Ask question

Ask question

All categories

4 years ago

8

Why does the area around the equator stay the same temperature year-round

1 answer:

Morgarella [4.7K]4 years ago

5 0

I would say because it is at the same point from the sun all year. It doesn’t turn away from the sun or towards it.

You might be interested in

In a series circuit, when batteries are added to the circuit the voltage will _____ and the current will _____.

Eva8 [605]

Answer: voltage, explode is the answers

Explanation:

Bc of the device for each server

5 0

3 years ago

HELP ME I WILL GIVE BRAINLYEST!!

DaniilM [7]

Answer:

Planting of trees on sloped surfaces is a method of conservation that utilizes the roots of plants. Planting trees on such areas would prevent hazards and maintain the soil formation future since the roots would hold the soil together preventing or minimizing any soil erosion.

6 0

3 years ago

Read 2 more answers

An open organ pipe of length 0.47328 m and another pipe closed at one end of length 0.702821 m are sounded together. What beat f

sineoko [7]

Answer:

fb = 240.35 Hz

Explanation:

In order to calculate the beat frequency generated by the first modes of each, organ and tube, you use the following formulas for the fundamental frequencies.

Open tube:

$f=\frac{v_s}{2L}$ (1)

vs: speed of sound = 343m/s

L: length of the open tube = 0.47328m

You replace in the equation (1):

$f=\frac{343m/s}{2(0.47228m)}=362.36Hz$

Closed tube:

$f'=\frac{v_s}{4L'}$

L': length of the closed tube = 0.702821m

$f'=\frac{343m/s}{4(0.702821m)}=122.00Hz$

Next, you use the following formula for the beat frequency:

$f_b=|f-f'|=|362.36Hz-122.00Hz|=240.35Hz$

The beat frequency generated by the first overtone pf the closed pipe and the fundamental of the open pipe is 240.35Hz

7 0

3 years ago

How much work would be needed to raise the payload from the surface of the moon (i.e., x = r) to an altitude of 5r miles above t

Cloud [144]

Let the data is as following

mass of payload = "m"

mass of Moon = "M"

now we know that we place the payload from the position on the surface of moon to the position of 5r from the surface

So in this case we can say that change in the gravitational potential energy is equal to the work done to move the mass from one position to other

so it is given by

$W = U_f - U_i$

we know that

$U_f = -\frac{GMm}{6r}$

$U_i = -\frac{GMm}{r}$

now from above formula

$W = -\frac{GMm}{6r} + \frac{GMm}{r}$

$W = \frac{5GMm}{6r}$

so above is the work done to move the mass from surface to given altitude

7 0

3 years ago

The moon orbits the earth at a distance of 3.85 x 10^8 m. Assume that this distance is between the centers of the earth and the

Inga [223]

Answer:

27.5 days

0.92 month

Explanation:

$r$ = radius of the orbit of moon around the earth = $3.85\times10^{8} m$

$M$ = Mass of earth = $5.98\times10^{24} m$

$T$ = Time period of moon's motion

According to Kepler's third law, Time period is related to radius of orbit as

$T^{2} = \frac{4\pi ^{2} r^{3} }{GM}$

inserting the values, we get

$T^{2} = \frac{4(3.14)^{2} (3.85\times10^{8})^{3} }{(6.67\times10^{-11})(5.98\times10^{24})}\\T = 2.3754\times10^{6} sec$

we know that

1 day = 24 hours = 24 x 3600 sec = 86400 s

$T = 2.3754\times10^{6} sec \frac{1 day}{86400 sec} \\T = 27.5 days$

1 month = 30 days

$T = 27.5 days \frac{1 month}{30 days} \\T = 0.92 month$

6 0

3 years ago