From ideal gas law, PV=nRT
where P is the pressure, V is the volume of the container, n is number of moles, R is the gas constant and T is the temperature.
Hence,
T= 110.65 k
Kinetic Energy =
K.E=
<h3>What is a kinetic energy? </h3>
The energy an object has as a result of motion is known as kinetic energy.
A force must be applied to an object in order to accelerate it. We must put in effort in order to apply a force. After the work is finished, energy is transferred to the item, which then moves at a new, constant speed. Kinetic energy is the type of energy that is transferred and is dependent on the mass and speed attained.
Kinetic energy can be converted into other types of energy and transported between objects. A flying squirrel may run into a chipmunk that is standing still, for instance. Some of the squirrel's initial kinetic energy may have been transferred to the chipmunk or changed into another kind of energy after the collision.
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Answer:
3 times louder
Explanation:
The Loudness in decibel Db L = 10㏒(I/I₀) where I = sound intensity level and I₀ = threshold of hearing = 10⁻¹² W/m².
Now, for Jessica, I₁ = sound intensity level of Jessica's music = 10⁻⁹
and I₂ = sound intensity level of Braylee's music = 10⁻³
So, substituting the variables into the equation, we have
L₁ = 10㏒(I₁/I₀)
L₁ = 10㏒(10⁻⁹/10⁻¹²)
L₁ = 10㏒(10³)
L₁ = 3 × 10㏒10
L₁ = 30㏒10
L₁ = 30 dB
Now, for Braylee, I₂ = sound intensity level of Braylee's music = 10⁻³
So, substituting the variables into the equation, we have
L₂ = 10㏒(I₁/I₀)
L₂ = 10㏒(10⁻³/10⁻¹²)
L₂ = 10㏒(10⁹)
L₂ = 9 × 10㏒10
L₂ =90㏒10
L₂ = 90 dB
So, the number of times Braylee's music is louder than Jessica's music is L₂/L₁ = 90 dB/30 dB = 3
So, Braylee's music is 3 times louder than Jessica's music
Answer:
A ball hits the ground and the ground pushes up on it
Explanation:
Newton's third law basically states that for every action, there's a reaction.
a ball hitting the ground would be the action. the ground pushing up on it with the same force is the reaction.
Hope this Helps!!! :)
The equation for electrical power is<span>P=VI</span>where V is the voltage and I is the current. This can be rearranged to solve for I in 6(a).
6(b) can be solved with Ohm's Law<span>V=IR</span>or if you'd like, from power, after substituting Ohm's law in for I<span>P=<span><span>V2</span>R</span></span>
For 7, realize that because they are in parallel, their voltages are the same.
We can find the resistance of each lamp from<span>P=<span><span>V2</span>R</span></span>Then the equivalent resistance as<span><span>1<span>R∗</span></span>=<span>1<span>R1</span></span>+<span>1<span>R2</span></span></span>Then the total power as<span><span>Pt</span>=<span><span>V2</span><span>R∗</span></span></span>However, this will reveal that (with a bit of algebra)<span><span>Pt</span>=<span>P1</span>+<span>P2</span></span>
For 8, again the resistance can be found as<span>P=<span><span>V2</span>R</span></span>The energy usage is simply<span><span>E=P⋅t</span></span>