Answer:
(A) As it moves farther and farther from Q, its speed will keep increasing.
Explanation:
When a positive charge Q is fixed on a horizontal frictionless tabletop and a second charge q is released near to it then according to the Coulombs law the force acting on it decreases with the square of the distance between them.
Mathematically:
where:
r = distance between the charges
permittivity of free space
By the Newtons' second law of motion if the we know that the acceleration is directly proportional to the force applied. So as the distance between the charges increases the its acceleration also decreases therefore now the charge feels less acceleration but still continues to accelerate with a fading magnitude.
Answer:
0.6 Ω
Explanation:
As shown in the diagram below,
Since the resistance and the ammeter are connected in series,
(i) The same amount of current flows through them.
(ii) The sum of their individual individual voltage is equal to the total voltage of the circuit.
Applying ohm's law,
V = IR................ Equation 1
Where V = Voltage across the ammeter, I = current flowing through the ammeter, R = resistance of the ammeter.
make R the subject of the equation
R = V/I............... Equation 2
Given: V = 1.2-0.9 = 0.3 V, I = 0.5 A.
Substitute into equation 2
R = 0.3/0.5
R = 0.6 Ω
The answer to this question is going to be False
Answer:
5.5 km
Explanation:
First, we convert the distance from km/h to m/s
910 * 1000/3600
= 252.78 m/s
Now, we use the formula v²/r = gtanθ to get our needed radius
making r the subject of the formula, we have
r = v²/gtanθ, where
r = radius of curvature needed
g = acceleration due to gravity
θ = angle of banking
r = 252.78² / (9.8 * tan 50)
r = 63897.73 / (9.8 * 1.19)
r = 63897.73 / 11.662
r = 5479 m or 5.5 km
Thus, we conclude that the minimum curvature radius needed for the turn is 5.5 km
Downward movement under the force of gravity only.
