Answer: -33.3 * 10^9 C/m^2( nC/m^2)
Explanation: In order to solve this problem we have to use the gaussian law, the we have:
Eoutside =0 so teh Q inside==
the Q inside= 4.6 nC/m*L + σ *2*π*b*L where L is the large of the Gaussian surface and b the radius of the shell.
Then we simplify and get
σ= -4.6/(2*π*b)= -33.3 nC/m^2
Answer:
a) ![W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J](https://tex.z-dn.net/?f=W_%7Bg%7D%3Dmdx%20%3D%200.21%20kg%20%2A9.8%5Cfrac%7Bm%7D%7Bs%5E2%7D%200.10m%3D0.2058%20J)
b) ![W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J](https://tex.z-dn.net/?f=W_%7Bspring%7D%3D%20-%5Cfrac%7B1%7D%7B2%7D%20Kx%5E2%20%3D-%5Cfrac%7B1%7D%7B2%7D%20200%20N%2Fm%20%280.1m%29%5E2%3D-1%20J)
c) ![V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s](https://tex.z-dn.net/?f=V_i%20%3D%5Csqrt%7B2%20%5Cfrac%7BW_g%20%2B%20W_%7Bspring%7D%7D%7B0.21%20kg%7D%7D%7D%3D%5Csqrt%7B2%20%5Cfrac%7B%281-0.2058%29%7D%7B0.21%20kg%7D%7D%7D%3D2.75m%2Fs)
d)
or 18.3 cm
Explanation:
For this case we have the following system with the forces on the figure attached.
We know that the spring compresses a total distance of x=0.10 m
Part a
The gravitational force is defined as mg so on this case the work donde by the gravity is:
![W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J](https://tex.z-dn.net/?f=W_%7Bg%7D%3Dmdx%20%3D%200.21%20kg%20%2A9.8%5Cfrac%7Bm%7D%7Bs%5E2%7D%200.10m%3D0.2058%20J)
Part b
For this case first we can convert the spring constant to N/m like this:
![2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}](https://tex.z-dn.net/?f=2%20%5Cfrac%7BN%7D%7Bcm%7D%20%5Cfrac%7B100cm%7D%7B1m%7D%3D200%20%5Cfrac%7BN%7D%7Bm%7D)
And the work donde by the spring on this case is given by:
![W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J](https://tex.z-dn.net/?f=W_%7Bspring%7D%3D%20-%5Cfrac%7B1%7D%7B2%7D%20Kx%5E2%20%3D-%5Cfrac%7B1%7D%7B2%7D%20200%20N%2Fm%20%280.1m%29%5E2%3D-1%20J)
Part c
We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:
![W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i](https://tex.z-dn.net/?f=%20W_%7Bg%7D%20%2BW_%7Bspring%7D%20%3D%20K_%7Bf%7D%20-K_%7Bi%7D%3D0-%20%5Cfrac%7B1%7D%7B2%7D%20m%20v%5E2_i)
And if we solve for the initial velocity we got:
![V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s](https://tex.z-dn.net/?f=V_i%20%3D%5Csqrt%7B2%20%5Cfrac%7BW_g%20%2B%20W_%7Bspring%7D%7D%7B0.21%20kg%7D%7D%7D%3D%5Csqrt%7B2%20%5Cfrac%7B%281-0.2058%29%7D%7B0.21%20kg%7D%7D%7D%3D2.75m%2Fs)
Part d
Let d1 represent the new maximum distance, in order to find it we know that :
![-1/2mV^2_i = W_g + W_{spring}](https://tex.z-dn.net/?f=-1%2F2mV%5E2_i%20%3D%20W_g%20%2B%20W_%7Bspring%7D)
And replacing we got:
![-1/2mV^2_i =mg d_1 -1/2 k d^2_1](https://tex.z-dn.net/?f=-1%2F2mV%5E2_i%20%3Dmg%20d_1%20-1%2F2%20k%20d%5E2_1)
And we can put the terms like this:
![\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20k%20d%5E2_1%20-mg%20d_1%20-1%2F2%20m%20V%5E2_i%20%3D0)
If we multiply all the equation by 2 we got:
![k d^2_1 -2 mg d_1 -m V^2_i =0](https://tex.z-dn.net/?f=%20k%20d%5E2_1%20-2%20mg%20d_1%20-m%20V%5E2_i%20%3D0)
Now we can replace the values and we got:
![200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0](https://tex.z-dn.net/?f=200N%2Fm%20d%5E2_1%20-0.21kg%289.8m%2Fs%5E2%29d_1%20-0.21%20kg%285.50%20m%2Fs%29%5E2%29%20%3D0)
![200 d^2_1 -2.058 d_1 -6.3525=0](https://tex.z-dn.net/?f=200%20d%5E2_1%20-2.058%20d_1%20-6.3525%3D0)
And solving the quadratic equation we got that the solution for
or 18.3 cm because the negative solution not make sense.
Force is a vector quantity
so pulling from opposite side will be negative
so
750+(-500)= 250N
C is the right answer
becauseause the man on the right applies greater force.
Answer:
![r=1.14m](https://tex.z-dn.net/?f=r%3D1.14m)
Explanation:
is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:
![F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB](https://tex.z-dn.net/?f=F_m%3DqvBsen%5Ctheta%5C%5CF_m%3DqvBsen%2890%5E%5Ccirc%29%5C%5CF_m%3DqvB)
A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:
![F_m=F_c\\qvB=F_c(1)](https://tex.z-dn.net/?f=F_m%3DF_c%5C%5CqvB%3DF_c%281%29)
is the centripetal force and is defined as:
![F_c=m\frac{v^2}{r}](https://tex.z-dn.net/?f=F_c%3Dm%5Cfrac%7Bv%5E2%7D%7Br%7D)
Here
is the proton's speed and
is the radius of the circular motion. Replacing this in (1) and solving for r:
![qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}](https://tex.z-dn.net/?f=qvB%3D%5Cfrac%7Bmv%5E2%7D%7Br%7D%5C%5Cr%3D%5Cfrac%7Bmv%5E2%7D%7BqvB%7D%5C%5Cr%3D%5Cfrac%7Bmv%7D%7BqB%7D)
Recall that 1 J is equal to
, so:
![4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J](https://tex.z-dn.net/?f=4.9MeV%2A%5Cfrac%7B1J%7D%7B6.242%2A10%5E%7B12%7DMeV%7D%3D7.85%2A10%5E%7B-13%7DJ)
We can calculate
from the kinetic energy of the proton:
![K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7Bmv%5E2%7D%7B2%7D%5C%5C%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B2K%7D%7Bm%7D%7D%5C%5Cv%3D%5Csqrt%7B%5Cfrac%7B2%287.85%2A10%5E%7B-13%7DJ%29%7D%7B1.67%2A10%5E%7B-27%7Dkg%7D%7D%5C%5Cv%3D3.06%2A10%5E%7B7%7D%5Cfrac%7Bm%7D%7Bs%7D)
Finally, we calculate the radius of the proton path:
![r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m](https://tex.z-dn.net/?f=r%3D%5Cfrac%7Bmv%7D%7BqB%7D%5C%5Cr%3D%5Cfrac%7B1.67%2A10%5E%7B-27%7Dkg%283.06%2A10%5E%7B7%7D%5Cfrac%7Bm%7D%7Bs%7D%29%7D%7B1.6%2A10%5E%7B-19%7DC%280.28T%29%7D%5C%5Cr%3D1.14m)
NOTE: The given question is incomplete.
<u>The complete question is given below.</u>
A student measures the speed of yellow light in water to be 2.00 x 10⁸ m/s. Calculate the speed of light in air.
Solution:
Speed of yellow light in water (v) = 2.00 x 10⁸ m/s
Refractive Index of water with respect to air (μ) = 4/3
Refractive Index = Speed of yellow light in air / Speed of yellow light in water
Or, The speed of yellow light in air = Refractive Index × Speed of yellow light in water
or, = (4/3) × 2.00 x 10⁸ m/s
or, = 2.67 × 10⁸ m/s ≈ 3.0 × 10⁸ m/s
Hence, the required speed of yellow light in the air will be 3.0 × 10⁸ m/s.