Answer:
θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²
Explanation:
This is an angular kinematic exercise the equation for the angular position
the particle A
θ = θ₀ + ω₀ t + ½ α t²
They say for the particle B
w₀B = ½ w₀
αB = 2 α
In addition, the particle begins at a time t_1 after particle A, in order to use the same timer, we must subtract this time from the initial
t´ = t - t_1
l
et's write the equation of particle B
θ = θ₀ + w₀B t´ + ½ αB t´2
replace
θ = θ₀ + ½ w₀ (t -t_1) + ½ 2α (t -t_1)²
θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²
Answer:
b. 0.6m/s, 0.7m/s, 0.61m/s, 0.62m/s
Explanation:
Precision of a measurement is the closeness of the experimental values to one another. Hence, experimental measurements are said to be precise if they are close to each other irrespective of how close they are to the accepted value. Precision can be determined by finding the range of each experimental value. The measurement with the LOWEST RANGE represents the MOST PRECISE.
Note: Range is the highest value - lowest value
Set A: 1.5 - 0.8 = 0.7
Set B: 0.7 - 0.6 = 0.1
Set C: 2.4 - 2.0 = 0.4
Set D: 3.1 - 2.9 = 0.2
Set B has the lowest range (0.1), hence, represent the most precise value.
Answer:
d = 4 d₀o
Explanation:
We can solve this exercise using the relationship between work and the variation of kinetic energy
W = ΔK
In that case as the car stops v_f = 0
the work is
W = -fr d
we substitute
- fr d₀ = 0 - ½ m v₀²
d₀ = ½ m v₀² / fr
now they indicate that the vehicle is coming at twice the speed
v = 2 v₀
using the same expressions we find
d = ½ m (2v₀)² / fr
d = 4 (½ m v₀² / fr)
d = 4 d₀o
Answer:
the displacement of the object is 5 units
Explanation:
The computation of the displacement of the object is shown below:
= Move to the right + move to the right - move to the left
= 6 units + 3 units - 4 units
= 9 units - 4 units
= 5 units
Hence, the displacement of the object is 5 units
Answer:
D. O2 is removed from the atmosphere, and CO2 is released
Explanation:
Hope it helps you
