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3 years ago

How does distance from the equator affect the climate of an area?

Physics

1 answer:

AVprozaik [17]3 years ago

6 0

Explanation :

There are different factors that affect the climate of an area.The distance from the equator is one of them.

Equator is closest to the sun. The rays from the sun strikes mostly near equator and gets spread over the larger area. Hence, equators are much hot as compared to that of poles.

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Convert 15 litre into cubic metre

atroni [7]

Answer:

0.015m^3

Explanation:

1 m^3 = 1000 liters

x m^3 = 15 liters

Cross multiply

xm^3 x 1000 l = 15 l

Divide both sides by 1000

xm^3 x1000/1000 = 15/1000

xm^3 = 0.015m^3

Therefore 15 liter = 0.015m^3

5 0

2 years ago

You push a table 8 meters for 16 minuutes and do 6720 j of work how much power do ypu use

Yuri [45]

P=W/t

P=Power
W=Work
t=Time

Convert 16 minutes in seconds:
16 mins = 960 secs

P=6720/960=7.23 W [Watt]

3 0

3 years ago

Squid use jet propulsion for rapid escapes. A squid pulls water into its body and then rapidly ejects the water backward to prop

Setler [38]

Answer:

a. FTh = 30 N

b. Fw = 30 N

c. a = 200 m/s2

Explanation:

See full explanation in the picture. Please rate as brainliest

5 0

2 years ago

assume the suns total energy output is 4.0 * 10^26 watts, and 1 watt is 1 joule/second. assume 4.3 * 10^-12 J is released from e

Dmitry [639]

Answer:

$9.3\cdot 10^{37}$

Explanation:

Power is defined as the energy produced (E) per unit of time (t):

$P= \frac{E}{t}$

This means that the energy produced in the Sun each second (1 s), given the power $P=4.0\cdot 10^{26}W$ , is

$E=Pt=(4.0\cdot 10^{26}W)(1s )=4.0\cdot 10^{26} J$

Each p-p chain reaction produces an amount of energy of

$E_1 = 4.3\cdot 10^{-12} J$

in order to get the total number of p-p chain reactions per second, we need to divide the total energy produced per second by the energy produced by each reaction:

$n=\frac{E}{E_1}=\frac{4.0\cdot 10^{26} J}{4.3\cdot 10^{-12} J}=9.3\cdot 10^{37}$

3 0

3 years ago

A swan on a lake gets airborne by flapping its wings and running on top of the water. If the swan must reach a velocity of 6.00

ira [324]

Answer:

A. 51.42 m.

B. 17.14 s.

Explanation:

Using equations of motion:

vf^2 = vi^2 + 2 * aS

Where,

vf = final velocity

a = acceleration

S = distance to which swan traveled

vi = 0 m/s

6.00^2 = 2 * 0.350S

S = 36/0.7

= 51.42 m.

vf = vi + at

6 = 0 + 0.35t

t = 6/0.35

= 17.14 s.

5 0

2 years ago