A change in velocity or a change in direction, or both! Hope this helps
Answer:
The answer is "Option 5".
Explanation:
Jadeen claims that only by hitting the wire to make it thinner and wider, it can improve a copper wire's strength, which is why a copper wire's permeability doesn't quite improve and can reduce once it is pounded. Arnell says so by heating its wire, it can improve its strength, and when it is heated, the wire's permeability reduces.
<span>Answer:
KE = (11/2)mω²r²,
particle B must have mass of 2m, while A has mass m.
Then the moment of inertia of the system is
I = Σ md² = m*(3r)² + 2m*r² = 11mr²
and then
KE = ½Iω² = ½ * 11mr² * ω² = 11mr²ω² / 2
So I'll proceed under that assumption.
For particle A, translational KEa = ½mv²
but v = ω*d = ω*3r, so KEa = ½m(3ωr)² = (9/2)mω²r²
For particld B, translational KEb = ½(2m)v²
but v = ω*r, so KEb = ½(2m)ω²r²
so total translational KE = (9/2 + 2/2)mω²r² = 11mω²r² / 2
which is equal to our rotational KE.</span>
Answer:
107 m down the incline
Explanation:
Given:
v₀ₓ = 25 m/s
v₀ᵧ = 0 m/s
aₓ = 0 m/s²
aᵧ = -10 m/s²
-Δy/Δx = tan 35°
Find: d
First, find Δy and Δx in terms of t.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (0 m/s) t + ½ (-10 m/s²) t²
Δy = -5t²
Δx = v₀ₓ t + ½ aₓ t²
Δx = (25 m/s) t + ½ (0 m/s²) t²
Δx = 25t
Substitute:
-(-5t²) / (25t) = tan 35°
t/5 = tan 35°
t = 5 tan 35°
t ≈ 3.50 s
Now find Δy and Δx.
Δy ≈ -61.3 m
Δx ≈ 87.5 m
Therefore, the distance down the incline is:
d = √(x² + y²)
d ≈ 107 m
