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3 years ago

GIVING BRAINLIEST TO THE CORRECT ANSWER!!

Physics

2 answers:

Degger [83]3 years ago

7 0

Answer is A locations by the ocean typically get much colder in the winter and much hotter in the summer than locations that are located inland.

Basile [38]3 years ago

6 0

The answer is A. locations by the ocean typically do not get as cold in the winter or as hot in the summer as locations that are located inland.

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Calculate the total displacement of a mouse walking along a ruler, if it begins at the x=5cm, and then does the following: It wa

Lana71 [14]

To begin, the mouse walks from 5 to 12 cm, for a displacement of 7 cm. Next, it walks 8 cm in the opposite direction, for a total displacement of (7 + [-8]) or (-1) cm. This leaves the mouse on 4 cm, and then it walks from there to the 7cm location, for a displacement of 7-4 or +3 cm. Adding 3cm to -1cm gives a final displacement of +2cm.

6 0

3 years ago

A real gas will behave most like an ideal gas under conditions of ________.

KengaRu [80]

Answer: high temperature and low pressure

Explanation:

The Ideal Gas equation is:

$P.V=n.R.T$

Where:

$P$ is the pressure of the gas

$V$ is the volume of the gas

$n$ the number of moles of gas

$R=0.0821\frac{L.atm}{mol.K}$ is the gas constant

$T$ is the absolute temperature of the gas in Kelvin

According to this law, molecules in gaseous state do not exert any force among them (attraction or repulsion) and the volume of these molecules is small, therefore negligible in comparison with the volume of the container that contains them.

Now, real gases can behave approximately to an ideal gas, under the conditions described above and taking into account the following:

When temperature is high a real gas approximates to ideal gas, because the molecules move quickly, preventing the repulsion or attraction forces to take effect. In addition, at low pressures, the volume of molecules is negligible.

4 0

3 years ago

Suppose a 2.0kg bird is flying at a speed of of 1m/s. It’s kinetic energy would be what?

Ganezh [65]

As we know that the formula of kinetic energy will be

$KE = \frac{1}{2} mv^2$

now here we know that

m = 2 kg

v = 1 m/s

so from the above equation we have

$KE = \frac{1}{2}(2)(1^2)$

$KE = 1 J$

7 0

3 years ago

Please help, I do not understand

Anettt [7]

I think the key here is to be exquisitely careful at all times, and
any time we make any move, keep our units with it.

We're given two angular speeds, and we need to solve for a time.

Outer (slower) planet:
Angular speed = ω rad/sec
Time per unit angle = (1/ω) sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/ω sec/rad) · (2π rad) = 2π/ω seconds .

Inner (faster) planet:
Angular speed = 2ω rad/sec
Time per unit angle = (1/2ω) sec/rad
Angle per revolution = 2π rad
Time per revolution = (1/2ω sec/rad) · (2π rad) = 2π/2ω sec = π/ω seconds.

So far so good. We have the outer planet taking 2π/ω seconds for one
complete revolution, and the inner planet doing it in only π/ω seconds ...
half the time for double the angular speed. Perfect !

At this point, I know what I'm thinking, but it's hard to explain.
I'm pretty sure that the planets are in line on the same side whenever the
total elapsed time is something like a common multiple of their periods.
What I mean is:

They're in line, SOMEwhere on the circles, when

   (a fraction of one orbit) = (the same fraction of the other orbit)
AND
   the total elapsed time is a common multiple of their periods.

Wait ! Ignore all of that. I'm doing a good job of confusing myself, and
probably you too. It may be simpler than that. (I hope so.) Throw away
those last few paragraphs.

The planets are in line again as soon as the faster one has 'lapped'
the slower one ... gone around one more time.
So, however many of the longer period have passed, ONE MORE
of the shorter period have passed. We're just looking for the Least
Common Multiple of the two periods.

      K (2π/ω seconds) = (K+1) (π/ω seconds)

   2Kπ/ω   =    Kπ/ω + π/ω

Subtract Kπ/ω :    Kπ/ω = π/ω

Multiply by ω/π :    K = 1

(Now I have a feeling that I have just finished re-inventing the wheel.)

And there we have it:

   In the time it takes the slower planet to revolve once,
   the faster planet revolves twice, and catches up with it.

   It will be 2π/ω seconds before the planets line up again.

   When they do, they are again in the same position as shown
   in the drawing.

To describe it another way . . .

   When Kanye has completed its first revolution ...

   Bieber has made it halfway around.

   Bieber is crawling the rest of the way to the starting point while ...

   Kanye is doing another complete revolution.

   Kanye laps Bieber just as they both reach the starting point ...

   Bieber for the first time, Kanye for the second time.

You're welcome. The generous bounty of 5 points is very gracious,
and is appreciated. The warm cloudy water and green breadcrust
are also delicious.

5 0

3 years ago

A 10.0-kg box starts at rest on a level floor. An external, horizontal force of 2.00 × 102 N is applied to the box for a distanc

Harman [31]

Answer:

vf = 11.2 m/s

Explanation:

m = 10 Kg

F = 2*10² N

x = 4.00 m

μ = 0.44

vi = 0 m/s

vf = ?

We can apply Newton's 2nd Law

∑ Fx = m*a (→)

F - Ffriction = m*a ⇒ F - (μ*N) = F - (μ*m*g) = m*a ⇒ a = (F - μ*m*g)/m

⇒ a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²

then , we use the equation

vf² = vi² + 2*a*x ⇒ vf = √(vi² + 2*a*x)

⇒ vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s

7 0

3 years ago