0N. The net force acting on this firework is 0.
The key to solve this problem is using the net force formula based on the diagram shown in the image. Fnet = F1 + F2.....Fn.
Based on the free-body diagram, we have:
The force of gases is Fgases = 9,452N
The force of the rocket Frocket = -9452
Then, the net force acting is:
Fnet = Fgases + Frocket
Fnet = 9,452N - 9,452N = 0N
In this problem we have the electric field intensity E:
E = 6.5 × 
newtons/coulomb
We have the magnitude of the load:
q = 6.4 × 
coulombs
We also have the distance d that the load moved in a direction parallel to the field 1.2 × 
meters.
We know that the electric potential energy (PE) is:
PE = qEd
So:
PE = (6.4 × 
)(6.5 × )(1.2 × )
PE = 5.0 x 
joules
None of the options shown is correct.
Answer:
k = 4422.35 KN/m
Explanation:
Given that
Frequency ,f= 29 Hz
m = 7.5 g
Natural frequency ω
ω = 2 π f
We also know that for spring mass system
ω ² m =k
k=Spring constant
So we can say that
( 2 π f)² = m k
By putting the values
(2 x π x 29)² = 7.5 x 10⁻³ k
33167.69 = 7.5 x 10⁻³ k
k=4422.35 x 10³ N/m
k = 4422.35 KN/m
Therefore spring constant will be 4422.35 KN/m
Yeah sure someone else in answering rn
Answer:
The law of conservation of energy
Explanation:
