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2 years ago

PLEASE PLEASE HELP AND PUT A REAL ANSWER ;-;. ALSO WILL GIVE BRAINLEST!!

Physics

2 answers:

torisob [31]2 years ago

8 0

Troposphere is the right answer

uysha [10]2 years ago

3 0

Answer:

Troposphere

High-pressure areas form due to downward motion through the troposphere, the atmospheric layer where weather occurs.

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A gas-filled balloon is submerged in a fluid. As a result, the balloon is subject to inward forces from all sides. How does the

enyata [817]

As we presume that the fluid density is greater than the gas density based on common sense, the volume of the balloon decreases. The mass per unit volume is known as fluid density.

Greek letter stands in for and (rho). Mass per length squared, or M/L3, is the unit of measurement for density. Specific Weight vs. Weight Density: A fluid density, also known as specific density, is determined by dividing the fluid's weight by its volume. Weight per volume of a fluid is also referred to as weight density.

A mathematical term called "volume" describes how much three-dimensional space is occupied by an item or a closed surface. The measurement of volume is done in cubic units, like m3, cm3, in3, etc.

Learn more about fluid density here

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8 0

7 months ago

An object has a position given by the radius vector r = [2.0 m + (3.00 m/s)t](i)+ [3.0 m - (2.00 m/s^2)t^2](j). Here (i) and (j)

jenyasd209 [6]

Answer:

The speed of the object is ( $3i - 4.00tj$ )m/s

The magnitude of the acceleration is 4.00m/s²

Explanation:

Given - position vector;

r = (2.0 + 3.00t)i + (3.0 - 2.00t²)j -------------------(i)

To get the speed vector ( $v$ ), take the first derivative of equation (i) with respect to time t as follows;

$v$ = $\frac{dr}{dt}$

$v$ = $\frac{d[(2.0 + 3.00t)i + (3.0 - 2.00t^2)j] }{dt}$

$v$ = $3i - 4.00tj$ ------------------------(ii)

To get the acceleration vector ( $a$ ), take the first derivative of the speed vector in equation(ii) as follows;

$a = \frac{dv}{dt}$

$a = \frac{d(3i - 4.00tj)}{dt}$

$a = -4.00$ j

The magnitude of the acceleration |a| is therefore given by

|a| = |-4.00|

|a| = 4.00 m/s²

In conclusion;

the speed of the object is ( $3i - 4.00tj$ )m/s

the magnitude of the acceleration is 4.00m/s²

3 0

2 years ago

How much work does the electric field do in moving a proton from a point with a potential of +145 v to a point where it is -55 v

Aleks04 [339]

The work done by the electric field in moving a charge is the negative of the potential energy difference between the two locations, which is the product between the magnitude of the charge q and the potential difference $\Delta V$ :
$W=-\Delta U=-q\Delta V=-q (V_f -V_i)$
The proton charge is $q=1.6 \cdot 10^{-19}C$ , and the two locations have potential of $V_i = +145 V$ and $V_f=-55 V$ , therefore the work is
$W=-(1.6 \cdot 10^{-19}C)(-55 V-(+145 V))=3.2 \cdot 10^{-17}J$

5 0

2 years ago

If the magnetic field of an electromagnetic wave is in the +x-direction and the electric field of the wave is in the +y directio

alexandr1967 [171]

Answer:

Explanation:

The direction of propagation of electromagnetic wave

is given by the direction of vector E x B where E is electrical field , B is magnetic field .

Given Electric field = E i because it is along x axis

Magnetic field = Bj because it is along y axis

E x B = Ei x Bj

= EB k .

so direction of E x B is along k direction or z - axis so wave is propagating along z - axis .

7 0

2 years ago

Find the orbital speed v for a satellite in a circular orbit of radius R.Express the orbital speed in terms of G, M, and R.

AlekseyPX

<h2>Answer: $V=\sqrt{G\frac{M}{R}}$ </h2>

The velocity of a satellite describing a circular orbit is <u>constant</u> and defined by the following expression:

$V=\sqrt{G\frac{M}{R}}$ (1)

Where:

$G$ is the gravity constant

$M$ the mass of the massive body around which the satellite is orbiting

$R$ the radius of the orbit (measured from the center of the planet to the satellite).

Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. I<u>t depends on the mass of the massive body.</u>

In addition, this orbital speed is constant because at all times <u>both the kinetic energy and the potential remain constant</u> in a circular (closed) orbit.

5 0

3 years ago