Answer:
1.53 s
Explanation:
Initially vertical component of velocity of the ball, uy = 7.5 m/s
Net displacement is vertical direction is zero, Δy =0
Use second equation of motion:
Δy = uy t + 0.5 a t²
Here, acceleration a = -g (g =9.8 m/s²)
Substitute all the values and solve for g
0 = 7.5 t -0.5 (9.8)t²
7.5 t = 4.9 t²
t = 1.53 s
The current flowing in each resistor of the circuit is 4 A.
<h3>
Equivalent resistance of the series resistors</h3>
The equivalent resistance of the series circuit is calculated as follows;
6 Ω and 4 Ω are in series = 10 Ω
5 Ω and 10Ω are in series = 15 Ω
<h3>Effective resistance of the circuit</h3>
<h3>Current flowing in the circuit</h3>
V = IR
I = V/R
I = 24/6
I = 4 A
Learn more about resistors in parallel here: brainly.com/question/15121871
We have to add two vectors.
Vector #1: 0.15 m/s north
Vector #2: 1.50 m/s east
Their sum:
Magnitude: √(0.15² + 1.50²)
Magnitude = √(0.0225+2.25)
Magnitude = √2.2725
Magnitude = <em>1.5075 m/s</em>
Direction = arctan(0.15/1.50) north of east
Direction = <em>5.71° north of east</em>
"v0" means that there are no friction forces at that speed
<span>mgsinΘ = (mv0²/r)cosΘ → the variable m cancels </span>
<span>sinΘ/cosΘ = tanΘ = v0² / gr
</span><span>Θ = arctan(v0² / gr) </span>
<span>When v > v0, friction points downslope: </span>
<span>mgsinΘ + µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ + µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = ((v²/r)cosΘ - gsinΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
<span>When v > v0, friction points upslope: </span>
<span>mgsinΘ - µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ - µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = (gsinΘ - (v²/r)cosΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
