PHYSICS COLLEGE

A train accelerates at -1.5 m/s2 for 10 seconds. If the train had an initialspeed of 32 m/s, what is its new speed?
A. 17 m/s
B. 15 m/s
C. 47 m/s
D. 32 m/s

Answers

Answer 1

Answer:

Answer:

17 m/s

Explanation:

Using formula a = (v-u) /t

acceleration a = -1.5 m/s2

final velocity v = unknown

initial velocity u = 32 m/s

time t = 10s

-1.5 = (v-32)/10

-15 = v - 32

-15 + 32 = v

v = 17 m/s

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The word acid comes from the Latin word

Answers

Hi :)

The word acid comes from the Latin word acere, which means sour

Hope this helps!

It is acere but for future reference you can search of definition press more and google tells you its origin

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.

Learn more about standing waves here:

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Nitrogen (N2) undergoes an internally reversible process from 6 bar, 247°C during which pν1.2 = constant. The initial volume is 0.1 m3 and the work for the process is 50 kJ. Assuming ideal gas behavior, and neglecting kinetic and potential energy effects, determine heat transfer, in kJ, and the entropy change, in kJ/K.

Answers

Answer:

$Q=25\ kJ$

$\Delta s= 0.2885 J.K^(-1)$

Explanation:

Given:

Initial pressure of nitrogen, $P_1=6* 10^5\ Pa$

initial temperature, $T_1=247+273=520\ K$

polytropic index, $n=1.2$

initial volume, $V_1=0.1\ m^3$

work done in the process, $W=50000\ J$

For heat interaction during the polytropic process we have:

$Q=W[(\gamma -n)/(\gamma-1) ]$

$Q=50*[(1.4-1.2)/(1.4-1) ]$

$Q=25\ kJ$

For ideal gas we have the Gas Law:

$P_1.V_1=m.R.T_1$

$6* 10^5* 0.1=m.R* 520$

$m.R=115.385\ J.K^(-1)$

For work we have the relation:

$W=m.R.((T_1-T_2))/((n-1))$

putting respective values

$50000=115.385* ((520-T_2))/((1.2-1))$

$T_2=433.33\ K$

We know entropy change:

$\Delta s=(dQ)/(dT)$

$\Delta s=(25)/(520-433.33)$

$\Delta s= 0.2885 J.K^(-1)$

Final answer:

The given question involves the thermodynamic process of an internally reversible process of nitrogen gas (N2) at specific pressure and temperature with a constant value of pν1.2. However, the question does not provide enough information to calculate the heat transfer and entropy change accurately.

Explanation:

The given question involves the thermodynamic process of an internally reversible process of nitrogen gas (N2) at specific pressure and temperature with a constant value of pν1.2. In order to determine the heat transfer and entropy change, we need to use the first and second laws of thermodynamics. However, the question does not provide enough information to calculate the heat transfer and entropy change accurately.

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Chapter 38, Problem 001 Monochromatic light (that is, light of a single wavelength) is to be absorbed by a sheet of a certain material. Photon absorption will occur if the photon energy equals or exceeds 0.42 eV, the smallest amount of energy needed to dissociate a molecule of the material.

(a) What is the greatest wavelength of light that can be absorbed by the material?
(b) In what region of the electromagnetic spectrum is this wavelength located?

Answers

Answer:

a) $\lambda=2.95x10^(-6)m$

b) infrared region

Explanation:

Photon energy is the "energy carried by a single photon. This amount of energy is directly proportional to the photon's electromagnetic frequency and is inversely proportional to the wavelength. If we have higher the photon's frequency then we have higher its energy. Equivalently, with longer the photon's wavelength, we have lower energy".

Part a

Is provide that the smallest amount of energy that is needed to dissociate a molecule of a material on this case 0.42eV. We know that the energy of the photon is equal to:

$E=hf$

Where h is the Planck's Constant. By the other hand the know that $c=f\lambda$ and if we solve for f we have:

$f=(c)/(\lambda)$

If we replace the last equation into the E formula we got:

$E=h(c)/(\lambda)$

And if we solve for $\lambda$ we got:

$\lambda =(hc)/(E)$

Using the value of the constant $h=4.136x10^(-15) eVs$ we have this:

$\lambda=(4.136x10^(15)eVs (3x10^8 (m)/(s)))/(0.42eV)=2.95x10^(-6)m$

$\lambda=2.95x10^(-6)m$

Part b

If we see the figure attached, with the red arrow, the value for the wavelenght obtained from part a) is on the infrared region, since is in the order of $10^(-6)m$

So to deal with the irrational belief in REBT, we must Group of answer choices

A. Consult with a friend and get their feeback

B. Dispute the beliefs by asking if these are true and examining the evidence

C. Seek mental health counseling

D. It is just too hard so let's just forget it.

Answers

Answer:

i believe the answer is B

Explanation:

Seeking the right answer is the best thing to do

The 9-inch-long elephant nose fish in the Congo River generates a weak electric field around its body using an organ in its tail. When small prey, or even potential mates, swim within a few feet of the fish, they perturb the electric field. The change in the field is picked up by electric sensor cells in the skin of the elephant nose. These remarkable fish can detect changes in the electric field as small as 3.00 μN/C. How much charge, modeled as a point charge, in the fish would be needed to produce such a change in the electric field at a distance of 63.5 cm ?