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The temperature within a thin plate with thermal conductivity of 10 W/m/K depends on position as given by the following expression: TT=(100 K)????????−xx2/????????xx 2cos�yy/????????yy�+300 K Where, Lx = 1 m, and Ly = 2 m. At the point (0.4 m, 1 m), find: a. The magnitude of the heat flux b. The direction of the heat flux

Answers

Answer 1

Answer:

Heat flux = (598.3î + 204.3j) W/m²

a) Magnitude of the heat flux = 632.22 W/m²

b) Direction of the heat flux = 18.85°

Explanation:

- The correct question is the first image attached to this solution.

- The solution to this question is contained on the second and third images attached to this solution respectively.

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Madelin fires a bullet horizontally. The rifle is 1.4 meters above the ground. The bullet travels 168 meters horizont before it hits the ground. What speed did Madelin's bullet have when it exited the rifle?
A 62.0 kg skier is moving at 6.90 m/s on a frictionless, horizontal, snow-covered plateau when she encounters a rough patch 4.50 m long. The coefficient of kinetic friction between this patch and her skis is 0.300. After crossing the rough patch and returning to friction-free snow, she skis down an icy, frictionless hill 2.50 m high.How fast is the skier moving when she gets to the bottom ofthe hill?
The volume control on a stereo is designed so that three clicks of the dial increase the output by 10 dB. How many clicks are required to increase the power output of the loudspeakers by a factor of 100?
an object down, but this is not true. If you place a box of mass 8 kg on a moving horizontal conveyor belt, the friction force of the belt acting on the bottom of the box speeds up the box. At first there is some slipping, until the speed of the box catches up to the speed of the belt, which is 5 m/s. The coefficient of kinetic friction between box and belt is 0.6. (a) How much time does it take for the box to reach this final speed

The type of function that describes the amplitude of damped oscillatory motion is _______. The type of function that describes the amplitude of damped oscillatory motion is _______. quadratic sinusoidal inverse exponential linear

Answers

Answer:

exponential

Explanation:

type of function that describes the amplitude of damped oscillatory motion is exponential because as we know that here function is

y = A × $e^{(-bt)/(2m)}$ × cos(ωt + ∅ ) ..................................... ( 1 )

here function A × $e^{(-bt)/(2m)}$ is amplitude

as per equation ( 1 )it is exponential

so that we can say that amplitude of damped oscillatory motion is exponential

As you go up the y-axis, what happens to the number of sprouted bean seeds?A. Sprouted bean seeds decrease.

B. Sprouted bean seeds increase.

C. Sprouted bean seeds remain constant.

D. None of the above

Answers

As we go up the y-axis, the number of sproutedbean seeds increase (option B).

What is a graph?

Graph is a data chart intended to illustrate the relationship between a set (or sets) of numbers (quantities, measurements or indicative numbers) and a reference set.

In a graph, there are two axes as follows;

Y-axis or vertical axis
X-axis or horizontal axis

According to this question, a graph of number of sprouted bean seeds on the y-axis is plotted against temperature on the x-axis.

We can observe that as we go up the y-axis, the number of sprouted bean seeds increase.

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At a given instant, a 2.2 A current flows in the wires connected to a parallel-plate capacitor. What is the rate at which the electric field is changing between the plates if the square plates are 2.0 cm on a side? Express your answer using two significant figures

Answers

Answer:

Check attachment for better understanding

Explanation:

Given that,

Current in wire I =2.2A

Capacitor plate dimension is 2cm by 2cm

s=2cm=2/100 = 0.02m

Rate at which electric field Is changing dE/dt?

The current in the wires must also be the displacement current in the capacitor. We find the rate at which the electric field is changing from

ID = ε0•A•dE/dt

Where ε0 is a constant

ε0= 8.85×10^-12C²/Nm²

Area of the square plate is

A =s² =0.02² = 0.0004m²

Then,

Make dE/dt the subject of formula

dE/dt = ID/ε0A

dE/dt = 2.2 / (8.85×10^-12 ×4×10^-4)

dE/dt = 6.215×10^14 V/m-s

dE/dt = 6.215×10^14 N/C.s

The rate at which the electric field is changing between the plates is 6.215×10^14 N/C.s

Consider a vertical elevator whose cabin has a total mass of 800 kg when fully loaded and 150 kg when empty. The weight of the elevator cabin is partially balanced by a 400-kg counterweight that is connected to the top of the cabin by cables that pass through a pulley located on top of the elevator well. Neglecting the weight of the cables and assuming the guide rails and the pulleys to be frictionless, determine (a) the power required while the fully loaded cabin is rising at a constant speed of 1.2 m/s and (b) the power required while the empty cabin is descending at a constant speed of 1.2 m/s. What would your answer be to (a) if no counterweight were used? What would your answer be to (b) if a friction force of 800 N has developed between the cabin and the guide rails?

Answers

Answer:

Part a)

$P = 4.71 * 10^3 Watt$

Part b)

$P = 2.94 * 10^3 W$

Part c)

$P = 9.4 * 10^3 W$

Part d)

$P = 3.9 * 10^3 W$

Explanation:

Part a)

When cabin is fully loaded and it is carried upwards at constant speed

then we will have

net tension force in the rope = mg

$T = (800)(9.81)$

$T = 7848 N$

now it is partially counterbalanced by 400 kg weight

so net extra force required

$F = 7848 - (400 * 9.81)$

$F = 3924 N$

now power required is given as

$P = Fv$

$P = 3924 (1.2)$

$P = 4.71 * 10^3 Watt$

Part b)

When empty cabin is descending down with constant speed

so in that case the force balance is given as

$F + (150 * 9.8) = (400 * 9.8)$

$F = 2450 N$

now power required is

$P = F.v$

$P = (2450)(1.2)$

$P = 2.94 * 10^3 W$

Part c)

If no counter weight is used here then for part a)

$F = 7848 N$

now power required is

$P = F.v$

$P = 7848 (1.2)$

$P = 9.4 * 10^3 W$

Part d)

Now in part b) if friction force of 800 N act in opposite direction

then we have

$F + (150 * 9.8) = 800 +(400 * 9.8)$

$F = 3250 N$

now power is

$P = (3250)(1.2)$

$P = 3.9 * 10^3 W$

A diverging lens has a focal length of -30.0 cm. An object is placed 18.0 cm in front of this lens.(a) Calculate the image distance.

(b) Calculate the magnification.

Answers

Answer:

A) Calculate the distance

The physics of wind instruments is based on the concept of standing waves. When the player blows into the mouthpiece, the column of air inside the instrument vibrates, and standing waves are produced. Although the acoustics of wind instruments is complicated, a simple description in terms of open and closed tubes can help in understanding the physical phenomena related to these instruments. For example, a flute can be described as an open-open pipe because a flutist covers the mouthpiece of the flute only partially. Meanwhile, a clarinet can be described as an open-closed pipe because the mouthpiece of the clarinet is almost completely closed by the reed.1. Consider a pipe of length 80.0 cm open at both ends. What is the lowest frequency f of the sound wave produced when you blow into the pipe?
2. A hole is now drilled through the side of the pipe and air is blown again into the pipe through the same opening. The fundamental frequency of the sound wave generated in the pipe is now:______.
a. the same as before.
b. lower than before.
c. higher than before.
3. If you take the original pipe in Part A and drill a hole at a position half the length of the pipe, what is the fundamental frequency of the sound that can be produced in the pipe?
4. What frequencies, in terms of the fundamental frequency of the original pipe in Part A, can you create when blowing air into the pipe that has a hole halfway down its length?
4-1. Recall from the discussion in Part B that the standing wave produced in the pipe must have an antinode near the hole. Thus only the harmonics that have an antinode halfway down the pipe will still be present.
A. Only the odd multiples of the fundamental frequency.
B. Only the even multiples of the fundamental frequency.
C. All integer multiples of the fundamental frequency.
E. What length of open-closed pipe would you need to achieve the same fundamental frequency as the open pipe discussed in Part A?
A. Half the length of the open-open pipe.
B. Twice the length of the open-open pipe.
C. One-fourth the length of the open-open pipe.
D. Four times the length of the open-open pipe.
E. The same as the length of the open-open pipe.
F. What is the frequency of the first possible harmonic after the fundamental frequency in the open-closed pipe described in Part E?
F-1. Recall that possible frequencies of standing waves that can be generated in an open-closed pipe include only odd harmonics. Then the first possible harmonic after the fundamental frequency is the third
harmonic.

Answers

Final answer:

The physics of wind instruments is based on standing waves. The lowest frequency of a sound wave produced in an open-open pipe can be calculated. When a hole is drilled through the pipe, the fundamental frequency is lower than before. Only odd multiples of the fundamental frequency will be present in a pipe with a hole halfway down its length. An open-closed pipe needs to be twice the length of an open-open pipe to achieve the same fundamental frequency. The first possible harmonic after the fundamental frequency in an open-closed pipe is the third harmonic.

Explanation:

The lowest frequency f of the sound wave produced when blowing into an open-open pipe can be calculated using the formula f = v/2L, where v is the speed of sound and L is the length of the pipe. Plugging in the values, we get f = 343/(2*0.8), which equals 214.375 Hz.

When a hole is drilled through the side of the pipe, the fundamental frequency of the sound wave generated in the pipe is lower than before. This is because the effective length of the pipe has been changed, resulting in a lower frequency.

The fundamental frequency of the sound that can be produced in the original pipe with a hole drilled halfway down its length can be calculated as f = v/L, where L is the new effective length of the pipe. Since the hole is halfway down, the effective length becomes half of the original length, resulting in a frequency equal to the original fundamental frequency.

When blowing air into the pipe with a hole halfway down its length, only the odd multiples of the fundamental frequency will be present. Therefore, the frequencies that can be created are only the odd multiples of the fundamental frequency.

The length of an open-closed pipe needed to achieve the same fundamental frequency as an open-open pipe is twice the length of the open-open pipe. This is because an open-closed pipe has only odd harmonics, which are spaced twice as far apart as the harmonics in an open-open pipe.

The first possible harmonic after the fundamental frequency in an open-closed pipe is the third harmonic.

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