Answer:
See below
Explanation:
I will assume the force is in a DOWNWARD direction ( I believe it makes no answer difference)
Horizontal component is then 100 cos 36° =80.9 N
F = ma
80.9 = 25 kg *a
<u>a = 3.24 m/s^2 </u>
Answer:
The angle of refraction is option b: 17°.
Explanation:
We can find the angle of refraction by using Snell's law:

Where:
n₁: is the index of refraction of the medium 1 (air) = 1.0003
n₂: is the index of refraction of the medium 2 (diamond) = 2.42
θ₁: is the angle of incidence = 45°
θ₂: is the angle of refraction =?
Hence, the angle of refraction is:


Therefore, the correct option is b: 17°.
I hope it helps you!
Answer:
ω = ω₀ + α t
ω² = ω₀² + 2 α θ
θ = θ₀ + ω₀ t + ½ α t²
Explanation:
Rotational kinematics can be treated as equivalent to linear kinematics, for this change the displacement will change to the angular displacement, the velocity to the angular velocity and the acceleration to the angular relation, that is
x → θ
v → ω
a → α
with these changes the three linear kinematics relations change to
ω = ω₀ + α t
ω² = ω₀² + 2 α θ
θ = θ₀ + ω₀ t + ½ α t²
where it should be clarified that to use these equations the angles must be measured in radians
Answer:
3.34×10^-6m
Explanation:
The shear modulus can also be regarded as the rigidity. It is the ratio of shear stress and shear strain
can be expressed as
shear stress/(shear strain)
= (F/A)/(Lo/ . Δx)
Stress=Force/Area
The sheear stress can be expressed below as
F Lo /(A *Δx)
Where A=area of the disk= πd^2/4
F=shearing force force= 600N
Δx= distance
S= shear modulus= 1 x 109 N/m2
Lo= Lenght of the cylinder= 0.700 cm=7×10^-2m
If we make Δx subject of the formula we have
Δx= FLo/(SA)
If we substitute the Area A we have
Δx= FLo/[S(πd^2/4]
Δx=4FLo/(πd^2 *S)
If we input the values we have
(4×600×0.7×10^-2)/10^9 × 3.14 ×(4×10^-2)^2
= 3.35×10^-6m
Therefore, its shear deformation is 3.35×10^-6m
A=area of the disk= πd^2/4
= [3.142×(4×10^-2)^2]/4
Since we have , v=f×lambda (wavelength). Where v equals 350m/s and wavelength equals 3.80. so it will become f = v/lambda=350/3.80=92.1052Hz