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3 years ago

Global winds move warm air toward the _____.

Physics

2 answers:

Bezzdna [24]3 years ago

5 0

global winds move warm air toward the poles

Taya2010 [7]3 years ago

3 0

Explanation:

It is known that equator lies near the Sun as the globe or Earth revolves around the Sun. Therefore, air near the region of equator is warm due to the heat coming from the Sun.

As a result, this warm air rises towards the poles as warmth from the Sun helps the air molecules to gain kinetic energy and therefore, they move towards the pole which is a colder region as compared to the equator.

Therefore, we can conclude that global winds move warm air toward the pole.

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How much time does it take light from a flash camera to reach a subject 6.0 meters across a room in scientific notation?

wariber [46]

The time it takes light from a flash camera to reach a subject 6.0 meters across a room in scientific notation is 2.0 *10^-8 s.

Explanation:

Given

t=?

d=6m

v=3*10^8 m/s

we have, v=d/t

here t=d/v

t=6m/3*10^8 m/s

v=2*10^-8 m/s

The time it takes light from a flash camera to reach a subject 6.0 meters across a room in scientific notation is 2.0 *10^-8 s.

6 0

3 years ago

Two rocks are at the top of a building. Rock 1 is dropped from rest while Rock 2 is thrown horizontaly at a velocity of 5 ms.

LuckyWell [14K]

Answer:

I'm pretty sure the answer is 0 m/s²

Explanation:

The horizontal velocity of the second rock is 5 m/s, so if we pretend air resistance doesn't exist, it will maintain that horizontal velocity, meaning that there is no horizontal acceleration.

7 0

3 years ago

an object weighing 15 newtons is lifted from the ground to a height of 0.22 meter what is the increase in the object's gravitati

kicyunya [14]

GPE= weight•height= 15 N• 0.22meter= 3.3 Joules
I hope this helps ~~Charlotte~~

5 0

3 years ago

Read 2 more answers

A coyote can locate a sound source with good accuracy by comparing the arrival times of a sound wave at its two ears. Suppose a

il63 [147K]

Answer:

0.0000109261200583 s

0.0109261200583

Explanation:

$d_2$ = Distance from right ear = 3 m

s = Distance between ears = 15 cm

v = Speed of sound in air = 343 m/s

Distance between the left ear and the bird

$d_1=\sqrt{s^2+d_2^2}\\\Rightarrow d_1=\sqrt{0.15^2+3^2}\\\Rightarrow d_1=3.00374765918\ m=3.004\ m$

Time

$t=\dfrac{Distance}{Speed}$

Time difference would be

$\Delta T=\dfrac{d_1}{v}-\dfrac{d_2}{v}\\\Rightarrow \Delta T=\dfrac{3.00374765918}{343}-\dfrac{3}{343}\\\Rightarrow \Delta T=0.0000109261200583\ s$

The time difference is 0.0000109261200583 s

Time period is given by

$T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{1000}\\\Rightarrow T=10^{-3}\ s$

The ratio is

$\dfrac{\Delta T}{T}=\dfrac{0.0000109261200583}{10^{-3}}\\\Rightarrow \dfrac{\Delta T}{T}=0.0109261200583$

The ratio is 0.0109261200583

7 0

3 years ago

You and a friend each carry a 15 kg suitcase up two flights of stairs, walking at a constant speed. Take each suitcase to be the

AlekseyPX

Answer:

Both of you did the same work but you expended more power.

Explanation:

Work done by an object is calculated by force applied multiplied by the distance.

W=F*d

From the figure given below let us calculate force applied bith you and yopur friend.

Let us take the stairs in positive x direction,

Work done by you W₁ ,

The force applied Fₓ = F - mgsinθ =maₓ

here aₓ = 0, because both of you move with constant speed

F - mgsinθ = 0

F= mgsinθ

The work done by you on the suitcase is

W = F L cos0° , where L is he length of the staircase.

W = FL = mgsinθL , by substituting value of F

Work done by you is W₁ = mgLsinθ

Similarly work done by your friend is W₂ = mgLsinθ.

Because both of you carry suitcase of same weight and in staircase is in same angle the force applied is same .

Therefore work done by both of you is same . Both of you did equal work.

The power , is defined as amount of energy converted or transfered per second or rate at which work is done .

P = $\frac{W}{t}$ = $\frac{FL}{t}$

Power spend by you P₁ = mgLsinθ/t

P₁ = 15*9.8*Lsinθ/30

P₁ = 4.9L sinθ eqn 1

Power spend by your friend is P₂ = mgLsinθ/t

P₂ =15*9.8*Lsinθ/60

P₂ = 2.45Lsinθ eqn 2

Dividing eqn 1 and eqn 2

P₁ = 2P₂

You have spend more power than your friend .

Hence Both of you did equal work but you spend more power.

7 0

3 years ago