Answer: (a) t1 = omega1/alpha
(b) theta1 = 1/2 * alpha*theta1^2
(c) t2 = omega2/5*alpha
Explanation: see attachment
Because of how it's worded the answer would most likely be number four
Answer:
I know that T= kx where T is the tension which equaka the force og gravity = mg = 1.37 * 10 = 13.7 x is the elongation of the spring so the length after dangling the object minus the original length.
I hope it helps
plz let me know if it is wrong or right.
Answer:

Explanation:
Acceleration is defined as the change in velocity divided by the time it took to produce such change. The formula then reads:

Where Vf is the final velocity of the object, (in our case 80 m/s)
Vi is the initial velocity of the object (in our case 0 m/s because the object was at rest)
and t is the time it took to change from the Vi to the Vf (in our case 0.05 seconds.
Therefore we have:

Notice that the units of acceleration in the SI system are
(meters divided square seconds)
Complete question:
Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.
Answer:
The magnitude of this field is 826 N/C
Explanation:
Given;
The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m
PEsinθ = T
where;
E is the magnitude of the electric field
P is the dipole moment
First, we determine the magnitude dipole moment;
Magnitude of dipole moment = q*r
P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m
Finally, we determine the magnitude of this field;

E = 826 N/C (in three significant figures)
Therefore, the magnitude of this field is 826 N/C