Answer:
T = mg - 
×V×g
Explanation:
Here we have the mass of the ball is given as 
The density of the fluid =
The volume of the ball = V
Acceleration due to gravity = g
According to Archimedes's principle, the upthrust = Weight of fluid displaced
Tension in string = T = weight of ball - upthrust
T = mg - ×V×g
Answer:
(a). The electric potential at 1.650 cm is 
.
(b). The electric potential at 2.81 cm is 
.
Explanation:
Given that,
Radius of sphere R=2.81 cm
Charge = +2.35 fC
Potential at center of sphere
(a). We need to calculate the potential at a distance r = 1.60 cm
Using formula of potential difference
The electric potential at 1.650 cm is .
(b). We need to calculate the potential at a distance r = R
Using formula of potential difference
The electric potential at 2.81 cm is .
Hence, This is the required solution.
It channels erode wider fed by many tributaries and has more discharge and is less steep
We have the following equation for height:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
Where,
a: acceleration
vo: initial speed
h0: initial height.
The value of the acceleration is:
a = -g = -9.8 m / s ^ 2
For t = 0 we have:
h (0) = (1/2) * (a) * 0 ^ 2 + vo * 0 + h0
h (0) = h0
h0 = 0 (reference system equal to zero when the ball is hit).
For t = 5.8 we have:
h (5.8) = (1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0
(1/2) * (- 9.8) * (5.8) ^ 2 + vo * (5.8) + 0 = 0
vo = (1/2) * (9.8) * (5.8)
vo = 28.42
Substituting values we have:
h (t) = (1/2) * (a) * t ^ 2 + vo * t + h0
h (t) = (1/2) * (- 9.8) * t ^ 2 + 28.42 * t + 0
Rewriting:
h (t) = -4.9 * t ^ 2 + 28.42 * t
The maximum height occurs when:
h '(t) = -9.8 * t + 28.42
-9.8 * t + 28.42 = 0
t = 28.42 / 9.8
t = 2.9 seconds.
Answer:
The ball was at maximum elevation when:
t = 2.9 seconds.
The planet closest to the sun; Mercury.
