Answer:
x = 11.23 m
Explanation:
For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.
Let's reduce to SI system units
θ = 155 rev (2pi rad / rev) = 310π rad
α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²
Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)
w² = w₀² + 2 α θ
w =√ 2 α θ
w = √(2 4pi 310pi)
w = 156.45 rad / s
The relationship between angular and linear velocity
v = w r
v = 156.45 0.175
v = 27.38 m / s
In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive
y = 
t - ½ g t²
As it leaves the highest point its speed is horizontal
y = 0 - ½ g t²
t = √ (-2y / g)
t = √ (-2 (-0.820) /9.8)
t = 0.41 s
With this time we calculate the horizontal distance, because the constant horizontal speed
x = vox t
x = 27.38 0.41
x = 11.23 m
The correct answer for the question that is being presented above is this one: "a. only from an instructor or supervisor." Ideally, rewards should be given immediately and frequently but <span>only from an instructor or supervisor to show authority. </span>
The magnetic field or force seems to be associated with the lineup of electrons withim the magnet
Answer:
Explanation:
There will not be any internal reflection . it will be only refraction
critical angle = θ
Sinθ = 1 / μg
μg = 1.43 / 1.33 =
Sinθ = 1.33 / 1.43
= .93
θ = 68.44
angle of incidence i = 68.44 / 2
= 34.22
Sin i / Sin r = μw = 1.33 / 1.43
= .93
sin 34.22 / sinθ₁ = .93 , θ₁ is angle of refraction.
sinθ₁ = sin 34.22 / .93
= .5623 / .93
= .6047
θ₁ = 37 degree Ans
Answer:
The change in kinetic energy (KE) of the car is more in the second case.
Explanation:
Let the mass of the car = m
initial velocity of the first case, u = 22 km/h = 6.11 m/s
final velocity of the first case, v = 32 km/h = 8.89 m/s
change in kinetic energy (K.E) = ¹/₂m(v² - u²)
ΔK.E = ¹/₂m(8.89² - 6.11²)
= 20.85m J
initial velocity of the second case, u = 32 km/h = 8.89 m/s
final velocity of the second case, v = 42 km/h = 11.67 m/s
change in kinetic energy (K.E) = ¹/₂m(v² - u²)
ΔK.E = ¹/₂m(11.67² - 8.89²)
= 28.58m J
The change in kinetic energy (KE) of the car is more in the second case.
