Answer:
On removal to the glass rod the leaves of the electroscope will disperse away under the influence of the positive charge due to deficiency of electrons.
Explanation:
When a glass rod is rubbed against a nylon then the electrons from the surface of the glass rod get transferred to the nylon piece and as a result the glass rod attains positive electrostatic charge.
When this positively charged rod is brought close to the knob of an electroscope then the electrons from that metallic of the electroscope feel attraction and move toward the glass rod get concentrated on the knob.
- In this situation when any of our body part comes in direct contact to the knob these static concentrated charges get a path to flow from high concentration to a lower concentration under the influence of the electrostatic force of the rod.
- So these charges flow into our body from the knob.
- Our body acts as a reservoir for the charges on the metallic knob.
- Now after the electrons form the knob are transferred to the person's body the electroscope becomes electron deficient and hence positively charged.
- So now, on removal to the glass rod the leaves of the electroscope will disperse away under the influence of the positive charge.
Answer:
The magnitude of the acceleration ae of the earth due to the gravitational pull of the moon is 
Explanation:
By Newton's gravitational law, the magnitude of the gravitational force between two objects is:
(1)
With G the gravitational constant, M the mass of earth, m the mass of the moon and r the distance between the moon and the earth, a quick search on physics books or internet websites give us the values:
Using those values on (1)
Now, by Newton's second Law we can find the acceleration of earth ae due moon's pull:
Answer:
Explanation:
q = 2e = 3.2 x 10^-19 C
mass, m = 6.68 x 10^-27 kg
Kinetic energy, K = 22 MeV
Current, i = 0.27 micro Ampere = 0.27 x 10^-6 A
(a) time, t = 2.8 s
Let N be the alpha particles strike the surface.
N x 2e = q
N x 3.2 x 10^-19 = i t
N x 3.2 x 10^-19 = 0.27 x 10^-6 x 2.8
N = 2.36 x 10^12
(b) Length, L = 16 cm = 0.16 m
Let N be the alpha particles
K = 0.5 x mv²
22 x 1.6 x 10^-13 = 0.5 x 6.68 x 10^-27 x v²
v² = 1.054 x 10^15
v = 3.25 x 10^7 m/s
So, N x 2e = i x t
N x 2e = i x L / v
N x 3.2 x 10^-19 = 2.7 x 10^-7 x 0.16 / (3.25 x 10^7)
N = 4153.85
(c) Us ethe conservation of energy
Kinetic energy = Potential energy
K = q x V
22 x 1.6 x 10^-13 = 2 x 1.5 x 10^-19 x V
V = 1.17 x 10^7 V
Complete question is;
Lamar writes several equations trying to better understand potential energy. What conclusion is best supported by Lamar’s work?
A) The elastic potential energy is the same for any distance from a reference point.
B) The gravitational potential energy equals the work needed to lift the object.
C) The gravitational potential energy is the same for any distance from a reference point.
D) The elastic potential energy equals the work needed to stretch the object
Answer:
B) The gravitational potential energy equals the work needed to lift the object.
Explanation:
In physics, we know that potential energy is the energy of a body at rest while the energy of a body in motion is known as kinetic energy.
However,the work required to lift a body from it's position of rest is equal to the Gravitational potential energy of that body.
Elastic potential energy is the one that is stored as a result of force applied to deform an elastic object. Thus, it is not equal to the work needed to stretch the object and it is also not the same for any distance from reference point.
Thus, looking at the options, Option B is correct
Answer:
m = 
x,
graph of x vs m
Explanation:
For this exercise, the simplest way to determine the mass of the cylinder is to take a spring and hang the mass, measure how much the spring has stretched and calculate the mass, using the translational equilibrium equation
F_e -W = 0
k x = m g
m = x
We are assuming that you know the constant k of the spring, if it is not known you must carry out a previous step, calibrate the spring, for this a series of known masses are taken and hung by measuring the elongation (x) from the equilibrium position, with these data a graph of x vs m is made to serve as a spring calibration.
In the latter case, the elongation measured with the cylinder is found on the graph and the corresponding ordinate is the mass
