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3 years ago

Air is warmer and less dense than surrounding air at the equator because the equator receives more?

Physics

2 answers:

Hatshy [7]3 years ago

4 0

The answer is C: Solar energy

gladu [14]3 years ago

3 0

Solar energy

Given the fact that near the equator you have the sun at roughly the same level all year, you deal with very little to no temperature change, whereas North or South of the equator experiences temperature change due to increasing or decreasing distance from the sun when the earth tilts on its axis.

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What makes music different from noise?

Zepler [3.9K]

the answers going to be A

5 0

3 years ago

Read 2 more answers

Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

3 0

3 years ago

A ball is droped from a height of 16m how much time will pass before the ball hits the ground

sergey [27]

Answer:

The time is 1.8s

Explanation:

The ball droped, will freely fall under gravity.

Hence we use free fall formula to calculate the time by the ball to hit the ground

$h= \frac{1}{2}g{t}^{2}$

Where h is the height from which the ball is droped, g is the acceleration due to gravity that acted on the ball, and t is the time taken by the ball to hit the ground.

From the question,

h=16m

Also, let take

$g = 9.8m{s}^{-2}$

By substitution we obtain,

$16= \frac{1}{2}\times 9.8{t}^{2}$

$\implies32=9.8{t}^{2}$

Diving through by 9.8

$\frac{32}{9.8}= \frac{ 9.8{t}^{2} }{9.8}$

$\implies{t}^{2} =3.265$

square root both sides, we obtain

$\implies t= \sqrt{3.265}$

$t=1.8s$

4 0

3 years ago

What is the public storm warning signal that carries winds of 30 to 60 km/hr and can cause no damage to very light damage

Shkiper50 [21]

Answer:

I know u need this

Explanation:

You gave me the runaround

I really hate the runaround

You really got me paranoid

I always keep a gun around

You always give me butterflies

When you come around

7 0

2 years ago

If a ball is thrown vertically upward with a velocity of 80 ft/s, then its height after t seconds is s = 80t − 16t 2 . (a) what

MAXImum [283]

Max height occurs when v = 0.
v(t) = ds(t)/dt
v(t) = 80 - 32t
0 = 80 - 32t
t = 5/2

s(5/2) = 80(5/2) - 16(5/2)^2
s(5/2) = 100
Answer: 100 ft

96 = 80t - 16t²
t = 3, 2
(80 ± √256) / 32 using the quadratic equation.

v(2) = 16
v(3) = -16

4 0

2 years ago