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.
Answer:
The partial pressure of H2 is 0.375 atm.
The partial pressure of Ne also stands at 0.375 atm.
Explanation:
Mass of H2 = 1 g
Mass of Ne = 1 g
Mass of Ar = 1 g
Mass of Kr = 1 g
Overall mass of the gas mixture totals 4 g.
Pressure in the sealed container is 1.5 atm.
Calculating the partial pressure for H2 yields: (mass of H2/total mass of gas mixture) × pressure of sealed container = 1/4 × 1.5 = 0.375 atm.
Calculating the partial pressure for Ne similarly gives: (mass of Ne/total mass of gas mixture) × pressure of sealed container = 1/4 × 1.5 = 0.375 atm.
Answer:
2.45 m
Explanation:
To begin, we need to determine the book's time of flight, utilizing the equation for vertical motion:
with
h = 1.19 m representing the height traveled by the book
g = 9.8 m/s^2 being the gravitational acceleration
t symbolizing the time of flight
By solving for t,
Next, we calculate the horizontal distance the book travels, defined by
where
denotes the horizontal velocity
t = 0.49 s being the time of flight
By substituting,
Thus, the book lands at a distance of 2.45 m.
Answer:
= 3289.8 m/s
Explanation:
This problem can be approached using momentum definitions.
I = ∫ F dt
We substitute and compute.
I = ∫ (at - bt²) dt
Integrating gives us:
I = a t² / 2 - b t³ / 3
We will evaluate between the limits I=0 for t = 0 ms and higher I=I for t = 2.74 ms:
I = a (2.74² / 2- 0) - b (2.74³ / 3 -0)
I = a 3.754 - b 6.857
Substituting the values for a and b, we find:
I = 1500 3.754 - 20 6.857
I = 5,631 - 137.14
I = 5493.9 N s
Next, we engage the relationship between impulse and momentum:
I = Δp = m - m v₀o
I = m - 0
= I / m
= 5493.9 /1.67
= 3289.8 m/s
Answer:
-utilize precisely the same apparatus
-maintain identical measures (release height)
Explanation:
