Answer:
![Y = 2.27 \times 10^{10} N/m^2](https://tex.z-dn.net/?f=Y%20%3D%202.27%20%5Ctimes%2010%5E%7B10%7D%20N%2Fm%5E2)
Explanation:
Natural length of the string is given as
![L_o = 43 cm](https://tex.z-dn.net/?f=L_o%20%3D%2043%20cm)
length of the string while block is hanging on it
![L = 53 cm](https://tex.z-dn.net/?f=L%20%3D%2053%20cm)
extension in length is given as
![\Delta L = 10 cm](https://tex.z-dn.net/?f=%5CDelta%20L%20%3D%2010%20cm)
now we have strain in the string is given as
![strain = \frac{\Delta L}{L}](https://tex.z-dn.net/?f=strain%20%3D%20%5Cfrac%7B%5CDelta%20L%7D%7BL%7D)
![strain = \frac{{10 cm}{43 cm}](https://tex.z-dn.net/?f=strain%20%3D%20%5Cfrac%7B%7B10%20cm%7D%7B43%20cm%7D)
![strain = 0.23](https://tex.z-dn.net/?f=strain%20%3D%200.23)
similarly we will have cross-sectional area of the string is given as
![A = 40 \times 10^{-6} m^2](https://tex.z-dn.net/?f=A%20%3D%2040%20%5Ctimes%2010%5E%7B-6%7D%20m%5E2)
now the stress in the string is given as
![Stress = \frac{T}{A}](https://tex.z-dn.net/?f=Stress%20%3D%20%5Cfrac%7BT%7D%7BA%7D)
![Stress = \frac{mg}{A}](https://tex.z-dn.net/?f=Stress%20%3D%20%5Cfrac%7Bmg%7D%7BA%7D)
![Stress = \frac{3.7 \times 9.81}{40 \times 10^{-6}}](https://tex.z-dn.net/?f=Stress%20%3D%20%5Cfrac%7B3.7%20%5Ctimes%209.81%7D%7B40%20%5Ctimes%2010%5E%7B-6%7D%7D)
![stress = 9.07 \times 10^5 N/m^2](https://tex.z-dn.net/?f=stress%20%3D%209.07%20%5Ctimes%2010%5E5%20N%2Fm%5E2)
Now Young's Modulus is given as
![Y = \frac{stress}{strain}](https://tex.z-dn.net/?f=Y%20%3D%20%5Cfrac%7Bstress%7D%7Bstrain%7D)
![Y = \frac{9.07 \times 10^5}{40\times 10^{-6}}](https://tex.z-dn.net/?f=Y%20%3D%20%5Cfrac%7B9.07%20%5Ctimes%2010%5E5%7D%7B40%5Ctimes%2010%5E%7B-6%7D%7D)
![Y = 2.27 \times 10^{10} N/m^2](https://tex.z-dn.net/?f=Y%20%3D%202.27%20%5Ctimes%2010%5E%7B10%7D%20N%2Fm%5E2)
If a gun is sighted to hit targets that are at the same height as the gun and 85 m away at the same height, how low, as a positive number in meters, will the bullet hit if aimed directly at a target 180 m away
The muzzle velocity of the bullet is 275 m/s
Answer:
y = -1.1109 m
Explanation:
Range = ![\frac{v^2sin2\theta}{g}](https://tex.z-dn.net/?f=%5Cfrac%7Bv%5E2sin2%5Ctheta%7D%7Bg%7D)
range = 85 m
velocity = 275 m/s
g = 9.8 m/s²
![85 = \frac{(275)^2sin2 \theta}{9.8}](https://tex.z-dn.net/?f=85%20%3D%20%5Cfrac%7B%28275%29%5E2sin2%20%5Ctheta%7D%7B9.8%7D)
![833=(275)^2sin2\theta](https://tex.z-dn.net/?f=833%3D%28275%29%5E2sin2%5Ctheta)
![833= 75625sin2\theta](https://tex.z-dn.net/?f=833%3D%2075625sin2%5Ctheta)
![0.01101 =sin2 \theta](https://tex.z-dn.net/?f=0.01101%20%3Dsin2%20%5Ctheta)
![0.005505 =sin \theta](https://tex.z-dn.net/?f=0.005505%20%3Dsin%20%5Ctheta)
![\theta =sin^{-1}(0.005505)](https://tex.z-dn.net/?f=%5Ctheta%20%3Dsin%5E%7B-1%7D%280.005505%29)
![\theta = 0.3154^0](https://tex.z-dn.net/?f=%5Ctheta%20%3D%200.3154%5E0)
If the bullet is aimed at a target 180 m, time required to travel 180 m with horizontal component will be;
![t=\frac{180}{275cos0.3154}](https://tex.z-dn.net/?f=t%3D%5Cfrac%7B180%7D%7B275cos0.3154%7D)
t = 0.6546 sec
To determine, how low as a positive number in meters.
![y=vyt-\frac{gt^2}{2}](https://tex.z-dn.net/?f=y%3Dvyt-%5Cfrac%7Bgt%5E2%7D%7B2%7D)
y =
![\frac{9.8*(0.6546)^2}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B9.8%2A%280.6546%29%5E2%7D%7B2%7D)
y = 0.9909 - 2.1018
y = -1.1109 m
Answer:
E = 1580594.95 N/C
Explanation:
To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:
(1)
dS: differential of the Gaussian surface
Qin: charge inside the Gaussian surface
εo: dielectric permittivity of vacuum = 8.85 × 10-12 C2/N ∙ m2
The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:
(2)
Qin is calculate by using the charge density:
(3)
Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.
The charge density is given by:
![\rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}](https://tex.z-dn.net/?f=%5Crho%3D%5Cfrac%7BQ%7D%7BV%7D%3D%5Cfrac%7B13%2A10%5E%7B-6%7DC%7D%7B%5Cfrac%7B4%7D%7B3%7D%5Cpi%28%280.15m%29%5E3-%280.10m%29%5E3%29%7D%5C%5C%5C%5C%5Crho%3D1.30%2A10%5E%7B-3%7D%5Cfrac%7BC%7D%7Bm%5E3%7D)
Next, you use the results of (3), (2) and (1):
![E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})](https://tex.z-dn.net/?f=E%284%5Cpi%20r%5E2%29%3D%5Cfrac%7B4%7D%7B3%5Cepsilon_o%7D%28r%5E3-a%5E3%29%5Crho%5C%5C%5C%5CE%3D%5Cfrac%7B%5Crho%7D%7B3%5Cepsilo_o%7D%28r-%5Cfrac%7Ba%5E3%7D%7Br%5E2%7D%29)
Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:
![E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B1.30%2A10%5E%7B-3%7DC%2Fm%5E3%7D%7B3%288.85%2A10%5E%7B-12%7DC%5E2%2FNm%5E2%29%7D%28%280.112m%29-%5Cfrac%7B%280.10%29%5E3%7D%7B%280.112m%29%5E2%7D%29%5C%5C%5C%5CE%3D1%2C580%2C594.95%5Cfrac%7BN%7D%7BC%7D)
hence, the electric field is 1580594.95 N/C
Answer:
<em>a) 37.5N</em>
<em>b) 9.375Joules</em>
Explanation:
a) According to Hooke's law
F = ke
k is the spring constant
e is the extension;
F = 150 * 0.25
F = 37.5N
b) Work done on the spring = 1/2ke^2
Work done on the spring = 1/2 * 150 * 0.25^2
Work done on the spring = 75 * 0.0625
Work done on the spring = 9.375Joules
Answer:
The correct answer for A P E X is A. a neap tide
Explanation:
I took the test