A parallel circuit is sometimes called a current divider because current splits up among all the resistors in the parallel circuit. In addition, the current through the branches is inversely proportional to the resistance of the branch. If the resistance in each branch is kept constant but the voltage is decreased, the current will decrease.
<h2><em>So there is two truths given. After an amount of time Ttotal (lets call it ‘t’):
</em></h2><h2><em>
</em></h2><h2><em>The car’s speed is 25m/s
</em></h2><h2><em>The distance travelled is 75m
</em></h2><h2><em>Then we have the formulas for speed and distance:
</em></h2><h2><em>
</em></h2><h2><em>v = a x t -> 25 = a x t
</em></h2><h2><em>s = 0.5 x a x t^2 -> 75 = 0.5 x a x t^2
</em></h2><h2><em>Now, we know that both acceleration and time equal for both truths. So we can say:
</em></h2><h2><em>
</em></h2><h2><em>t = 25 / a
</em></h2><h2><em>t^2 = 75 / (0.5 x a) = 150 / a
</em></h2><h2><em>Since we don’t want to use square root at 2) we go squared for 1):
</em></h2><h2><em>
</em></h2><h2><em>t^2 = (25 / a) ^2 = 625 / a^2
</em></h2><h2><em>t^2 = 150 / a
</em></h2><h2><em>Since t has the same value for both truths we can say:
</em></h2><h2><em>
</em></h2><h2><em>625 / a^2 = 150 / a
</em></h2><h2><em>
</em></h2><h2><em>Thus multiply both sides with a^2:
</em></h2><h2><em>
</em></h2><h2><em>625 = 150 x a, so a = 625 / 150 = 4.17
</em></h2><h2><em>
</em></h2><h2><em>We can now calculate t as well t = 25 * 150 / 625 = 6</em></h2>
Answer:
They are...if I'm correct Chemically combined, sorry if I'm wrong.
Explanation:
Acceleration is the change in speed over change in time.
a = Δv / Δt
a. The car's acceleration is:
a = (80 km/h − 0 km/h) / 10 s
a = 8 km/h/s
So every second, the speed increases by 8 km/h.
b. The cyclist's acceleration is:
a = (16 m/s − 4.0 m/s) / 5.6 s
a = 2.1 m/s²
c. The stone's speed is:
10.0 m/s² = (v − 0 m/s) / 3.5 s
v = 35 m/s
d. The time is:
1.6 m/s² = (10 m/s − 0 m/s) / t
t = 6.3 s
D = distance between the cars at the start of time = 680 km
v₁ = speed of one car
v₂ = speed of other car = v₁ - 10
t = time taken to meet = 4 h
distance traveled by one car in time "t" + distance traveled by other car in time "t" = D
v₁ t + v₂ t = D
(v₁ + v₂) t = D
inserting the values
(v₁ + v₁ - 10) (4) = 680
v₁ = 90 km/h
rate of slower car is given as
v₂ = v₁ - 10
v₂ = 90 - 10 = 80 km/h
