Answer:

Explanation:
Given:
- mass of solid disk,

- radius of disk,

- force of push applied to disk,

- distance of application of force from the center,

<em>For the condition of no slip the force of static friction must be greater than the applied force so that there is no skidding between the contact surfaces at the contact point.</em>

where:
= static frictional force




The pressure of the water on the diver is given in an expression written as:
<span>p=15+15/33d
where p is the pressure and d is the distance of the diver </span><span>below the surface.
The pressure is calculated as follows:
</span>p=15+15/33(100) = 15.00 pounds per square feet
Therefore, the correct answer is option A.
Answer:
a. FALSE
b. FALSE
c. TRUTH
d. FALSE
e. FALSE
Explanation:
To determine which statements are truth or false you focus in the following formula, for the electric potential generated by a conducting sphere:
inside the sphere
for r > R (outside the sphere)
R: radius of the sphere
ε0: dielectric permittivity of vacuum
Q: charge of the sphere
As you can notice, inside the sphere the potential is constant. Inside the sphere, the potential is the same. Outside the surface the potential decreases as 1/r, being r the distance to the center of the sphere.
Hence, you can conclude:
a. The potential at the center of the sphere is zero. FALSE
b.The potential is lowest, but not zero, at the center of the sphere. FALSE
c. The potential at the center of the sphere is the same as the potential at the surface. TRUTH
d. The potential at the center is the same as the potential at infinity. FALSE
e. The potential at the surface is higher than the potential at the center. FALSE
Answer:
Pretty sure it is air mass C
Explanation:
It looks like yhe air has warm air where as the other 2 options have cold air.
Answer:
given,
radius of rod = 5 cm
charge per unit length (λ) = 30.0 n C/m = 30 × 10⁻⁹ C/m
electric field duel to long straight wire =
a) distance = 3 + 5 = 8 cm

= 18 × 10⁹

E = 6750 N/C
b) distance = 10 + 5 = 15 cm

= 18 × 10⁹

E = 3600 N/C
c) distance = 100 + 5 = 105 cm

= 18 × 10⁹

E =514.28 N/C