Explanation:
(a) First, we will calculate the number of moles as follows.
No. of moles = 
Molar mass of helium is 4 g/mol and mass is given as 0.1 kg or 100 g (as 1 kg = 1000 g).
Putting the given values into the above formula as follows.
No. of moles =
= 
= 25 mol
According to the ideal gas equation,
PV = nRT
or, 
= 336.17 K
Hence, temperature change will be 336.17 K.
(b) The total amount of heat required for this process will be calculated as follows.
q = 
= 
= 174573.081 J/K
or, = 174.57 kJ/K (as 1 kJ = 1000 J)
Therefore, the amount of total heat required is 174.57 kJ/K.
Answer:
Power, P = 722.96 watts
Explanation:
It is given that,
Voltage, V = 120 V
Length of nichrome wire, l = 8.9 m
Diameter of wire, d = 0.86 mm
Radius of wire, r = 0.43 mm = 0.00043 m
Resistivity of wire, 
We need to find the power drawn by this heater. Power is given by :
And, 
P = 722.96 watts
So, the power drawn by this heater element is 722.96 watts. Hence, this is the required solution.
Angular momentum is given by the length of the arm to the object, multiplied by the momentum of the object, times the cosine of the angle that the momentum vector makes with the arm. From your illustration, that will be:
<span>L = R * m * vi * cos(90 - theta) </span>
<span>cos(90 - theta) is just sin(theta) </span>
<span>and R is the distance the projectile traveled, which is vi^2 * sin(2*theta) / g </span>
<span>so, we have: L = vi^2 * sin(2*theta) * m * vi * sin(theta) / g </span>
<span>We can combine the two vi terms and get: </span>
<span>L = vi^3 * m * sin(theta) * sin(2*theta) / g </span>
<span>What's interesting is that angular momentum varies with the *cube* of the initial velocity. This is because, not only does increased velocity increase the translational momentum of the projectile, but it increase the *moment arm*, too. Also note that there might be a trig identity which lets you combine the two sin() terms, but nothing jumps out at me right at the moment. </span>
<span>Now, for the first part... </span>
<span>There are a few ways to attack this. Basically, you have to find the angle from the origin to the apogee (highest point) in the arc. Once we have that, we'll know what angle the momentum vector makes with the moment-arm because, at the apogee, we know that all of the motion is *horizontal*. </span>
<span>Okay, so let's get back to what we know: </span>
<span>L = d * m * v * cos(phi) </span>
<span>where d is the distance (length to the arm), m is mass, v is velocity, and phi is the angle the velocity vector makes with the arm. Let's take these one by one... </span>
<span>m is still m. </span>
<span>v is going to be the *hoizontal* component of the initial velocity (all the vertical component got eliminated by the acceleration of gravity). So, v = vi * cos(theta) </span>
<span>d is going to be half of our distance R in part two (because, ignoring friction, the path of the projectile is a perfect parabola). So, d = vi^2 * sin(2*theta) / 2g </span>
<span>That leaves us with phi, the angle the horizontal velocity vector makes with the moment arm. To find *that*, we need to know what the angle from the origin to the apogee is. We can find *that* by taking the arc-tangent of the slope, if we know that. Well, we know the "run" part of the slope (it's our "d" term), but not the rise. </span>
<span>The easy way to get the rise is by using conservation of energy. At the apogee, all of the *vertical* kinetic energy at the time of launch (1/2 * m * (vi * sin(theta))^2 ) has been turned into gravitational potential energy ( m * g * h ). Setting these equal, diving out the "m" and dividing "g" to the other side, we get: </span>
<span>h = 1/2 * (vi * sin(theta))^2 / g </span>
<span>So, there's the rise. So, our *slope* is rise/run, so </span>
<span>slope = [ 1/2 * (vi * sin(theta))^2 / g ] / [ vi^2 * sin(2*theta) / g ] </span>
<span>The "g"s cancel. Astoundingly the "vi"s cancel, too. So, we get: </span>
<span>slope = [ 1/2 * sin(theta)^2 ] / [ sin(2*theta) ] </span>
<span>(It's not too alarming that slope-at-apogee doesn't depend upon vi, since that only determines the "magnitude" of the arc, but not it's shape. Whether the overall flight of this thing is an inch or a mile, the arc "looks" the same). </span>
<span>Okay, so... using our double-angle trig identities, we know that sin(2*theta) = 2*sin(theta)*cos(theta), so... </span>
<span>slope = [ 1/2 * sin(theta)^2 ] / [ 2*sin(theta)*cos(theta) ] = tan(theta)/4 </span>
<span>Okay, so the *angle* (which I'll call "alpha") that this slope makes with the x-axis is just: arctan(slope), so... </span>
<span>alpha = arctan( tan(theta) / 4 ) </span>
<span>Alright... last bit. We need "phi", the angle the (now-horizontal) momentum vector makes with that slope. Draw it on paper and you'll see that phi = 180 - alpha </span>
<span>so, phi = 180 - arctan( tan(theta) / 4 ) </span>
<span>Now, we go back to our original formula and plug it ALL in... </span>
<span>L = d * m * v * cos(phi) </span>
<span>becomes... </span>
<span>L = [ vi^2 * sin(2*theta) / 2g ] * m * [ vi * cos(theta) ] * [ cos( 180 - arctan( tan(theta) / 4 ) ) ] </span>
<span>Now, cos(180 - something) = cos(something), so we can simplify a little bit... </span>
<span>L = [ vi^2 * sin(2*theta) / 2g ] * m * [ vi * cos(theta) ] * [ cos( arctan( tan(theta) / 4 ) ) ] </span>
The air pressure inside the balloon is: 0.1432 Pa
The formulas and procedures that we will use to solve this problem are:
Where:
- a = area of the sphere
- r = radius
- π = mathematical constant
- P = Pressure
- F = Force
- a = surface area
Information about the problem:
- r = 5.0 m
- F = 45 N
- 1 Pa = N/m²
- 1 N = kg * m/s²
- a=?
- P=?
Using the formula of the sphere area we get:
a = 4 * π * r²
a = 4 * 3.1416 * (5.0 m)²
a = 314.16 m²
Applying the pressure formula we get:
P = F/a
P = 45 N/314.16 m²
P = 0.1432 Pa
<h3>What is pressure?</h3>
It is a physical quantity that expresses the force applied on the area of a surface.
Learn more about pressure at: brainly.com/question/26269477
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