Answer:
Inertia is the property of mass that resists change. Therefore, it is safe to say that as the mass of an object increases so does its inertia.
Explanation:
Friction- the external force that acts on objects and causes them to slow down when no other external force acts upon them.
Answer:
D = 2.38 m
Explanation:
This exercise is a diffraction problem where we must be able to separate the license plate numbers, so we must use a criterion to know when two light sources are separated, let's use the Rayleigh criterion, according to this criterion two light sources are separated if The maximum diffraction of a point coincides with the first minimum of the second point, so we can use the diffraction equation for a slit
a sin θ = m λ
Where the first minimum occurs for m = 1, as in these experiments the angle is very small, we can approximate the sine to the angle
θ = λ / a
Also when we use a circular aperture instead of slits, we must use polar coordinates, which introduce a numerical constant
θ = 1.22 λ / D
Where D is the circular tightness
Let's apply this equation to our case
D = 1.22 λ / θ
To calculate the angles let's use trigonometry
tan θ = y / x
θ = tan⁻¹ y / x
θ = tan⁻¹ (4.30 10⁻² / 140 10³)
θ = tan⁻¹ (3.07 10⁻⁷)
θ = 3.07 10⁻⁷ rad
Let's calculate
D = 1.22 600 10⁻⁹ / 3.07 10⁻⁷
D = 2.38 m
Answer: 0.0146m
Explanation: The formula that defines the velocity of a simple harmonic motion is given as
v = ω√A² - x²
Where v = linear velocity, A = amplitude = 1.69cm = 0.0169m, x = displacement.
The maximum speed of a simple harmonic motion is derived when x = A, hence v = ωA
One half of maximum speed = speed of motion
3ωA/2 = ω√A² - x²
ω cancels out on both sides of the equation, hence we have that
A/2 = √A² - x²
(0.0169)/2 = √(0.0169² - x²)
0.00845 = √(0.0169² - x²)
By squaring both sides, we have that
0.00845² = 0.0169² - x²
x² = 0.0169² - 0.00845²
x² = 0.0002142
x = √0.0002142
x = 0.0146m
Answer:
Explanation:
a ) Let let the frictional force needed be F
Work done by frictional force = kinetic energy of car
F x 107 = 1/2 x 1400 x 35²
F = 8014 N
b )
maximum possible static friction
= μ mg
where μ is coefficient of static friction
= .5 x 1400 x 9.8
= 6860 N
c )
work done by friction for μ = .4
= .4 x 1400 x 9.8 x 107
= 587216 J
Initial Kinetic energy
= .5 x 1400 x 35 x 35
= 857500 J
Kinetic energy at the at of collision
= 857500 - 587216
= 270284 J
So , if v be the velocity at the time of collision
1/2 mv² = 270284
v = 19.65 m /s
d ) centripetal force required
= mv₀² / d which will be provided by frictional force
= (1400 x 35 x 35) / 107
= 16028 N
Maximum frictional force possible
= μmg
= .5 x 1400 x 9.8
= 6860 N
So this is not possible.