The solution for this problem is through this formula:Ø = w1 t + 1/2 ã t^2
where:Ø - angular displacement w1 - initial angular velocity t - time ã - angular acceleration
128 = w1 x 4 + ½ x 4.5 x 5^2 128 = 4w1 + 56.254w1 = -128 + 56.25 4w1 = 71.75w1 = 71.75/4
w1 = 17.94 or 18 rad s^-1
w1 = wo + ãt
w1 - final angular velocity
wo - initial angular velocity
18 = 0 + 4.5t t = 4 s
Answer/Explanation:
The weight of an object is defined as the force that is exerted due to the gravitational force.
Mathematically, it can be written as :
W = m g
Where
m is the mass of the object
g is the acceleration due to gravity
Also,
We know that the value of g varies with respect to the location. At the equator, the value of g is less as compared to the poles.
The feature of an object that affects its weight are :
Mass of the object
Location of the object
How much force Earth exerts on the object
Power is defined as
P = I*V
where I is the current and V is the voltage
Ohm's law gives us the relation betwen Voltage and current in a resistive component
V = I*R , Then
P = V² / R
We solve for R,
R = (110 V)²/ 75W = 161.33 ohms
Answer:
a.auditory ossicles
b.oval window
c.Round window
d.tympanic membrane
Answer is tympanic membrane
Explanation:
The tympanic membrane otherwise called the ear drum is a membrane shaped like a cone,it connects the outside to the inner ear,it serves to convert vibration from air into fluid membrane vibration a good example of mechanical waves for onwatd transmission into the cochlea of the inner ear through the oval window
Answer:
= 6.55cm
Explanation:
Given that,
distance = 1.26 m
distance between two fourth-order maxima = 53.6 cm
distance between central bright fringe and fourth order maxima
y = Y / 2
= 53.6cm / 2
= 26.8 cm
=0.268 m
tan θ = y / d
= 0.268 m / 1.26 m
= 0.2127
θ = 12°
4th maxima
d sinθ = 4λ
d / λ = 4 / sinθ
d / λ = 4 / sin 12°
d / λ = 19.239
for first (minimum)
d sinθ = λ / 2
sinθ = λ / 2d
= 1 / 2(19.239)
= 1 / 38.478
= 0.02599
θ = 1.489°
tan θ = y / d
y = d tan θ
= 1.26 tan 1.489°
= 0.03275
the total width of the central bright fringe
Y = 2y
= 2(0.03275)
= 0.0655m
= 6.55cm
