Answer:
(a) = 3.7 × 10⁻⁵
(b) = 4.1 × 10⁻⁶ N.M²/C
Explanation:
(a) Diameter of the sphere, d = 1.2 m
Radius of the sphere, r = 0.6 m
Surface charge density, = 8.1 mC/m2 = 8.1 × 10⁻⁶ C/m²
Total charge on the surface of the sphere,
Q = Charge density × Surface area
= 4πr²σ
= 4 (3.14) (0.12²) (8.1 × 10⁻⁶)
= 3.66 × 10⁻⁵C
≅ 3.7 × 10⁻⁵C
Therefore, the net charge on the sphere is 3.7 × 10⁻⁵C
(b)
Total electric flux (∅)
=Q / ε₀
ε₀ = 8.854 × 10⁻¹² N⁻¹C² m⁻²
Q = 3.66 × 10⁻⁵C
= 3.66 × 10⁻⁵ / 8.854 × 10⁻¹²
= 4.1 × 10⁻⁶ N.M²/C
Answer:
v_o= 12 m/s
Explanation:
The ball experienced a constant acceleration motion, so we need to apply the following formula:

Note: we set the acceleration of gravity as negative because it is going down.
the ball was thrown vertically upward with an initial velocity of 12 m/s
Answer:
The value is 
Explanation:
From the question we are told that
The first position of the fulcrum is x = 49.7 cm
The mass attached is
The position of the attachment is
The second position of the fulcrum is 
Generally the sum of clockwise torque = sum of anti - clockwise torque
So

Here CWT stands for clockwise torque
So

=> 
=> 
Answer:
Its called the hydrosphere UwU!
Explanation: