Answer:
1) Q ’= 8 Q
, 2) q ’= 16 q
, 3) r ’= ¾ r
Explanation:
For this exercise we will use Coulomb's law
F = k q Q / r²
It asks us to calculate the change of any of the parameters so that the force is always F
Original values
q, Q, r
Scenario 1
q ’= 2q
r ’= 4r
F = k q ’Q’ / r’²
we substitute
F = k 2q Q ’/ (4r)²
F = k 2q Q '/ 16r²
we substitute the value of F
k q Q / r² = k q Q '/ 8r²
Q ’= 8 Q
Scenario 2
Q ’= Q
r ’= 4r
we substitute
F = k q ’Q / 16r²
k q Q / r² = k q’ Q / 16 r²
q ’= 16 q
Scenario 3
q ’= 3/2 q
Q ’= ⅜ Q
we substitute
k q Q r² = k (3/2 q) (⅜ Q) / r’²
r’² = 9/16 r²
r ’= ¾ r
Answer:
When the voltage is at a maximum positive value, the the current is at a value that is maximum and positive
Explanation:
We know that the relation between the Voltage and the current is given using the Ohm's law, which states that the voltage (V) is directly proportional to the current (I)
Mathematically,
V ∝ I
Hence,
When the voltage is at a maximum positive value, the the current is at a value that is maximum and positive
The answer to this question is false
Answer:
= 3521m/s
The tangential speed is approximately 3500 m/s.
Explanation:
F = m * v² ÷ r
Fg = (G * M * m) ÷ r²
(m v²) / r = (G * M * m) / r²
v² = (G * M) / r
v = √( G * M ÷ r)
G * M = 6.67 * 10⁻¹¹ * 5.97 * 10²⁴ = 3.98199 * 10¹⁴
r = 32000km = 32 * 10⁶ meters
G * M / r = 3.98199 * 10¹⁴ ÷ 32 * 10⁶
v = √1.24 * 10⁷
v = 3521.36m/s
The tangential speed is approximately 3500 m/s.
Answer:
A, C, D
Explanation:
Newton's first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force.
and according to Newton's 4th law: An object that is at rest will stay at rest unless a force acts upon it. An object that is in motion will not change its velocity unless a force acts upon it.