Answer:
Explanation:
The time period of a simple pendulum: It is defined as the time taken by the pendulum to finish one full oscillation and is denoted by “T”.
The period of simple pendulum is given as
T=2π√l/g
Increasing the density of the object implies that the mass has increased 
Because density is mass/volume
The period is independent on the density of the object, it depends on the length of the string and the acceleration due to gravity. So the period remains the same.
2. Acceleration 
The acceleration of a simple pendulum is given as
a=-gSin θ
Therefore the acceleration is independent on the density of the object, it depends on the acceleration due to gravity and the angle of displacement. 
Then, the acceleration remains the same.
So the answer is Option C,
Same 
Same
 
        
             
        
        
        
The time it takes the baton to complete one spin will be 0.56 s. Option B is correct.
<h3>What is centripetal acceleration?</h3>
The acceleration needed to move a body in a curved way is understood as centripetal acceleration. 
The direction of centripetal acceleration is always in the path of the center of the course. The total acceleration is the result of tangential and centripetal acceleration. 
The entire question is;
"How long does it take the baton to complete one spin?
A twirler’s baton is 0.76 m long and spins around its center. The end of the baton has a centripetal acceleration of 47.8 m/s2.
a.0.31 s
b.0.56 s
c.4.3 s
d.70 s"
The given data in the problem;
Length of baton,L = 0.76 m
Centripetal acceleration,
The centripetal acceleration is found by;

Substitute the given value:

The time it takes the baton to complete one spin will be 0.56 s.
Hence option B is correct.
To learn more about centripetal acceleration, refer to the link;
brainly.com/question/17689540
#SPJ1
 
        
             
        
        
        
Answer:
A)   V = -136.36 V
, B)  V = 4.85 10³ V
, C)  V = 1.62 10⁴ V
Explanation:
To calculate the potential at an external point of the spheres we use Gauss's law that the charge can be considered at the center of the sphere, therefore the potential for an external point is
           V = k ∑ 
where  and
 and  are the loads and the point distances.
 are the loads and the point distances.
A) We apply this equation to our case
           V = k (q₁ / r₁ + q₂ / r₂)
They ask us for the potential at the midpoint of separation
          r = 3.30 / 2 = 1.65 m
this distance is much greater than the radius of the spheres
let's calculate
          V = 9 10⁹ (1.1 10⁻⁸ / 1.65  + (-3.6 10⁻⁸) / 1.65)
          V = 9 10¹ / 1.65 (1.10 - 3.60)
          V = -136.36 V
B) The potential at the surface sphere A
r₂ is the distance of sphere B above the surface of sphere A
               r₂ = 3.30 -0.02 = 3.28 m
               r₁ = 0.02 m
we calculate
              V = 9 10⁹ (1.1 10⁻⁸ / 0.02  - 3.6 10⁻⁸ / 3.28)
              V = 9 10¹ (55 - 1,098)
              V = 4.85 10³ V
C) The potential on the surface of sphere B
       r₂ = 0.02 m
       r₁ = 3.3 -0.02 = 3.28 m
       V = 9 10⁹ (1.10 10⁻⁸ / 3.28  - 3.6 10⁻⁸ / 0.02)
        V = 9 10¹ (0.335 - 180)
        V = 1.62 10⁴ V
 
        
             
        
        
        
Answer:
Explanation:
Consider another special case in which the inclined plane is vertical (θ=π/2). In this case, for what value of m1 would the acceleration of the two blocks be equal to zero
F - Force
T = Tension
m = mass
a = acceleration
g = gravitational force
Let the  given Normal on block 2 = N 
and 
 and the tension in the given string is said to be 
When the acceleration 
 for the said block 1. 
It will definite be zero only when Force is zero , F=0. 
Here by Force, F
 I refer net force on block 1.
Now we know

It is known that if the said 
 ,
 , 
then Tension  
 ![[since \sin(\pi/2) = 1]](https://tex.z-dn.net/?f=%5Bsince%20%5Csin%28%5Cpi%2F2%29%20%3D%201%5D) ,
, 
Now making 
So If we are to make Force equal to zero 
