Answer:
diffraction
Explanation:
diffraction occurs when light passes sharp edges or goes through narrow slits the rays are deflected and produce fringes of light and dark bands
Answer:
so angular velocity is 7.13128 sec−1
Explanation:
velocity v = 2.2 m/s
displacement s = 220 mm = 0.220 m
distance d = 510 mm = 0.510 m
to find out
angular velocity
solution
we know that
angular velocity will be velocity ( v) / (displacement² + distance²) .....1
now put all these value in equation 1 and we get angular velocity i.e.
angular velocity = velocity ( v) / (displacement² + distance²)
angular velocity = 2.2 / (0.22² + 0.51²)
angular velocity = 2.2 / 0.3085
angular velocity = 7.13128
so angular velocity is 7.13128 sec−1
By definition,
Momentum = Mass * Velocity
Let v = the velocity of the truck, m/s
The mass of the truck is 36,287 kg.
The momentum is 907,175 (kg-m)/s.
Therefore
907,175 (kg-m)/s = (36287 kg)*(v m/s)
v = 907175/36287 = 25 m/s
Answer: 25 m/s
Carla draws two circuit diagrams that connect the same components in different ways, as shown. The correct statement is "the total resistance in circuit a is greater than that in-circuit b.". Option A. This is further explained below.
<h3>What is a
circuit?</h3>
Generally, a circuit is simply defined as a fixed schedule of activities or places for a certain activity, usually including athletics or public performance.
In conclusion, Carla creates two circuit schematics, as illustrated, that link the same components in various ways. "The overall resistance in circuit an is larger than that in b," is the correct statement.
Read more about resistance
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Answer:
μk = 0.26885
Explanation:
Conceptual analysis
We apply Newton's second law:
∑Fx = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Data:
a= -0.9 m/s²,
g = 9.81 m/s² : acceleration due to gravity
W= 75 N : Block weight
W= m*g
m = W/g = 75/9.8= 7.65 kg : Block mass
Friction force : Ff
Ff= μk*N
μk: coefficient of kinetic friction
N : Normal force (N)
Problem development
We apply the formula (1)
∑Fy = m*ay , ay=0
N-W-25 = 0
N = 75
+25
N= 100N
∑Fx = m*ax
20-Ff= m*ax
20-μk*100
= 7.65*(-0.90 )
20+7.65*(0.90) = μk*100
μk = ( 20+7.65*(0.90)) / (100)
μk = 0.26885
