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Harlamova29_29 [7]
2 years ago
11

_________ changes from day to day, while ____________ changes from region to region. A) climate, weather B) weather, climate C)

wind, atmosphere D) atmosphere, wind
Physics
2 answers:
vladimir2022 [97]2 years ago
7 0

B) Weather changes day to day, while climate changes region to region.

Climate is the weather in a certain area. It's usually the average weather over a long period of time

Weather is in shorter terms then climate

Hope this helped!

~Just a girl in love with Shawn Mendes

PtichkaEL [24]2 years ago
3 0

Answer:

Choice b: Weather changes day to day, while climate changes region to region.

Explanation:

Weather only refers to the short-term conditions of the atmosphere while climate measures average daily weather for a long period of time at a certain location.

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A person pushes a refrigerator across a horizontal floor. The mass of the refrigerator is 110 kg, the coefficient of static fric
Law Incorporation [45]

Answer:

Explanation:

mass of refrigerator, m = 110 kg

coefficient of static friction, μs = 0.85

coefficient of kinetic friction, μk = 0.59

(a) the minimum force required to just start the motion in refrigerator

F = μs x mg

F = 0.85 x 110 x 9.8

F = 916.3 N

(b) The force required to move the refrigerator with constant speed

F' = μk x mg

F' = 0.59 x 110 x 9.8

F' = 636.02 N

(c) Let a be the acceleration.

Net force = Applied force - friction force

F net = 950 - 636.02

F net = 313.98 N

a = F net / mass

a = 313.98 / 110

a = 2.85 m/s²

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3. The pressure at the bottom of the ocean is great enough to crush submarines with steel walls that are 10 centimeters thick. S
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Read 2 more answers
A car slows down uniformly from a speed of 30.0 m/s to rest in 7.20 s
Ede4ka [16]

When acceleration is constant, the average velocity is given by

\bar v=\dfrac{v+v_0}2

where v and v_0 are the final and initial velocities, respectively. By definition, we also have that the average velocity is given by

\bar v=\dfrac{\Delta x}{\Delta t}=\dfrac{x-x_0}{t-t_0}

where x,x_0 are the final/initial displacements, and t,t_0 are the final/initial times, respectively.

Take the car's starting position to be at t_0=0\,\mathrm s. Then

\dfrac{v+v_0}2=\dfrac{x-x_0}t\implies x=x_0+\dfrac12(v+v_0)t

So we have

x=0\,\mathrm m+\dfrac12\left(0\,\dfrac{\mathrm m}{\mathrm s}+30.0\,\dfrac{\mathrm m}{\mathrm s}\right)(7.20\,\mathrm s)=108\,\mathrm m

You also could have first found the acceleration using the equation

v=v_0+at

then solve for x via

x=x_0+v_0t+\dfrac12at^2

but that would have involved a bit more work, and it turns out we didn't need to know the precise value of a anyway.

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