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3 years ago

What would happen if earth was further from the sun

Physics

2 answers:

Colt1911 [192]3 years ago

3 0

Earth would have different climates and would be more colder.

AURORKA [14]3 years ago

3 0

Our Earth would be colder. Just like when moving a heating lamp farther away from you, you get a tad colder. Hope this helps!

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Which statement describes the relationship between bond strength and the melting and boiling points of a substance? A. As the fo

icang [17]

Answer:

Explanation:

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3 years ago

A coin is dropped off of a building landing on its side. It hits with a pressure of 400/2 It hits with a force of 0.1 Calculate

Arte-miy333 [17]

Complete Question:

A coin is dropped off of a building landing on its side. It hits with a pressure of 400 N/m². It hits with a force of 0.1N. Calculate the area of the coin?

Answer:

Area = 0.00025 m²

Explanation:

Given the following data;

Pressure = 400N/m²

Force = 0.1N

To find the area of the coin;

Pressure = Force/area

Area = Force/pressure

Substituting into the equation, we have;

Area = 0.1/400

Area = 0.00025 m²

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3 years ago

What 1columb=?<br> please help

Juli2301 [7.4K]

Explanation:

it's a unit used to measure charge (C)

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2 years ago

Read 2 more answers

a man drags a 8.10 kg bag of mulch at a constant speed, applying a 29.5 N at 38°. what is the coefficient of friction?

lesantik [10]

Answer:

The coefficient of friction is 0.38.

Explanation:

The free body diagram is drawn below.

Let $f$ be frictional force acting in the backward direction as shown. Let the coefficient of friction be $\mu$ . Let $N$ be the normal reaction force acting on the bag.

Given:

Mass of the bag is, $m=8.10\textrm{ kg}$

Force acting at $\theta = 38$ ° is $F= 29.5\textrm{ N}$

Acceleration due to gravity is, $g=9.8\textrm{ }m/s^{2}$

The force F can be resolved into its components as $F_{x}=F \cos \theta$ and $F_{y}=F \sin \theta$

Therefore,

$F_{x}=29.5\cos(38)=23.25\textrm{ N}\\F_{y}=29.5\sin(38)=18.16\textrm{ N}$

Now, as there is no acceleration in vertical direction, therefore,

Sum of upward forces = Sum of downward forces

$N+F_{y}=mg\\N=mg-F_{y}=8.10\times 9.8-18.16\\N=79.38-18.16=61.22\textrm{ N}$

Now, as the bag is moving at a constant speed, so acceleration in the horizontal direction is also zero as acceleration is the rate of change of velocity.

Therefore, backward force = forward force.

$f=F_{x}\\f=23.25\textrm{ N}$