Answer:
Explanation:
We define the linear charge density as:
Where L is the length of the rod, in this scenario the semicircle's length is L = πr
The potential at the center created by a differential element of charge is:
where k denotes Coulomb's constant
r signifies the distance from dq to the center of the circle
Thus.
The potential at the semicircle's center
Conclusion:
The total net force acting on the objects is 16 N, directed towards the right.
Clarification:
It is stated that,
The force exerted by the dog, (to the right)
The force exerted by Simone, (backward)
Here, assume the backward direction is negative and the right direction is positive.
The net force will move in the direction where the larger force is present. The net force can be calculated as:
F = 16 N
Thus, the net force amounts to 16 N, acting towards the right.
Answer:
Explanation:
a )
Each blade resembles a rod with its axis positioned near one end.
The moment of inertia for one blade is:
= 1/3 x m l²
where m stands for the mass of the blade
l represents the length of each blade.
Total moment of inertia for 3 blades is:
= 3 x x m l²
ml²
2 )
Details provided include:
m = 5500 kg
l = 45 m
Substituting these values produces:
moment of inertia of one blade:
= 1/3 x 5500 x 45 x 45
= 37.125 x 10⁵ kg.m²
Moment of inertia for 3 blades:
= 3 x 37.125 x 10⁵ kg.m²
= 111.375 x 10⁵ kg.m²
c )
Angular momentum
= I x ω
I denotes the moment of inertia of the turbine
ω symbolizes angular velocity
ω = 2π f
f indicates the rotational frequency of the blades
d )
We have I = 111.375 x 10⁵ kg.m² (Calculated)
f = 11 rpm (revolutions per minute)
= 11 / 60 revolutions per second
ω = 2π f
= 2π x 11 / 60 rad / s
Calculating angular momentum yields
= I x ω
111.375 x 10⁵ kg.m² x 2π x 11 / 60 rad / s
= 128.23 x 10⁵ kgm² s⁻¹.
Answer:
35.79 meters
Explanation:
We have an archer, and there is a target. Denote the distance between them as d.
The bowman releases the arrow, which travels the distance d at a velocity of 40 m/s until it hits the target. We establish the equation as:
Right after this, the arrow produces a muffled noise, traveling the same distance d at a speed of 340 m/s in time . Thus, we can derive:
.
Consequently, the sound reaches the archer, precisely 1 second post-firing the bow, resulting in:
.
Using this relationship in the distance formula for sound allows us to write:
.
Substituting the value of d from the first equation yields:
.
Now, after some calculations, we can proceed further:
.
Finally, the value is inserted into the initial equation: