Answer:
a.6.5025 J
b.6.5025 J
Explanation:
We are given that
Mass of pellet,m=0.27 g=
1 kg=1000 g
Spring constant,k=1800 N/m
x=8.5 cm=
1m=100 cm
a.Potential energy stored in the compressed spring is given by
P.E=
b.By using law of conservation of energy
P.E of spring=K.E of the pellet
K.E of the pellet=6.5025 J
The question is incomplete. I can help you by adding the information missing. They want you to calculate a) the radius of the cyclotron orbit for an electron with speed 1.0 * 10^6 m/s^2 and b) the radius of a cyclotron orbit for a proton with speed 5.0 * 10^4 m/s.
The two tasks involve combining the equations of the magnectic force and the centripetal force in a circular motion.
When you do that, you will obtain an expression to find the radius of the circular motion, which is the radius of the cyclotron that impulses the particles.
a)
Magentic force, F = q*v*B
q is the charge of the electron = 1.6 * 10^ -19 C
v is the speed = 1.0 * 10 ^ 6 m/s
B is the magentic field = 5.0 * 10 ^-5 T
Centripetal force, F = m*Ac = m * v^2 / R
where,
Ac = centripetal acceleration
m = mass of the electron = 9.11 * 10 ^-31 kg
R = the radius of the orbit
Now equal the two forces: q*v*B = m * v^2 / R => R = m*v / (q*B)
=> R = (9.11 * 10^31 kg) (1.0*10^6m/s) / [ (1.6 * 10^-19C)* (5.0 * 10^-5T) ]
=> R = 0.114 m
b) The equations are the same, just now use the speed, charge and mass of the proton instead of those of the electron.
R = m*v / (qB) = (1.66*10^-27 kg)(5.0*10^4 m/s) / [(1.6*10^-19C)(5*10^-5T)]
=> R = 10.4 m
The answer is C. replacement
This would be an example of a replacement type of reaction. As the iron is replaced with the aluminum in the reaction.
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Answer:
(a) f= 622.79 Hz
(b) f= 578.82 Hz
Explanation:
Given Data
Frequency= 600 Hz
Distance=1.0 m
n=120 rpm
Temperature =20 degree
Before solve this problem we need to find The sound generator moves on a circular with tangential velocity
So
Speed of sound is given by
c = √(γ·R·T/M)
............in an ideal gas
where γ heat capacity ratio
R universal gas constant
T absolute temperature
M molar mass
The speed of sound at 20°C is
c = √(1.40 ×8.314472J/molK ×293.15K / 0.0289645kg/mol)
c= 343.24m/s
The sound moves on a circular with tangential velocity
vt = ω·r.................where
ω=2·π·n
vt= 2·π·n·r
vt= 2·π · 120min⁻¹ · 1m
vt= 753.6 m/min
convert m/min to m/sec
vt= 12.56 m/s
Part A
For maximum frequency is observed
v = vt
f = f₀/(1 - vt/c )
f= 600Hz / (1 - (12.56m/s / 343.24m/s) )
f= 622.789 Hz
Part B
For minimum frequency is observed
v = -vt
f = f₀/(1 + vt/c )
f= 600Hz / (1 + (12.56m/s / 343.24m/s) )
f= 578.82 Hz
Answer:
There is no actual question attached to this, to get a real answer be sure to include the documents/question that is provided on your work.
Explanation:
