Answer:
question5: F=74312.5N
question6: charge at the end of antenna=0.37N
Explanation:
Coulomb's law: the magnitude of the force of attraction or repulsion due to two charges is proportional to the product of the magnitude of the charges and inversely proportional to the square of distance between the charges.
⇒
∴
where
is the force of attraction or repulsion
is Coulumb's constant=
and
are the magnitude of the charges
is the distance between two charges
The force between the two charges is attractive if they are of different polarity
The force between the two charges is repulsive if they are of same polarity
Question5:
Given: q1=0.041 C, q2=0.029 C, r=12 m
therefore by Coulumb's law,


Question6:
Given: q1=
, r=5 m, F=
therefore by Coulumb's law,

⇒
the greater the <u>mass</u> of an object the more force is needed to cause acceleration
Answer:
In a velocity selector, there are two forces namely;
» Electric field Intensity
» Magnetic field density
<u>Relationship</u><u>:</u>

E is the electric field intensity
B is the magnetic flux density
For part a)
Since the conical surface is not exposed to the radiation coming from the walls only from the circular plate and assuming steady state, the temperature of the conical surface is also equal to the temperature of the circular plate. T2 = 600 K
For part b)
To maintain the temperature of the circular plate, the power required would be calculated using:
Q = Aσ(T₁⁴ - Tw⁴)
Q = π(500x10^-3)²/4 (5.67x10^-8)(600⁴ - 300⁴)
Q = 5410.65 W
Answer:
Explanation:
The charges will repel each other and go away with increasing velocity , their kinetic energy coming from their potential energy .
Their potential energy at distance d
= kq₁q₂ / d
= 9 x 10⁹ x 36 x 10⁻¹² / 2 x 10⁻² J
= 16.2 J
Their total kinetic energy will be equal to this potential energy.
2 x 1/2 x mv² = 16.2
= 3 x 10⁻⁶ v² = 16.2
v = 5.4 x 10⁶
v = 2.32 x 10³ m/s
When masses are different , total P.E, will be divided between them as follows
K E of 3 μ = (16.2 / 30+3) x 30
= 14.73 J
1/2 X 3 X 10⁻⁶ v₁² = 14.73
v₁ = 3.13 x 10³
K E of 30 μ = (16.2 / 30+3) x 3
= 1.47 J
1/2 x 30 x 10⁻⁶ x v₂² = 1.47
v₂ = .313 x 10³ m/s