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3 years ago

Why does it take several minutes for the smell of a freshly cut lemon to be detectable on the opposite side of the kitchen?

1 answer:

riadik2000 [5.3K]3 years ago

5 0

When the lemon is cut, it releases gaseous molecules into the surrounding air which are responsible for the lemon's smell. These molecules move slowly at room temperature and the distance that they must cover, in relevance to their size, is immense. These reasons contribute to the fact that the smell takes a while to reach the other side of the room.

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For no apparent reason, a poodle is running at a constant speed of 5.00 m/s in a circle with radius 2.9 m . Let v⃗ 1 be the velo

Nostrana [21]

At a constant speed of 5.00 m/s, the speed at which the poodle completes a full revolution is

$\left(5.00\,\dfrac{\mathrm m}{\mathrm s}\right)\left(\dfrac{1\,\mathrm{rev}}{2\pi(2.9\,\mathrm m)}\right)\approx0.2744\,\dfrac{\mathrm{rev}}{\mathrm s}$

so that its period is $T=3.644\,\frac{\mathrm s}{\mathrm{rev}}$ (where 1 revolution corresponds exactly to 360 degrees). We use this to determine how much of the circular path the poodle traverses in each given time interval with duration $\Delta t$ . Denote by $\theta$ the angle between the velocity vectors (same as the angle subtended by the arc the poodle traverses), then

$\Delta t=0.4\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.4\,\mathrm s}\theta\implies\theta\approx39.56^\circ$

$\Delta t=0.2\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.2\,\mathrm s}\theta\implies\theta\approx19.78^\circ$

$\Delta t=7\times10^{-2}\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{7\times10^{-2}\,\mathrm s}\theta\implies\theta\approx6.923^\circ$

We can then compute the magnitude of the velocity vector differences $\Delta\vec v$ for each time interval by using the law of cosines:

$|\Delta\vec v|^2=|\vec v_1|^2+|\vec v_2|^2-2|\vec v_1||\vec v_2|\cos\theta$

$\implies|\Delta\vec v|=\begin{cases}3.384\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.4\,\mathrm s\\1.718\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.2\,\mathrm s\\0.6038\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}$

and in turn we find the magnitude of the average acceleration vectors to be

$\implies|\vec a|=\begin{cases}8.460\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.4\,\mathrm s\\8.588\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.2\,\mathrm s\\8.625\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}$

So that takes care of parts A, C, and E. Unfortunately, without knowing the poodle's starting position, it's impossible to tell precisely in what directions each average acceleration vector points.

5 0

3 years ago

How resistance, current, and voltage behave in a series circuit.?

mylen [45]

b.the covalent circuit is held throw magnetic waves

4 0

4 years ago

A cart moves along a track at a velocity of 3.5 cm/s. When a force is applied to the cart, its velocity increases to 8.2 cm/s. I

Lorico [155]

Answer:

3.13cm/s²

Explanation:

Given

Initial velocity u = 3.5cm/s

Final velocity v = 8.2cm/s

Time t = 1.5secs

Required

Acceleration of the cart a

To get that, we will use the equation of motion

v = u+at

Substitute the given parameters

8.2 = 3.5+1.5a

1.5a = 8.2-3.5

1.5a = 4.7

a = 4.7/1.5

a = 3.13cm/s²

Hence the acceleration to the cart is 3.13cm/s²

3 0

3 years ago

How does a pith ball electroscope work

Mice21 [21]

Answer:

look at the explanation

Explanation:

This pith-ball electroscope is used to detect the presence of a static electricity charge. The two lightweight “pith” balls suspended from the strings are attracted to objects with a static electric charge. The pith balls can also be charged by touching them to an object with a static electric charge.

3 0

3 years ago

The following represents a mass attached to a spring oscillating in simple harmonic motion. X(t) = 4.0 cos(3.0t +0.10) units of

kolbaska11 [484]

Answer:

a) A = 4.0 m , b) w = 3.0 rad / s , c) f = 0.477 Hz , d) T = 20.94 s

Explanation:

The equation that describes the oscillatory motion is

x = A cos (wt + fi)

In the exercise we are told that the expression is

x = 4.0 cos (3.0 t + 0.10)

let's answer the different questions

a) the amplitude is

A = 4.0 m

b) the frequency or angular velocity

w = 3.0 rad / s

c) angular velocity and frequency are related

w = 2π f

f = w / 2π

f = 3 / 2π

f = 0.477 Hz

d) the period

frequency and period are related

T = 1 / f

T = 1 / 0.477

T = 20.94 s

e) the phase constant

Ф = 0.10 rad

f) velocity is defined by

v = dx / dt

v = - A w sin (wt + Ф)

speed is maximum when sine is + -1

v = A w

v = 4 3

v = 12 m / s

g) the angular velocity is

w² = k / m

k = m w²

k = 1.2 3²

k = 10.8 N / m

h) the total energy of the oscillator is

Em = ½ k A²

Em = ½ 10.8 4²

Em = 43.2 J

i) the potential energy is

Ke = ½ k x²

for t = 0 x = 4 cos (0 + 0.1)

x = 3.98 m

j) kinetic energy

K = ½ m v²

for t = 00.1 ²

v = A w sin 0.10

v = 4 3 sin 0.10

v = 1.98 m / s

3 0

3 years ago