Answer:
(a): 
(b): 
(c): 
Explanation:
Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053
m.
Part (a):
According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges
and
respectively is given by

where,
= Coulomb's constant = 
= distance of separation between the charges.
For the given system,
The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, 
The charge on the electron, 
These two are separated by the distance, 
Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

Part (b):
The gravitational force of attraction between two objects of masses
and
respectively is given by

where,
= Universal Gravitational constant = 
= distance of separation between the masses.
For the given system,
The mass of proton, 
The mass of the electron, 
Distance between the two, 
Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

The ratio
:

Answer:
The semi truck travels at an initial speed of 69.545 meters per second downwards.
Explanation:
In this exercise we see a case of an entirely inellastic collision between the semi truck and the car, which can be described by the following equation derived from Principle of Linear Momentum Conservation: (We assume that velocity oriented northwards is positive)
(1)
Where:
,
- Masses of the semi truck and the car, measured in kilograms.
,
- Initial velocities of the semi truck and the car, measured in meters per second.
- Final speed of the system after collision, measured in meters per second.
If we know that
,
,
and
, then the initial velocity of the semi truck is:





The semi truck travels at an initial speed of 69.545 meters per second downwards.
Answer: When you touch wet canvas, surface tension will draw water to your finger. However, the drop left behind where you touched, like any irregular point on an overhead surface, will draw condensation from inside the tent if it is humid.
Answer:
She can swing 1.0 m high.
Explanation:
Hi there!
The mechanical energy of Jane (ME) can be calculated by adding her gravitational potential (PE) plus her kinetic energy (KE).
The kinetic energy is calculated as follows:
KE = 1/2 · m · v²
And the potential energy:
PE = m · g · h
Where:
m = mass of Jane.
v = velocity.
g = acceleration due to gravity (9.8 m/s²).
h = height.
Then:
ME = KE + PE
Initially, Jane is running on the surface on which we assume that the gravitational potential energy of Jane is zero (the height is zero). Then:
ME = KE + PE (PE = 0)
ME = KE
ME = 1/2 · m · (4.5 m/s)²
ME = m · 10.125 m²/s²
When Jane reaches the maximum height, its velocity is zero (all the kinetic energy was converted into potential energy). Then, the mechanical energy will be:
ME = KE + PE (KE = 0)
ME = PE
ME = m · 9.8 m/s² · h
Then, equallizing both expressions of ME and solving for h:
m · 10.125 m²/s² = m · 9.8 m/s² · h
10.125 m²/s² / 9.8 m/s² = h
h = 1.0 m
She can swing 1.0 m high (if we neglect dissipative forces such as air resistance).
Answer:
C) 7.35*10⁶ N/C radially outward
Explanation:
- If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
- So, we can write the following equation:

- As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
- So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
- So, on the outer surface of the shell there must be a charge that be the difference between them:

- Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

- As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.