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3 years ago

Defention of reversibily

1 answer:

5 0

Capable of being reversed or of reversing: as a : capable of going through a series of actions (as changes) either backward or forward
Example : water ----> ice
melts into water again

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Explain your problem of Federalism

telo118 [61]

That is more of a History of English question.

6 0

3 years ago

Two forces,

serg [7]

First compute the resultant force F:

$\mathbf F_1=(5.90\,\mathbf i-5.60\,\mathbf j)\,\mathrm N$

$\mathbf F_2=(4.65\,\mathbf i-5.55\,\mathbf j)\,\mathrm N$

$\implies\mathbf F=\mathbf F_1+\mathbf F_2=(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N$

Then use Newton's second law to determine the acceleration vector $\mathbf a$ for the particle:

$\mathbf F=m\mathbf a$

$(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N=(2.10\,\mathrm{kg})\mathbf a$

$\mathbf a\approx(5.02\,\mathbf i-5.31\,\mathbf j)\dfrac{\rm m}{\mathrm s^2}$

Let $\mathbf x(t)$ and $\mathbf v(t)$ denote the particle's position and velocity vectors, respectively.

(a) Use the fundamental theorem of calculus. The particle starts at rest, so $\mathbf v(0)=0$ . Then the particle's velocity vector at t = 10.4 s is

$\mathbf v(10.4\,\mathrm s)=\mathbf v(0)+\displaystyle\int_0^{10}\mathbf a(u)\,\mathrm du$

$\mathbf v(10.4\,\mathrm s)=\left((5.02\,\mathbf i-5.31\,\mathbf j)u\,\dfrac{\rm m}{\mathrm s^2}\right)\bigg|_{u=0}^{u=10.4}$

$\mathbf v(10.4\,\mathrm s)\approx(52.2\,\mathbf i-55.2\,\mathbf j)\dfrac{\rm m}{\rm s}$

If you don't know calculus, then just use the formula,

$v_f=v_i+at$

So, for instance, the velocity vector at t = 10.4 s has x-component

$v_{f,x}=0+\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)(10.4\,\mathrm s)=52.2\dfrac{\rm m}{\mathrm s^2}$

(b) Compute the angle $\theta$ for $\mathbf v(10.4\,\mathrm s)$ :

$\tan\theta=\dfrac{-55.2}{52.2}\implies\theta\approx-46.6^\circ$

so that the particle is moving at an angle of about 313º counterclockwise from the positive x axis.

(c) We can find the velocity at any time t by generalizing the integral in part (a):

$\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du$

$\implies\mathbf v(t)=\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j$

Then using the fundamental theorem of calculus again, we have

$\mathbf x(10.4\,\mathrm s)=\mathbf x(0)+\displaystyle\int_0^{10.4}\mathbf v(u)\,\mathrm du$

where $\mathbf x(0)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m$ is the particle's initial position. So we get

$\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\displaystyle\int_0^{10.4}\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\right)\,\mathrm du$

$\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\dfrac12\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=10.4}$

$\mathbf x(10.4\,\mathrm s)\approx(542\,\mathbf i-570\,\mathbf j)\,\mathrm m$

So over the first 10.4 s, the particle is displaced by the vector

$\mathbf x(10.4\,\mathrm s)-\mathbf x(0)\approx(270\,\mathbf i-283\,\mathbf j)\,\mathrm m-(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m\approx(272\,\mathbf i-287\,\mathbf j)\,\mathrm m$

or a net distance of about 395 m away from its starting position, in the same direction as found in part (b).

(d) See part (c).

3 0

3 years ago

2. El sonido de una ballena es en especial de frecuencia baja, pero existe una especie de ballena la Whalien cuya frecuencia es

ikadub [295]

Answer:

The sound of a whale is especially low frequency, but there is a species of whale the Whalien whose frequency is 52 Hz, if the propagation speed of the wave is 1400m / s What will be its period in the water and the air? And what will be the wavelength in each medium? Remember that the propagation speed in air is 340m / s

Explanation:

From wave equation, the speed, wavelength and frequency is related using

V = fλ

Where

V is the speed

f is the frequency

And λ is the wavelength

So,

The frequency of the whale is

f = 52Hz

The speed in water is V_w = 1400m/s

The speed in air is V_a = 340m/s

We want to find the period in each medium, the period is related to the frequency and since the frequency is constant.

Then, period in equal in both medium

T = 1 / f

T_w = T_a = 1 / f

T = 1 / 52

T = 0.0192 seconds

We want to find the wavelength in each medium

For water,

V = fλ

V_w = f × λ_w

Then,

λ_w = V_w / f.

λ_w = 1400 / 52 = 26.92 m

The wavelength in water is 26.92m

Now, in air

V = fλ

V_a = f × λ_a

Then,

λ_a = V_a / f.

λ_a = 340 / 52 = 6.54 m

The wavelength in air is 6.54 m

In Spanish

De la ecuación de onda, la velocidad, la longitud de onda y la frecuencia se relacionan usando

V = fλ

Dónde

V es la velocidad

f es la frecuencia

Y λ es la longitud de onda

Entonces,

La frecuencia de la ballena es

f = 52Hz

La velocidad en el agua es V_w = 1400m / s

La velocidad en el aire es V_a = 340m / s

Queremos encontrar el período en cada medio, el período está relacionado con la frecuencia y dado que la frecuencia es constante.

Luego, período igual en ambos medios

T = 1 / f

T_w = T_a = 1 / f

T = 1/52

T = 0.0192 segundos

Queremos encontrar la longitud de onda en cada medio

Para agua,

V = fλ

V_w = f × λ_w

Entonces,

λ_w = V_w / f.

λ_w = 1400/52 = 26,92 m

La longitud de onda en el agua es de 26,92 m.

Ahora en el aire

V = fλ

V_a = f × λ_a

Entonces,

λ_a = V_a / f.

λ_a = 340/52 = 6,54 m

La longitud de onda en el aire es de 6.54 m.

8 0

4 years ago

Which statement is true about the type of light used by plants to make sugar during photosynthesis? Group of answer choices

denis-greek [22]

Answer:

4. It is infrared.

Explanation:

the photosystems on the chlorophyll absorb light at 600-700nm

3 0

3 years ago

The gravitational acceleration on the moon is about one sixth the size of the gravitational on earth. According to newton's seco

alexandr402 [8]

Weight of a person on the earth is given by,

W= mg,

Where, g is the gravitational acceleration of the earth and m is the mass of the person.

Gravitational acceleration of the moon is one-sixth of the earth.

Therefore, the weight of a astronaut who goes to the moon will be one sixth of that on earth. Hence, the astronaut will weight less.

7 0

3 years ago