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3 years ago

Why is the land significantly cooler at night when compared to the ocean?

1 answer:

5 0

The reason that land is significantly cooler than the ocean is because of land breeze. Sea breeze is what makes oceans colder than land during the day, but when night happens, the heat from the land goes into the ocean, changing temperatures of both sides.

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"Pluto Is Changing" talks about sheets of frozen nitrogen on Pluto. Tell what is happening to these sheets of ice

drek231 [11]

What is Pluto is changing exactly <span />

7 0

4 years ago

Identify the labeled parts in the figure.

ad-work [718]

Answer:

Chemical symbols Al, O, H, O or a,c,f,h.

Coefficient 2 or e.

Number of atoms of the element - b,d ,g.

One atom of the element - i.

5 0

4 years ago

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the

Margaret [11]

Hi there!

$\boxed{\omega = 0.38 rad/sec}$

We can use the conservation of angular momentum to solve.

$\large\boxed{L_i = L_f}$

Recall the equation for angular momentum:

$L = I\omega$

We can begin by writing out the scenario as a conservation of angular momentum:

$I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)$

$I_m$ = moment of inertia of the merry-go-round (kgm²)

$\omega_m$ = angular velocity of merry go round (rad/sec)

$\omega_f$ = final angular velocity of COMBINED objects (rad/sec)

$I_b$ = moment of inertia of boy (kgm²)

$\omega_b$ = angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

$I = MR^2$

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

$\omega = \frac{v}{r}$

$L_b = MR^2(\frac{v}{R}) = MRv$

Plug in the given values:

$L_b = (20)(3)(5) = 300 kgm^2/s$

Now, we must solve for the boy's moment of inertia:

$I = MR^2\\I = 20(3^2) = 180 kgm^2$

Use the above equation for conservation of momentum:

$600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}$

8 0

3 years ago

Standing waves can be established within a cylindrical volume. However, it is not known if the volume is closed at both ends, op

Evgen [1.6K]

The frequencies are missing in the question. The three successive resonance frequencies within the volume are $f_1, f_2 \text{ and}\ f_3$ .

Solution :

Let the volume be : v

The frequency for one end open and one end closed is given by :

So, $f_n = \frac{(2n-1)v}{4L}$

Therefore,

$f_1 = \frac{v}{4L}$ , $f_2 = \frac{3v}{4L}$ , $f_3 = \frac{5v}{4L}$

So, $f_2 - 3f_1$ and $f_3=\frac{5}{3}f_2$

Therefore, the ratio of $\frac{f_3}{f_2}=\frac{5}{3}$ which is not a whole number.

Now the frequency for open volume and closed at both end

$f_n=\frac{nv}{4L}$

So, $f_1=\frac{v}{4L}$ , $f_2 = 2f_1$ , $f_2=3f_1$

From above formulae we can see that ratio of is not a whole number that is a identification for the frequency of the volume at one end open and one end closed.

Also ratio of the consecutive frequency of the volume at open from both side and closed from both side is always a whole number.

4 0

3 years ago

This is for 6.02 in comprehensive science for flvs

Katen [24]

Answer:

Please provide an image to help clarify

Explanation:

thanks :)

4 0

3 years ago