Answer:
Explanation:
Transformation of Energy
Also known as energy conversion, this refers to the process in which energy shifts from one type to another. In this context, three energy forms are involved. When the object is stationary at the ramp's peak, it possesses gravitational potential energy, calculated as
As the object descends the frictionless ramp, it converts all its potential energy into kinetic energy, represented as
Thus,
Ultimately, when the object encounters a rough surface, all energy converts to thermal energy. The work performed by the friction force corresponds to the alteration in kinetic energy, as all velocity is lost in this process:
Given the kinetic energy equals the initial potential energy:
The negative sign indicates that the work acted against the direction of movement, meaning the force and displacement are 180° apart.
This outcome is independent of the distance D needed to halt the block or the kinetic friction coefficient.
B) 14.0 N
To address this inquiry, we need to evaluate the kinetic energy of the box before and after crossing the rough section. The kinetic energy is given by the formula:
E = 0.5 M V^2
where
E = Energy
M = Mass
V = velocity
Now, utilizing the known data, we compute the energy prior and post.
Before:
E = 0.5 M V^2
E = 0.5 * 13.5kg * (2.25 m/s)^2
E = 6.75 kg * 5.0625 m^2/s^2
E = 34.17188 kg*m^2/s^2 = 34.17188 joules
After:
E = 0.5 M V^2
E = 0.5 * 13.5kg * (1.2 m/s)^2
E = 6.75 kg * 1.44 m^2/s^2
E = 9.72 kg*m^2/s^2 = 9.72 Joules
Hence, the box consumed energy equal to 34.17188 J - 9.72 J = 24.451875 J over a length of 1.75 meters. Next, we will calculate the loss per meter by dividing the energy loss by the distance traversed.
24.451875 J / 1.75 m = 13.9725 J/m = 13.9725 N
When we round to one decimal point, we arrive at 14.0 N, which corresponds with option “B.”
a) 3.56 x 10^22 N. b) 3.56 x 10^22 N. The sun’s mass is M = 2 x 10^30 kg, while the Earth's mass is m = 6 x 10^24 kg, with a distance of R = 1.5 x 10^11 m separating them. Applying Newton's law for gravitational force F = G (mM / R²), where G = 6.67 × 10^-11 m^3 kg^-1 s^-2 gives us F = 3.56 x 10^22 N. A) The gravitational force by the sun on Earth equates to the force exerted by Earth on the sun, which is also 3.56 x 10^22 N.
Answer:
3.5 cm
Explanation:
mass, m = 50 kg
diameter = 1 mm
radius, r = half the diameter = 0.5 mm = 0.5 x 10^-3 m
L = 11.2 m
Y = 2 x 10^11 Pa
Cross-sectional area of the wire = π r² = 3.14 x 0.5 x 10^-3 x 0.5 x 10^-3
= 7.85 x 10^-7 m^2
Let the change in length of the wire be ΔL.
The equation for Young's modulus is given by
ΔL = 0.035 m = 3.5 cm
Thus, the wire stretches by 3.5 cm.
Answer:
a) Ф = 0.016 N / C m, b) q_{int} = 0.14 10⁻¹² C
Explanation:
a) For this scenario, we rely on Gauss's law
Ф = E.ds = 
/ε₀
As the field points in the x direction, there is no flux through the cylinder walls.
Ф = E A
The area of a circle is
A = π r
Ф = E π r
Ф = (x- 3.6) r
Now, let's compute
Ф = (3.7 -3.6) 0.16
Ф = 0.016 N / C m
b) Using Gauss's law, we have
q_{int} = Ф ε₀
Where the flow is present on both sides, at the face corresponding to x = 0, the flow is zero
q_{int} = 0.016 8.85 10⁻¹²
q_{int} = 0.14 10⁻¹² C
