When I drive to the office, I drive through two school zones,
and four intersections that are controlled by traffic lights.
My average speed for the trip is higher than my instantaneous
speed is at any point in the school zones, or at any time when
I'm waiting for a red light to change.
Answer:
B) electrons transferred from sphere to rod.
(2) 1.248 x 10¹¹ electrons were transferred
Explanation:
Given;
initial charge on the plastic rod, q₁ = 15nC
final charge on the plastic rod, q₂ = - 5nC
let the charge acquired by the plastic rod = q
q + 15nC = -5nC
q = -5nC - 15nC
q = -20 nC
Thus, the plastic rod acquired excess negative charge from the metal sphere.
Hence, electrons transferred from sphere to rod
B) electrons transferred from sphere to rod.
2) How many charged particles were transferred?
1.602 x 10⁻¹⁹ C = 1 electron
20 x 10⁻⁹ C = ?
= 1.248 x 10¹¹ electrons
Thus,1.248 x 10¹¹ electrons were transferred
The final velocity of the red barge in the collision elastic is 0.311 m/s when it collides with blue barge pf mass 1000000 kg.
Final velocity(v3) of the red barge is calculated by following formula
m1×v1+ m2×v2= (m1+m2)v3
Substituting the value of m1= 150000 kg, v1= 0.25 m/s, m2= 1000000 kg, v2= 0.32 m/s
150000 × 0.25+ 1000000×0.32= (150000+1000000)×v3
37500+ 320000= 1150000×v3
357500= 1150000×v3
v3= 0.311 m/s
<h3>What is elastic collision velocity? </h3>
- The velocity of the target particle after a head-on elastic impact in which the projectile is significantly more massive than the target will be roughly double that of the projectile, but the projectile velocity will remain virtually unaltered.
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Since g is constant, the force the escaping gas exerts on the rocket will be 10.4 N
<h3>
What is Escape Velocity ?</h3>
This is the minimum velocity required for an object to just escape the gravitational influence of an astronomical body.
Given that the velocity of a 0.25kg model rocket changes from 15m/s [up] to 40m/s [up] in 0.60s. The gravitational field intensity is 9.8N/kg.
To calculate the force the escaping gas exerts of the rocket, let first highlight all the given parameters
- Mass (m) of the rocket 0.25 Kg
- Initial velocity u = 15 m/s
- Final Velocity v = 40 m/s
- Gravitational field intensity g = 9.8N/kg
The force the gas exerts of the rocket = The force on the rocket
The rate change in momentum of the rocket = force applied
F = ma
F = m(v - u)/t
F = 0.25 x (40 - 15)/0.6
F = 0.25 x 41.667
F = 10.42 N
Since g is constant, the force the escaping gas exerts on the rocket is therefore 10.4 N approximately.
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