You performed 0 work for the fact that work means the distance of movement made on an object not the amount of force it is exposed to. 0 work because it didn't move
Answer:
(1) Sure, the frequency is 1000 Hz.
Explanation:
Frequency = wave speed ÷ wave distance
wave speed = 100 m/s
wave distance = 10 cm = 10/100 = 0.1 m
Frequency = 100 ÷ 0.1 = 1000 Hz
Consider velocity to the right as positive.
First mass:
m₁ = 4.0 kg
v₁ = 2.0 m/s to the right
Second mass:
m₂ = 8.0 kg
v₂ = -3.0 m/s to the left
Total momentum of the system is
P = m₁v₁ + m₂v₂
= 4*2 + 8*(-3)
= -16 (kg-m)/s
Let v (m/s) be the velocity of the center of mass of the 2-block system.
Because momentum of the system is preserved, therefore
(m₁+m₂)v= -16
(4+8 kg)*(v m/s) = -16 (kg-m)/s
v = -1.333 m/s
Answer:
The center of mass is moving at 1.33 m/s to the left.
Answer:
A. 1.64 J
Explanation:
First of all, we need to find how many moles correspond to 1.4 mg of mercury. We have:
![n=\frac{m}{M_m}](https://tex.z-dn.net/?f=n%3D%5Cfrac%7Bm%7D%7BM_m%7D)
where
n is the number of moles
m = 1.4 mg = 0.0014 g is the mass of mercury
Mm = 200.6 g/mol is the molar mass of mercury
Substituting, we find
![n=\frac{0.0014 g}{200.6 g/mol}=7.0\cdot 10^{-6} mol](https://tex.z-dn.net/?f=n%3D%5Cfrac%7B0.0014%20g%7D%7B200.6%20g%2Fmol%7D%3D7.0%5Ccdot%2010%5E%7B-6%7D%20mol)
Now we have to find the number of atoms contained in this sample of mercury, which is given by:
![N=n N_A](https://tex.z-dn.net/?f=N%3Dn%20N_A)
where
n is the number of moles
is the Avogadro number
Substituting,
atoms
The energy emitted by each atom (the energy of one photon) is
![E_1 = \frac{hc}{\lambda}](https://tex.z-dn.net/?f=E_1%20%3D%20%5Cfrac%7Bhc%7D%7B%5Clambda%7D)
where
h is the Planck constant
c is the speed of light
is the wavelength
Substituting,
![E_1 = \frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{5.08\cdot 10^{-7} m}=3.92\cdot 10^{-19} J](https://tex.z-dn.net/?f=E_1%20%3D%20%5Cfrac%7B%286.63%5Ccdot%2010%5E%7B-34%7D%20Js%29%283%5Ccdot%2010%5E8%20m%2Fs%29%7D%7B5.08%5Ccdot%2010%5E%7B-7%7D%20m%7D%3D3.92%5Ccdot%2010%5E%7B-19%7D%20J)
And so, the total energy emitted by the sample is
![E=nE_1 = (4.22\cdot 10^{18} )(3.92\cdot 10^{-19}J)=1.64 J](https://tex.z-dn.net/?f=E%3DnE_1%20%3D%20%284.22%5Ccdot%2010%5E%7B18%7D%20%29%283.92%5Ccdot%2010%5E%7B-19%7DJ%29%3D1.64%20J)
Answer:
Explanation:
To find Sammy's course you have to add the two velocities (vectors), 18 mph 327º and 4 mph 60º.
To add the two vectors analytically you decompose each vector into their vertical and horizontal components.
<u>1. 18 mph 327º</u>
- Horizontal component: 18 mph × cos (327º) = 15.10 mph
- Vertical component: 18 mph × sin (327º) = - 9.80 mph
![15.10\hat i-9.80\hat j](https://tex.z-dn.net/?f=15.10%5Chat%20i-9.80%5Chat%20j)
<u>2. 4 mph 60º</u>
- Horizontal component: 4 mph × cos (60º) = 2.00 mph
- Vertical component: 4 mph × sin (60º) = 3.46 mph
![2.00\hat i+3.46\hat j](https://tex.z-dn.net/?f=2.00%5Chat%20i%2B3.46%5Chat%20j)
<u>3. Addition:</u>
You add the corresponding components:
![15.10\hat i-9.80\hat j+2.00\hat i+3.46\hat j\\ \\ 17.10\hat i-6.34\hat j](https://tex.z-dn.net/?f=15.10%5Chat%20i-9.80%5Chat%20j%2B2.00%5Chat%20i%2B3.46%5Chat%20j%5C%5C%20%5C%5C%2017.10%5Chat%20i-6.34%5Chat%20j)
To find the magnitude use Pythagorean theorem:
<u>4. Direction:</u>
Use the tangent ratio:
Find the inverse: