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.
If a man pushes on a wall with some force then according to Newton's third law, wall will also apply force on man with same magnitude but opposite in direction.
To solve this problem we will use the definition of the period in a simple pendulum, which warns that it is dependent on its length and gravity as follows:
![T =2\pi \sqrt{\frac{L}{g}}](https://tex.z-dn.net/?f=T%20%3D2%5Cpi%20%5Csqrt%7B%5Cfrac%7BL%7D%7Bg%7D%7D)
Here,
L = Length
g = Acceleration due to gravity
We can realize that
is a constant so it is proportional to the square root of its length over its gravity,
![T \propto \sqrt{\frac{L}{g}}](https://tex.z-dn.net/?f=T%20%5Cpropto%20%5Csqrt%7B%5Cfrac%7BL%7D%7Bg%7D%7D)
Since the body is in constant free fall, that is, a point where gravity tends to be zero:
![g \rightarrow 0 \Rightarrow T \rightarrow \infty](https://tex.z-dn.net/?f=g%20%5Crightarrow%200%20%5CRightarrow%20T%20%5Crightarrow%20%5Cinfty)
The value of the period will tend to infinity. This indicates that the pendulum will no longer oscillate because both the pendulum and the point to which it is attached are in free fall.
Answer:
The minimum frequency required to ionize the photon is 111.31 ×
Hertz
Given:
Energy = 378 ![\frac{kJ}{mol}](https://tex.z-dn.net/?f=%5Cfrac%7BkJ%7D%7Bmol%7D)
To find:
Minimum frequency of light required to ionize magnesium = ?
Formula used:
The energy of photon of light is given by,
E = h v
Where E = Energy of magnesium
h = planks constant
v = minimum frequency of photon
Solution:
The energy of photon of light is given by,
E = h v
Where E = Energy of magnesium
h = planks constant
v = minimum frequency of photon
738 ×
= 6.63 ×
× v
v = 111.31 ×
Hertz
The minimum frequency required to ionize the photon is 111.31 ×
Hertz