How does the frequency of gamma rays compare to the frequency of microwaves?
1 answer:
Answer:
Frequency of gamma rays is more than microwave.
Explanation:
The number of vibrations per unit time is called frequency of a wave. The SI unit of frequency is Hertz. It is equal to .
The frequency of gamma rays is of the order of . The frequency of microwave is of the order of
.
It is clear that gamma rays have more frequency that of the microwave.
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Answer:
The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.
Explanation:
Given that,
Weight of metal A = 12.5%
Weight of metal B = 87.5%
Length of unit cell = 0.395 nm
Density of A = 4.27 g/cm³
Density of B= 6.35 g/cm³
Weight of A = 61.4 g/mol
Weight of B = 125.7 g/mol
We need to calculate the density of the alloy
Using formula of density
....(I)
Where, n = number of atoms per unit cells
m = Mass of the alloy
V=Volume of the unit cell
N = Avogadro number
We calculate the density of alloy
We calculate the mass of the alloy
Put the value into the equation (I)
Hence, The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.
Answer:
(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface
(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:
Wf= -20.4 J is negative
(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane:
Wg= 58.8 J is positive
Explanation:
Nomenclature
vf: final velocity
v₀ :initial velocity
a: acceleleration
d: distance
Ff: Friction force
W: weight
m:mass
g: acceleration due to gravity
Graphic attached
The attached graph describes the variables related to the kinetics of the object (forces and accelerations)
Calculation de of the components of W in the inclined plane
W=m*g
Wx₁ = m*g*sin30°
Wy₁= m*g*cos30°
Object kinematics on the inclined plane
vf₁²=v₀₁²+2*a₁*d₁
v₀₁=0
vf₁²=2*a₁*d₁
Equation (1)
Object kinetics on the inclined plane (μ= 0.2)
∑Fx₁=ma₁ :Newton's second law
-Ff₁+Wx₁ = ma₁ , Ff₁=μN₁
-μ₁N₁+Wx₁ = ma₁ Equation (2)
∑Fy₁=0 : Newton's first law
N₁-Wy₁= 0
N₁- m*g*cos30°=0
N₁ = m*g*cos30°
We replace N₁ = m*g*cos30 and Wx₁ = m*g*sin30° in the equation (2)
-μ₁m*g*cos30₁+m*g*sin30° = ma₁ : We divide by m
-μ₁*g*cos30°+g*sin30° = a₁
g*(-μ₁*cos30°+sin30°) = a₁
a₁ =9.8(-0.2*cos30°+sin30°)=3.2 m/s²
We replace a₁ =3.2 m/s² and d₁= 3m in the equation (1)
Rough surface kinematics
vf₂²=v₀₂²+2*a₂*d₂ v₀₂=vf₁=4.38 m/s
0 =4.38²+2*a₂*d₂ Equation (3)
Rough surface kinetics (μ= 0.3)
∑Fx₂=ma₂ :Newton's second law
-Ff₂=ma₂
--μ₂*N₂ = ma₂ Equation (4)
∑Fy₂= 0 :Newton's first law
N₂-W=0
N₂=W=m*g
We replace N₂=m*g inthe equation (4)
--μ₂*m*g = ma₂ We divide by m
--μ₂*g = a₂
a₂ =-0.2*9.8= -1.96m/s²
We replace a₂ = -1.96m/s² in the equation (3)
0 =4.38²+2*-1.96*d₂
3.92*d₂ = 4.38²
d₂=4.38²/3.92
d₂=4.38²/3.92
(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface
(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:
Wf = - Ff₁*d₁
Ff₁= μ₁N₁= μ₁*m*g*cos30°= -0.2*4*9.8*cos30° = 6,79 N
Wf= - 6.79*3 = 20.4 N*m
Wf= -20.4 J is negative
(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane
Wg=W₁x*d= m*g*sin30*3=4*9.8*0.5*3= 58.8 N*m
Wg= 58.8 J is positive
