Answer:
Therefore the surface area of the balloon is increased at 4 cm³/s.
Explanation:
The balloon is being filled with air at a rate of 10 cm³/s
It means the volume of the balloon is increased at a rate 10 cm³/s.
i.e ![\frac{dv}{dt} =10 cm^3/s](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bdt%7D%20%3D10%20cm%5E3%2Fs)
Consider r be the radius of the balloon.
The volume of of a sphere is
![v=\frac{4}{3} \pi r^3](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20r%5E3)
Differentiate with respect to t
![\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bdt%7D%20%3D%5Cfrac%7B4%7D%7B3%7D%20%5Cpi%20%5Ctimes%203r%5E2%5Cfrac%7Bdr%7D%7Bdt%7D)
![\Rightarrow 10 =4\pi r^2\frac{dr}{dt}](https://tex.z-dn.net/?f=%5CRightarrow%2010%20%3D4%5Cpi%20r%5E2%5Cfrac%7Bdr%7D%7Bdt%7D)
![\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7Bdr%7D%7Bdt%7D%3D%5Cfrac%7B10%7D%7B4%5Cpi%20r%5E2%7D)
The surface of area of the balloon is(S) = ![4\pi r^2](https://tex.z-dn.net/?f=4%5Cpi%20r%5E2)
![S=4\pi r^2](https://tex.z-dn.net/?f=S%3D4%5Cpi%20r%5E2)
Differentiate with respect to t
![\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7BdS%7D%7Bdt%7D%20%3D4%5Cpi%5Ctimes2r%5Cfrac%7Bdr%7D%7Bdt%7D)
![\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7BdS%7D%7Bdt%7D%20%3D8%5Cpi%20r%5Cfrac%7Bdr%7D%7Bdt%7D)
Putting the value of
![\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7BdS%7D%7Bdt%7D%20%3D8%5Cpi%20r%5Ctimes%5Cfrac%7B10%7D%7B4%5Cpi%20r%5E2%7D)
![\Rightarrow \frac{dS}{dt} =\frac{20}{ r}](https://tex.z-dn.net/?f=%5CRightarrow%20%5Cfrac%7BdS%7D%7Bdt%7D%20%3D%5Cfrac%7B20%7D%7B%20r%7D)
Given that r = 5 cm
=4 cm³/s
Therefore the surface area of the balloon is increased at 4 cm³/s.
The applied force is different for the two cases
The case A with a greater force involves the greatest momentum change
The case A involves the greatest force.
<h3>What is collision?</h3>
- This is the head-on impact between two object moving in opposite or same direction.
The initial momentum of the two ball is the same.
P = mv
where;
- m is the mass of each
- v is the initial velocity of each ball
Since the force applied by the arm is different, the final velocity of the balls before stopping will be different.
Thus, the final momentum of each ball will be different
The impulse experienced by each ball is different since impulse is the change in momentum of the balls.
J = ΔP
The force applied by the rigid arm is greater than the force applied by the relaxed arm because the force applied by the rigid arm will cause the ball to be brought to rest faster.
Thus, we can conclude the following;
- The applied force is different for the two cases
- The case A with a greater force involves the greatest momentum change
- The case A involves the greatest force.
Learn more about impulse here: brainly.com/question/25700778
Answer:
Hey!
_________________
Voltage (V) = 0.8V
Current (I) = 200 mA = 200/10^3 = 2/10
Resistance = ?
Resistance = Voltage / Current
Voltage = Current × Resistance
0.8 = 2/10 × Resistance
0.8×10/2 = Resistance
8/2 = Resistance
Resistance = 4 ohm
_________________
Hope it helps...!!!
Explanation:
Answer:
Matter
Pure substances Mixture
Element compound Homogenous Heterogenous
<span>3.36x10^5 Pascals
The ideal gas law is
PV=nRT
where
P = Pressure
V = Volume
n = number of moles of gas particles
R = Ideal gas constant
T = Absolute temperature
Since n and R will remain constant, let's divide both sides of the equation by T, getting
PV=nRT
PV/T=nR
Since the initial value of PV/T will be equal to the final value of PV/T let's set them equal to each other with the equation
P1V1/T1 = P2V2/T2
where
P1, V1, T1 = Initial pressure, volume, temperature
P2, V2, T2 = Final pressure, volume, temperature
Now convert the temperatures to absolute temperature by adding 273.15 to both of them.
T1 = 27 + 273.15 = 300.15
T2 = 157 + 273.15 = 430.15
Substitute the known values into the equation
1.5E5*0.75/300.15 = P2*0.48/430.15
And solve for P2
1.5E5*0.75/300.15 = P2*0.48/430.15
430.15 * 1.5E5*0.75/300.15 = P2*0.48
64522500*0.75/300.15 = P2*0.48
48391875/300.15 = P2*0.48
161225.6372 = P2*0.48
161225.6372/0.48 = P2
335886.7441 = P2
Rounding to 3 significant figures gives 3.36x10^5 Pascals.
(technically, I should round to 2 significant figures for the result of 3.4x10^5 Pascals, but given the precision of the volumes, I suspect that the extra 0 in the initial pressure was accidentally omitted. It should have been 1.50e5 instead of 1.5e5).</span>