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3 years ago

8

A dust particle floats in front of a silent loudspeaker as shown in the figure. The loudspeaker is turned on and plays a constan

t low frequency note.

2 answers:

dybincka [34]3 years ago

8 0

Answer:tbh idk

Explanation:

tbh idk

notka56 [123]3 years ago

4 0

Explanation:

question not clear. please upload full equation

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During an isothermal process, 10 j of heat is removed from an ideal gas. What is the work done by the gas in the process?

Black_prince [1.1K]

The work done in the isothermal process is 10 joule.

We need to know about the isotherm process to solve this problem. The isotherm process can be described as a process where the initial temperature system will be the same as the final temperature. Hence, the internal energy change will be zero.

ΔU = 0

Hence,

ΔU = Q - W

0 = Q - W

Q = W

It means that the heat transferred is the same as the work done.

From the question above, we know that the heat transferred is 10 joule. Thus, the work done in the isothermal process is 10 joule.

Find more on isothermal at: brainly.com/question/17097259

#SPJ4

8 0

1 year ago

Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 1.38 m and finds that it makes 441

Tasya [4]

The period of a simple pendulum is given by:
$T=2 \pi \sqrt{ \frac{L}{g} }$
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
$g= \frac{4 \pi^2}{T^2}L$ (1)

We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.

We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
$f= \frac{N}{t}= \frac{441}{1090 s}=0.40 Hz$
And its period is the reciprocal of its frequency:
$T= \frac{1}{f}= \frac{1}{0.40 Hz}=2.47 s$

So now we can use eq.(1) to find the gravitational acceleration of the planet:
$g= \frac{4 \pi^2}{T^2}L = \frac{4 \pi^2}{(2.47 s)^2} (1.38 m) =8.92 m/s^2$

3 0

3 years ago

State the law of conservation of energy

storchak [24]

Answer:

In physics and chemistry, the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time. ... For instance, chemical energy is converted to kinetic energy when a stick of dynamite explodes.

7 0

2 years ago

An automobile with a standard differential turns sharply to the left. The left driving wheel turns on a 20-m radius. Distance be

Inessa05 [86]

Explanation:

The given data is as follows.

Inner wheel Radius = 20 m,

Distance between left and right wheel = 1.5m,

Let us assume speed of drive shaft is N rpm.

Formula to calculate angular velocity is as follows.

Angular velocity of automobile = w = $\frac{V}{R}$

where, V = linear velocity of automobile m/min,

R = turning radius from automobile center in meter

In the given case, angular velocity remains same for inner and outer wheel but there is change in linear velocity of inner wheel and outer wheel.

Now, we assume that

u = linear velocity of inner wheel

and, u' = linear velocity of outer wheel.

Formula for angular velocity of inner wheel w = ,

Formula for angular velocity of outer wheel w =

Now, for inner wheels

w =

= $\frac{u}{(R - d)}$

u = $V \times \frac{(R - d)}{R}$

= $V \times (1 - \frac{d}{R})$

If radius of wheel is r it will cover distance in one min.

Since, velocity of wheel is u it will cover distance u in unit time(min)

Thus, u = $2\pi rn$ = $V \times (1 - \frac{d}{R})$

Now, rotation per minute of inner wheel is calculated as follows.

n = $\frac{V}{2 \pi r \times (1 - \frac{d}{R})}$

= $\frac{V}{2 \pi r \times (1 - \frac{0.75}{20})}$ (since 2d = 1.5m given, d = 0.75m),

= $\frac{V}{r} \times 0.1532$

So, rotation per minute of outer wheel; n' =

= $\frac{V}{2 \pi r \times (1 + \frac{0.75}{20})}$

= $\frac{V}{r} \times 0.1651$

5 0

3 years ago

12. A frain moves from rest to a speed of 25 m/s in 30.0 seconds. What is its acceleration?

ziro4ka [17]

initial velocity=u=0m/s
Final velocity=v=25m/s
Time=t=30s

$\\ \tt\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \tt\longmapsto Acceleration=\dfrac{25-0}{30}$

$\\ \tt\longmapsto Acceleration=\dfrac{25}{30}$

$\\ \tt\longmapsto Acceleration=0.8m/s^2$

6 0

3 years ago