The question is incomplete. The complete question is :
In a certain underdamped RLC circuit, the voltage across the capacitor decreases in one cycle from 5.0 V to 3.8 V. The period of the oscillations is 1.2 microseconds (1.2*10^-6). What is Q?
Solution :
The underdamped RLC circuit
We know in one time period, v = 2v, at t = T, 
so, 
Now, Q value 
∴ 
= 11.45
Answer: The force of attraction that holds two molecules is a chemical bond
Explanation:
What is Chemical bonds?
Chemical bonds are forces that hold atoms together to make compounds or molecules.
Types of chemical bonds
Chemical bonds include
1.covalent,
2. polar covalent, and
3. ionic bonds.
Atoms with relatively similar electronegativities share electrons between them and are connected by covalent bonds.
Answer:
Force exerted, F = 1.5 N
Explanation:
It is given that, a boxer punches a sheet of paper in midair and brings it from rest up to a speed of 30 m/s in 0.060 s.
i.e. u = 0
v = 30 m/s
Time taken, t = 0.06 s
Mass of the paper, m = 0.003 kg
We need to find the force the boxer exert on it. The force can be calculated using second law of motion as :
F = 1.5 N
So, the force the boxer exert on the paper is 1.5 N. Hence, this is the required solution.
Answer:
a. 8.96 m/s b. 1.81 m
Explanation:
Here is the complete question.
a) A long jumper leaves the ground at 45° above the horizontal and lands 8.2 m away.
What is her "takeoff" speed v
0
?
b) Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 m away horizontally and 2.5 m, vertically below.
If she long jumps from the edge of the left bank at 45° with the speed calculated in part a), how long, or short, of the opposite bank will she land?
a. Since she lands 8.2 m away and leaves at an angle of 45 above the horizontal, this is a case of projectile motion. We calculate the takeoff speed v₀ from R = v₀²sin2θ/g. where R = range = 8.2 m.
So, v₀ = √gR/sin2θ = √9.8 × 8.2/sin(2×45) = √80.36/sin90 = √80.36 = 8.96 m/s.
b. We use R = v₀²sin2θ/g to calculate how long or short of the opposite bank she will land. With v₀ = 8.96 m/s and θ = 45
R = 8.96²sin(2 × 45)/9.8 = 80.2816/9.8 = 8.192 m.
So she land 8.192 m away from her bank. The distance away from the opposite bank she lands is 10 - 8.192 m = 1.808 m ≅ 1.81 m
negative acceleration- deceleration
