Crescent, gibbous, waxing, and waning.
Answer
Assuming
east is the positive x direction
north is the positive y direction
initial velocity , u = 19 j m/s
a)
acceleration , a = 1.6 j m/s^2
Using first equation of motion
v = u + a × t
v = 19 + 5.6× 1.6
v = 28 j m/s
the velocity of the car after 5.6 s is 28 m/s north
b)
acceleration , a = -1.5 j m/s^2
Using first equation of motion
v = u + a × t
v = 19 - 5.6 ×1.5
v = 10.6 j m/s
the velocity of the car after 5.6 s is 10.6 m/s north
The statement that best describes how the windmill technology benefits the environment is this: WINDMILLS DO NOT POLLUTE THE ENVIRONMENT.
Constructing a windmill involves harnessing the power of the wind to generate electricity. This type of electricity generation has no side effects whatsoever. It is environmental friendly and does not pollute the environment.
The de Broglie wavelength of a 0.56 kg ball moving with a constant velocity of 26 m/s is 4.55×10⁻³⁵ m.
<h3>De Broglie wavelength:</h3>
The wavelength that is incorporated with the moving object and it has the relation with the momentum of that object and mass of that object. It is inversely proportional to the momentum of that moving object.
λ=h/p
Where, λ is the de Broglie wavelength, h is the Plank constant, p is the momentum of the moving object.
Whereas, p=mv, m is the mass of the object and v is the velocity of the moving object.
Therefore, λ=h/(mv)
λ=(6.63×10⁻³⁴)/(0.56×26)
λ=4.55×10⁻³⁵ m.
The de Broglie wavelength associated with the object weight 0.56 kg moving with the velocity of 26 m/s is λ=4.55×10⁻³⁵ m.
Learn more about de Broglie wavelength on
brainly.com/question/15330461
#SPJ1
Answer:
The peak-to-peak ripple voltage = 2V
Explanation:
120V and 60 Hz is the input of an unfiltered full-wave rectifier
Peak value of output voltage = 15V
load connected = 1.0kV
dc output voltage = 14V
dc value of the output voltage of capacitor-input filter
where
V(dc value of output voltage) represent V₀
V(peak value of output voltage) represent V₁
V₀ = 1 - ( 
)V₁
make C the subject of formula
V₀/V₁ = 1 - (1 / 2fRC)
1 / 2fRC = 1 - (v₀/V₁)
C = 2fR ((1 - (v₀/V₁))⁻¹
Substitute for,
f = 240Hz , R = 1.0Ω, V₀ = 14V , V₁ = 15V
C = 2 * 240 * 1 (( 1 - (14/15))⁻¹
C = 62.2μf
The peak-to-peak ripple voltage
= (1 / fRC)V₁
= 1 / ( (120 * 1 * 62.2) )15V
= 2V
The peak-to-peak ripple voltage = 2V
