Response:R=1607556m
θ=180degrees
Clarification:
d1=74.8m
d2=160.7km=160.7km*1000
d2=160700m
d3=80m
d4=198.1m
Utilizing an analytical approach:
Rx=-(160700+75*cos(41.8))= -160755.9m
Ry= -(74.8+75sin(41.8))-198.1=73m
Magnitude, R:
R=√Rx+Ry
R=√160755.9^2+20^2=160755.916
R=160756m
Direction,θ:
θ=arctan(Rx/Ry)
θ=arctan(-73/160755.9)
θ=-7.9256*10^-6
It is worth noting that since θ is in the second quadrant, 180 is added
θ=180-7.9256*10^6=180degrees
Answer:
Explanation:
The equation used to determine the maximum height of the bowling pin during its trajectory is given by;
H = u²/2g
where u, the initial speed/velocity, equals 10m/s
g stands for gravitational acceleration = 9.81m/s²
Substituting in the values gives us
H = 10²/2(9.81)
H = 100/19.62
Consequently, the highest point of the bowling pin's center of mass is approximately 5.0m.
Answer:
Explanation:
The position of the charge q₁ is established at (0,0)
Meanwhile, the charge q₂ is located at (x₁,0)
Thus, the electric potential energy between these two charges is determined by:
Now, the location of charge q₂ shifts from (x₁,0) to (x₂,y₂). The updated electric potential energy between the charges can be represented as:
According to the work-energy theorem, the alteration in potential energy corresponds to the work performed. This is expressed mathematically as:
Consequently, the work done by the electrostatic force on the moving charge is 
. Therefore, this concludes the solution.
Answer:
1.5 m/s²
Explanation:
Begin by sketching a free body diagram. Three forces are at play on the sea lion: the force of gravity acting downwards, the normal force that is perpendicular to the ramp, and the frictional force parallel to the ramp.
Considering the forces perpendicular to the incline:
∑F = ma
N − mg cos θ = 0
This gives us N = mg cos θ
Next, examining the forces parallel to the incline:
∑F = ma
mg sin θ − Nμ = ma
Substituting for N yields:
mg sin θ − (mg cos θ) μ = ma
g sin θ − g cos θ μ = a
hence a = g (sin θ − μ cos θ)
If we set θ = 23° and μ = 0.26:
a = 9.8 (sin 23 − 0.26 cos 23)
this results in a = 1.48
When rounded to two significant figures, the acceleration of the sea lion is 1.5 m/s².
