Answer:
- <em>Abbie’s acceleration is (1/2) Zak’s acceleration.</em>
Explanation
1. <u>Data</u>:
a) ω = constant
b) Abbie: r₁ = 1 m
c) Zak: r₂ = 2 m
d) Ac₁ = ? Ac₂
2. <u>Formulae</u>
3. <u>Solution</u>:
a) Abbie:
b) Zack:
c) Divide Ac₁ / Ac₂
- Ac₁ / Ac₂ = ω² (1m) / [ω² (2m) ] = 1/2
⇒ Ac₁ = (1/2) Ac₂ = Ac₂ / 2 = 0.5 Ac₂
Answer:
93 km/h
Explanation:
Given that a bus took 8 hours to travel 639 km. For the first 5 hours, it travelled at an average speed of 72 km/h
Let the first 5 hours journey distance = F
From the formula of speed,
Speed = distance/time
Substitute speed and time
72 = F/5
F = 72 × 5 = 360 km
The remaining distance will be:
639 - 360 = 279km
The remaining time will be:
8 - 5 = 3 hours
Speed = 279/3
Speed = 93 km/h
Therefore, the average speed for the remaining time of the journey is equal to 93 km/h
Answer:
x = 1.6 + 1.7 t^2 omitting signs
a) at t = 0 x = 1.6 m
b) V = d x / d t = 3.4 t
at t = 0 V = 0
c) A = d^2 x / d t^2 = 3.4 (at t = 0 A = 3.4 m/s^2)
d) x = 1.6 + 1.7 * (4.4)^2 = 34.5 (position at 4.4 sec = 34.5 m)
Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y = 
t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45
)
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² = 
- 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct
The springs stored energy is transferred to the cube as kinetic energy and then by the slop the KE is converted to height energy.
<span>0.5 . k . x^2 = 0.5 . m . v^2 = m . g . ∆h </span>
<span>0.5 . 50 . (0.1^2) = 0.05 . 9.8 . ∆h </span>
<span>∆h = 0.51 m = 51 cm </span>
<span>This is the height gained </span>
<span>Distance along the slope = ∆h / sin 60 = 0.589 = 59 cm </span>
<span>In the second case, the stored spring energy is converted into height energy AND frictional heat energy. </span>
<span>The height energy is m . g . d sin 60 where d is the distance the cube moves along the slope. </span>
<span>The Frictional energy converted is F . d </span>
<span>F ( the frictional force ) = µ . N </span>
<span>N ( the reaction to the component of the gravity force perpendicular to the surface of the slope ) = m . g . cos60 </span>
<span>Total energy converted </span>
<span>0.5 . k . x^2 = (m . g . dsin60) + (µ . m . g . cos60 . d ) </span>
<span>Solve for d </span>
<span>d = 0.528 = 53 cm</span>
