Answer:
Explanation:
q₁ = 8 x 10⁻⁶ C , q₂ = 3.5 x 10⁻⁶ C , q₃ = -2.5 x 10⁻⁶ C
Force between charge = 9 x 10⁹ x q₁ q₂ / d²
Force on q₁ due to q₂
= 9 x 10⁹ x 8 x 10⁻⁶ x 3.5 x 10⁻⁶ /( .10 )²
= 25.2 N . This force will be negative as it will point towards left ( q₂ will repel q₁ )
Force on q₁ due to q₃
= 9 x 10⁹ x 8 x 10⁻⁶ x 2.5 x 10⁻⁶ /( .15 )²
= 8 N . This force will be negative as it will point towards left ( q₃ will attract q₁ )
Net force = 25.2 N - 8 N
= 17.2 N .
Answer: option 1. thermal expansion and contraction.
The thermostat uses the thermal expansion and contraction of metals: when the metal expands it acts as a switch, a coiled ribbon expands and the electric circuit is off, then the current stops. When the temperatures decrease, the coiled ribbon contracts and the circuit is on, then the current starts to flow and the resistance starts to radiate heat.
Answer:
442.5 rad
Explanation:
w₀ = initial angular velocity of the disk = 7.0 rad/s
α = Constant angular acceleration = 3.0 rad/s²
t = time period of rotation of the disk = 15 s
θ = angular displacement of the point on the rim
Angular displacement of the point on the rim is given as
θ = w₀ t + (0.5) α t²
inserting the values
θ = (7.0) (15) + (0.5) (3.0) (15)²
θ = 442.5 rad
Answer:
a) y= 3.5 10³ m, b) t = 64 s
Explanation:
a) For this exercise we use the vertical launch kinematics equation
Stage 1
y₁ = y₀ + v₀ t + ½ a t²
y₁ = 0 + 0 + ½ a₁ t²
Let's calculate
y₁ = ½ 16 10²
y₁ = 800 m
At the end of this stage it has a speed
v₁ = vo + a₁ t₁
v₁ = 0 + 16 10
v₁ = 160 m / s
Stage 2
y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²
y₂ = 800 + 150 5 + ½ 11 5²
y₂ = 1092.5 m
Speed is
v₂ = v₁ + a₂ t
v₂ = 160 + 11 5
v₂ = 215 m / s
The rocket continues to follow until the speed reaches zero (v₃ = 0)
v₃² = v₂² - 2 g y₃
0 = v₂² - 2g y₃
y₃ = v₂² / 2g
y₃ = 215²/2 9.8
y₃ = 2358.4 m
The total height is
y = y₃ + y₂
y = 2358.4 + 1092.5
y = 3450.9 m
y= 3.5 10³ m
b) Flight time is the time to go up plus the time to go down
Let's look for the time of stage 3
v₃ = v₂ - g t₃
v₃ = 0
t₃ = v₂ / g
t₃ = 215 / 9.8
t₃ = 21.94 s
The time to climb is
= t₁ + t₂ + t₃
t_{s} = 10+ 5+ 21.94
t_{s} = 36.94 s
The time to descend from the maximum height is
y = v₀ t - ½ g t²
When it starts to slow down it's zero
y = - ½ g t_{b}²
t_{b} = √-2y / g
t_{b} = √(- 2 (-3450.9) /9.8)
t_{b} = 26.54 s
Flight time is the rise time plus the descent date
t = t_{s} + t_{b}
t = 36.94 + 26.54
t =63.84 s
t = 64 s