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4 years ago

Why are average temperatures higher at the equator than near the poles?

2 answers:

8 0

The poles are on either side of the sun so therefore the equator is constantly the closest to the sun and gets the most heat

elena55 [62]4 years ago

3 0

Bc the equator is constantly facing the sun while the earth rotates

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Define force and provide an example

ollegr [7]

Answer:

force-strength,power or energy as an attribute of motion, movement or action. Example: Frictional force.

4 0

3 years ago

The dimensions of the disk of our milky way galaxy are:.

Katen [24]

Answer:

According to studies, the milky way is approximately, "170,000–200,000 light-years (52–61 kpc) in diameter and, on average, approximately 1,000 ly (0.3 kpc) thick."

With that being said, it is safe to say that the dimensions are somewhere around 100,000 by 1,000

6 0

2 years ago

A city planner is working on the redesign of a hilly portion of a city. An important consideration is how steep the roads can be

Artyom0805 [142]

Explanation:

It is known that relation between force and acceleration is as follows.

F = $m \times a$

I is given that, mass is 1090 kg and acceleration is 21 m/s. Therefore, we will calculate force as follows.

F = $m \times a$

= $1090 \times (\frac{21}{16})$

= 1430.625 N

Also, it is known that

$sin(\theta) = \frac{\text{Force car can exert}}{\text{Force gravity pulls car}}$

$sin(\theta) = \frac{1430.625 N}{(1090 \times 9.8) N}$

$\theta$ = 7.70 degrees

Thus, we can conclude that the maximum steepness for the car to still be able to accelerate is 7.70 degrees.

8 0

3 years ago

Calcular la aceleración que produce una fuerza de 40 N sobre un cuerpo con 88 Kg de masa. Expresar el resultado en m⁄s^2 *

tigry1 [53]

Answer:

a = 0.45 m/s²

Explanation:

The given question is ''Calculate the acceleration that produces a force of 40 N on a body with 88 kg of mass".

Given that,

Force, F = 40 N

Mass of the body, m = 88 kg

The net force acting on the body is given by :

F = ma

Where

a is the acceleration of the body

$a=\dfrac{F}{m}\\\\a=\dfrac{40\ N}{88\ kg}\\\\a=0.45\ m/s^2$

So, the required acceleration is 0.45 m/s².

4 0

3 years ago

How do motion and Newton's laws apply to your everyday life?

Marina86 [1]

When you jumó ur legs put force on ground

6 0

3 years ago