A body travel from rest and acceleration to a speed of 120kg/hours in 10 seconds . calculate acceleration
1 answer:
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Answer:
Explanation:
Fa - u*m*g = m*a
Fa = u*m*g + m*a
Fa - m*a = u*m*g
u =
A) 750 m
First of all, let's find the wavelength of the microwave. We have
is the frequency
is the speed of light
So the wavelength of the beam is
Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:
where
m = 1 since we are interested only in the central fringe
D = 30 km = 30,000 m
a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)
Substituting, we find
and so, the diameter is
B) 0.23 W/m^2
First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is
So the area is
And since the power is
The average intensity is
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
