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Anton [14]
4 years ago
14

What is the transition from a gas to a liquid?

Physics
1 answer:
postnew [5]4 years ago
6 0
The transition from gas to liquid is called condensation. An example would be water droplets forming on an ice cold glass placed in room temperature.
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A baseball is hit nearly straight up into the air with a speed of 22 m/s. (a) how high does it go ?
AnnZ [28]

So, this is a problem where the accleration is not provided, since it is implied.  The only acceleration is acceleration due to gravity (9.8 m/s)


The equation we will use for this problem is V^2 =V_{0}^2 + 2a (X-X_0)

V is the final velocity, V₀ is the initial velocity, a is the acceleration, X is the final height, and X₀ is the starting height.


We can assume that the ball starts on the ground since no height is given, so now we plug our numbers in.

We will use 0 as the final velocity, since the ball will stop moving upwards when it is the highest.  We will use -9.8 since that is the acceleration due to gravity and we will use 22m/s as V₀ since that is the starting velocity.

V^2 =V_{0}^2 + 2a (X-X_0)\\0^2 = 22^2 + 2\times-9.8(X-0)\\0=484-19.6x\\-484=-19.6x\\24.69387755 = x\\x\approx24.69


So, the ball will go 24.69 meters up


4 0
3 years ago
A 12-inch object is placed 30 inches in front of a plane mirror. A ray of light from the object strikes the mirror at a 45-degre
gtnhenbr [62]

I believe that it is C.24 inches

3 0
3 years ago
Read 2 more answers
Choose all of the true statements regarding the relationship between voltage, resistance, and current. Current is measured in am
AveGali [126]

-- <u><em>Current is measured in amps.</em></u>  (You can use any symbol you want to represent current, but the most common one is " I ", not "Δ".)

-- <u><em>The relationship between current, voltage, and resistance is mathematically defined by Ohm's Law. </em></u>

-- <u><em>Current is the flow of electrons through a circuit.</em></u>

-- (Ohm's Law is NOT mathematically represented by the equation V=I/R.)  <u><em>It should be V = I · R</em></u> .  

(When solving for Resistance in a circuit and both voltage and current are known values, the equation I =V*R is not true, and not the way to solve it.)  <u><em>If the resistance is what you're looking for, then the equation to use is  </em></u><u><em>R = V / I</em></u><u><em> .  </em></u>

<em>-- </em><u><em>If the voltage in a circuit is increased, the current will also increase.</em></u>

6 0
4 years ago
Read 2 more answers
The forces in (Figure 1) are acting on a 1.0 kg object.What is ax , the x -component of the object's acceleration
a_sh-v [17]

The x -component of the object's acceleration is 2 m/s².

<h3>What's the resultant force along x- direction?</h3>
  • Forces along x axis direction are as follows
  1. 4N along +x axis, so it's taken as +4 N
  2. 2N along -x axis , so it's taken as -2N.
  • Resultant force along x direction = 4N - 2N = 2 N which is along + ve x direction.

<h3>What's the acceleration along x axis direction?</h3>
  • As per Newton's second law, Force = mass × acceleration of the object
  • Force along x axis= mass × acceleration along x axis= 2N
  • Acceleration = 2/ mass = 2/1 = 2 m/s²

Thus, we can conclude that the acceleration along x axis is 2 m/s².

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: The forces in (Figure 1) are acting on a 1.0 kg object. What is ax, the x-component of the object's acceleration?

Learn more about the acceleration here:

brainly.com/question/460763

#SPJ1

3 0
2 years ago
Exoplanets (planets outside our solar system) are an active area of modern research. Suppose astronomers find such a planet that
Leno4ka [110]

Answer:

8.829 m/s²

Explanation:

M = Mass of Earth

m = Mass of Exoplanet

g_e = Acceleration due to gravity on Earth = 9.81 m/s²

g = Acceleration due to gravity on Exoplanet

m=M-0.1M\\\Rightarrow m=0.9M

g_e=G\frac{M}{r^2}

g=G\frac{0.9M}{r^2}

Dividing the equations we get

\frac{g}{g_e}=\frac{G\frac{0.9M}{r^2}}{G\frac{M}{r^2}}\\\Rightarrow \frac{g}{g_e}=0.9\\\Rightarrow g=0.9g_e\\\Rightarrow g=0.9\times 9.81\\\Rightarrow g=8.829\ m/s^2

Acceleration due to gravity on the surface of the Exoplanet is 8.829 m/s²

3 0
3 years ago
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