed of 12 m/s

Initial Energy $=mgh=5\times 9.8\times 10=490 J$

Final Energy $=\frac{1}{2}mv^2+W_{f}$

$=\frac{1}{2}\cdot 5\cdot 12^2+W_{f}$

where $W_{f}$ is the work done by friction, if any

$490=360+W_{f}$

$W_{f}=130 J$

As friction is present, this indicates an open system with a net external force of zero.

An open system allows for the exchange of energy and mass, and the presence of friction indicates that it is indeed an open system.

4 0

2 months ago