Answer:
v = 8.09 m/s
Explanation:
For this exercise we use that the work done by the friction force plus the potential energy equals the change in the body's energy.
Let's calculate the energy
starting point. Higher
Em₀ = U = m gh
final point. To go down the slope
Em_f = K = ½ m v²
The work of the friction force is
W = fr L cos 180
to find the friction force let's use Newton's second law
Axis y
N - W_y = 0
N = W_y
X axis
Wₓ - fr = ma
let's use trigonometry
sin θ = y / L
sin θ = 11/110 = 0.1
θ = sin⁻¹ 0.1
θ = 5.74º
sin 5.74 = Wₓ / W
cos 5.74 = W_y / W
Wₓ = W sin 5.74
W_y = W cos 5.74
the formula for the friction force is
fr = μ N
fr = μ W cos θ
Work is friction force is
W_fr = - μ W L cos θ
Let's use the relationship of work with energy
W + ΔU = ΔK
-μ mg L cos 5.74 + (mgh - 0) = 0 - ½ m v²
v² = - 2 μ g L cos 5.74 +2 (gh)
v² = 2gh - 2 μ gL cos 5.74
let's calculate
v² = 2 9.8 11 - 2 0.07 9.8 110 cos 5.74
v² = 215.6 -150.16
v = √65.44
v = 8.09 m/s
Answer:
nm
Explanation:
= Index of refraction of soap bubble = 1.33
= thickness of the soap bubble = 115 nm = 115 x 10⁻⁹ m
= wavelength of light = ?
= order = 0
For reflection , the necessary condition is
nm
The answer is C. F=ma basically says that force is a function of mass multiplied by acceleration. The first two answers don’t make sense because there’s no necessary relationship between mass and acceleration. And for the last two, the higher the mass, the higher the force needed, therefore C is the correct answer.
Answer:
The value of gauge pressure at outlet = -38557.224 pascal
Explanation:
Apply Bernoulli' s Equation
+ + = + + --------------(1)
Where
= Gauge pressure at inlet = 3.70105 pascal
= velocity at inlet = 2.4
= Gauge pressure at outlet = we have to calculate
= velocity at outlet = 3.5
= 3.6 m
Put all the values in equation (1) we get,
⇒ + = + + 3.6
⇒ 0.294 = + 0.6244 + 3.6
⇒ = 0.294 - 0.6244 - 3.6
⇒ = - 3.9304
⇒ = - 38557.224 pascal
This is the value of gauge pressure at outlet.