Answer: 0%
Explanation: hope this help
B4 the tackle:
<span>The linebacker's momentum = 115 x 8.5 = 977.5 kg m/s north </span>
<span>and the halfback's momentum = 89 x 6.7 = 596.3 kg m/s east </span>
<span>After the tackle they move together with a momentum equal to the vector sum of their separate momentums b4 the tackle </span>
<span>The vector triangle is right angled: </span>
<span>magnitude of final momentum = √(977.5² + 596.3²) = 1145.034 kg m/s </span>
<span>so (115 + 89)v(f) = 1145.034 ←←[b/c p = mv] </span>
<span>v(f) = 5.6 m/s (to 2 sig figs) </span>
<span>direction of v(f) is the same as the direction of the final momentum </span>
<span>so direction of v(f) = arctan (596.3 / 977.5) = N 31° E (to 2 sig figs) </span>
<span>so the velocity of the two players after the tackle is 5.6 m/s in the direction N 31° E </span>
<span>btw ... The direction can be given heaps of different ways ... N 31° E is probably the easiest way to express it when using the vector triangle to find it</span>
<h3>
Answer:</h3>
24.94 L
<h3>
Explanation:</h3>
<u>We are given;</u>
- Moles of Helium, n= 1 mole
- Pressure of the gas, P = 100 kPa
- Temperature of the gas, T = 300 K
We are required to calculate the volume of the balloon
- We are going to use the ideal gas equation;
- According to the ideal gas equation;
PV = nRT , where R is the ideal gas constant, 8.314 L.kPa/mol.K
V = nRT ÷ P
= ( 1 × 8.314 × 300 K) ÷ 100 kPa
= 24.94 L
Therefore, the volume of the balloon is 24.94 L
I believe it is, All of the above.
Answer:
Explanation:
From the given information:
We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then
The volume charge distribution relates to the radial direction at r = R
∴
![\rho (r) \ \alpha \ \delta (r -R)](https://tex.z-dn.net/?f=%5Crho%20%28r%29%20%5C%20%20%5Calpha%20%20%5C%20%20%5Cdelta%20%28r%20-R%29)
![\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)](https://tex.z-dn.net/?f=%5Crho%20%28r%29%20%3D%20k%20%5C%20%20%5Cdelta%20%28r%20-R%29%20%5C%20%5C%20%20at%20%5C%20%5C%20%20%28r%20%3D%20R%29)
![\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)](https://tex.z-dn.net/?f=%5Crho%20%28r%29%20%3D%200%5C%20%5C%20since%20%5C%20r%3C%20R%20%20%5C%20%5C%20or%20%20%5C%20%5C%20r%3ER----%20%281%29)
To find the constant k, we examine the total charge Q which is:
![Q = \int \rho (r) \ dV = \int \sigma \times dA](https://tex.z-dn.net/?f=Q%20%3D%20%5Cint%20%5Crho%20%28r%29%20%5C%20dV%20%3D%20%5Cint%20%5Csigma%20%5Ctimes%20dA)
![Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2](https://tex.z-dn.net/?f=Q%20%3D%20%5Cint%20%5Crho%20%28r%29%20%5C%20dV%20%3D%20%5Csigma%20%5Ctimes4%20%5Cpi%20R%5E2)
∴
![\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2](https://tex.z-dn.net/?f=%5Cint%20%5E%7B2%20%5Cpi%7D_%7B0%7D%20%5Cint%20%5E%7B%5Cpi%7D_%7B0%7D%20%5Cint%20%5E%7BR%7D_%7B0%7D%20%5Crho%20%28r%29%20r%5E2sin%20%5Ctheta%20%20%5C%20dr%20%5C%20d%5Ctheta%20%5C%20d%5Cphi%20%3D%20%5Csigma%20%5Ctimes%204%20%5Cpi%20R%5E2)
![\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2](https://tex.z-dn.net/?f=%5Cint%5E%7B2%20%5Cpi%7D_%7B0%7D%20d%20%5Cphi%2A%20%5Cint%20%5E%7B%5Cpi%7D_%7B0%7D%20%5C%20sin%20%5Ctheta%20d%20%5Ctheta%20%2A%20%5Cint%20%5E%7BR%7D_%7B0%7D%20k%20%5Cdelta%20%28r%20-R%29%20%2A%20r%5E2dr%20%3D%20%5Csigma%20%5Ctimes%204%20%5Cpi%20R%5E2)
![(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2](https://tex.z-dn.net/?f=%282%20%5Cpi%29%282%29%20%2A%20%5Cint%20%5E%7BR%7D_%7B0%7D%20k%20%5Cdelta%20%28r%20-R%29%20%2A%20r%5E2dr%20%3D%20%5Csigma%20%5Ctimes%204%20%5Cpi%20R%5E2)
Thus;
![k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2](https://tex.z-dn.net/?f=k%20%2A%204%20%5Cpi%20%20%5Cint%20%5E%7BR%7D_%7B0%7D%20%20%5Cdelta%20%28r%20-R%29%20%2A%20r%5E2dr%20%3D%20%5Csigma%20%5Ctimes%20%20R%5E2)
![k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2](https://tex.z-dn.net/?f=k%20%2A%20%5Cint%20%5E%7BR%7D_%7B0%7D%20%20%5Cdelta%20%28r%20-R%29%20%20r%5E2dr%20%3D%20%5Csigma%20%5Ctimes%20%20R%5E2)
![k * R^2= \sigma \times R^2](https://tex.z-dn.net/?f=k%20%2A%20R%5E2%3D%20%5Csigma%20%5Ctimes%20%20R%5E2)
![k = R^2 --- (2)](https://tex.z-dn.net/?f=k%20%20%3D%20%20%20R%5E2%20---%20%282%29)
Hence, from equation (1), if k = ![\sigma](https://tex.z-dn.net/?f=%5Csigma)
![\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Crho%20%28r%29%20%3D%20%5Cdelta%2A%20%5Cdelta%20%28r%20-R%29%20%20%5C%20%5C%20%20at%20%20%20%5C%20%5C%20%20%28r%3DR%29%7D)
![\mathbf{\rho (r) =0 \ \ at \ \ rR}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Crho%20%28r%29%20%3D0%20%5C%20%5C%20%20at%20%20%20%5C%20%5C%20%20r%3CR%20%20%5C%20%5C%20%20or%20%5C%20%20%5C%20r%3ER%7D)
To verify the units:
![\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Crho%20%28r%29%20%3D%5Csigma%20%5C%20%2A%20%20%5C%20%5Cdelta%20%28r-R%29%7D)
↓ ↓ ↓
c/m³ c/m³ × 1/m
Thus, the units are verified.
The integrated charge Q
![Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr](https://tex.z-dn.net/?f=Q%20%3D%20%5Cint%20%5Crho%20%28r%29%20%5C%20dV%20%5C%5C%20%5C%5C%20Q%20%3D%20%5Cint%20%5E%7B2%20%5C%20%5Cpi%7D_%7B0%7D%20%5Cint%20%5E%7B%5Cpi%7D_%7B0%7D%20%5Cint%20%5ER_0%20%5Crho%20%28r%29%20%5C%20%5C%20r%5E2%20%5C%20%5C%20%20sin%20%5Ctheta%20%20%5C%20dr%20%5C%20d%5Ctheta%20%5C%20%20d%20%5Cphi%20%20%5C%5C%20%5C%5C%20%20Q%20%3D%20%5Cint%20%5E%7B2%20%5Cpi%7D_%7B0%7D%20%5C%20%20d%20%5Cphi%20%20%5Cint%20%5E%7B%5Cpi%7D_%7B0%7D%20%5C%20sin%20%5Ctheta%20%20%5Cint%20%5ER_%7B0%7D%20%5Crho%20%28r%29%20r%5E2%20%5C%20dr)
![Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr](https://tex.z-dn.net/?f=Q%20%3D%20%282%20%5Cpi%29%20%282%29%20%5Cint%20%5ER_0%20%5Csigma%20%2A%20%5Cdelta%20%28r-R%29%20r%5E2%20%5C%20dr)
![Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr](https://tex.z-dn.net/?f=Q%20%3D%204%20%5Cpi%20%20%5Csigma%20%20%5Cint%20%5ER_0%20%20%2A%20%5Cdelta%20%28r-R%29%20r%5E2%20%5C%20dr)
since ![( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )](https://tex.z-dn.net/?f=%28%20%5Cint%20%5E%7Bxo%7D_%7B0%7D%20%28x%20-x_o%29%20f%28x%29%20%5C%20dx%20%3D%20f%28x_o%29%20%29)
![\mathbf{Q = 4 \pi R^2 \sigma }](https://tex.z-dn.net/?f=%5Cmathbf%7BQ%20%3D%204%20%5Cpi%20R%5E2%20%20%5Csigma%20%20%7D)