It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating).
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
<span>Just set F=the drag force and the electric field strength is (K*Q2)/r^2, plugging those values in gives you </span>
<span>(7.25*10^-11 N) = (1*10^5 N/C)*Q1 ---> Q1 = 7.25*10^-16 C </span>
a. 0.5 T
- The amplitude A of a simple harmonic motion is the maximum displacement of the system with respect to the equilibrium position
- The period T is the time the system takes to complete one oscillation
During a full time period T, the mass on the spring oscillates back and forth, returning to its original position. This means that the total distance covered by the mass during a period T is 4 times the amplitude (4A), because the amplitude is just half the distance between the maximum and the minimum position, and during a time period the mass goes from the maximum to the minimum, and then back to the maximum.
So, the time t that the mass takes to move through a distance of 2 A can be found by using the proportion
and solving for t we find
b. 1.25T
Now we want to know the time t that the mass takes to move through a total distance of 5 A. SInce we know that
- the mass takes a time of 1 T to cover a distance of 4A
we can set the following proportion:
And by solving for t, we find
<span>The unknown substance is silver.
I don't see a list of available substances, but let's see if there's something reasonable available that will match. First, let's calculate the density of the unknown substance. Density is mass per volume, so
273 g / 26 mL = 10.5 g/mL
Looking up a list of elements sorted by density, I see the following:
10.07 Actinium
10.22 Molybdenum
10.5 Silver
11.35 Lead
And silver at 10.5 g/ml is a very nice match for the unknown substances' density of 10.5 g/ml.</span>
Answer:
Stress = F / A force per unit area
A = 3.00 cm^2 = 3 E-4 m^2
F = 2.4E8 N/m^2 * 3E-4 m^2 = 7.2E4 N max force applied
F/3 = 2.4E4 N if force not to exceed limit (= f)
f = M a
a = 2.4 E4 N / 1.2 E3 kg = 20 m / s^2 about 2 g
Answer:
B. less
Explanation:
acceleration due to gravity on Earth, g = 9.8 m/s²
acceleration due to gravity on Moon, g = 1.6 m/s²
Given mass of the object as, m = 5 kg
Weight of an object is given as, W = mg
Weight of the object on Earth, W = 5 x 9.8 = 49 N
Weight of the object on Moon, W = 5 x 1.6 = 8 N
Therefore, the object weighs less on the moon compared to its weight on Earth.
The correct option is "B. less"
