Select from the following a scenario that does not have a net force acting on it.
2 answers:
Answer:
B - Driving straight down the highway at 60 mph
Explanation:
As we know by Newton's II law that net force on an object will follow the relation given as
so here all the cases where object is accelerating then it means it is followed by some external force on it
so here we have
A - Accelerating a car from a stop sign
C - Lifting a chess pics off a board
D - Throwing a ball
all above cases objects are accelerating so it shows net force on them
while correct answer for no force on the object is
B - Driving straight down the highway at 60 mph
You might be interested in
Answer:
(a) The potential difference between the spheres is 750 kVA
(b) The charge on the smaller sphere is 6. μC
(c) The charge on the smaller sphere, Q₁ = 13. μC
Explanation:
(a) The given parameters are;
The charge on the inner sphere, q = 5.00 μC
The radius of the inner sphere, r = 3.00 cm = 0.03 m
The charge on the larger sphere, Q = 15.0 μμC
The radius of the larger sphere, R = 6.00 cm = 0.06 m
The potential difference between two concentric spheres is given according to the following equation;
Where;
R = The radius of the larger sphere = 0.06 m
r = The radius of the inner sphere = 0.03 m
q = The charge of the inner sphere = 5.00 × 10⁻⁶ C
Q = The charge of the outer sphere = 15.00 × 10⁻⁶ C
k = 9 × 10⁹ N·m²/C²
Therefore, by plugging in the value of the variables, we have;
The potential difference between the spheres, = 750,000 N·m/C = 750 kVA
(b) When the spheres are connected with a wire, the charge, 'q', on the smaller sphere will be added to the charge, 'Q', on the larger sphere which as follows;
= Q + q = (5 + 15) × 10⁻⁶ C = 20 × 10⁻⁶ C
= 20 × 10⁻⁶ C
From which we have;
Q₁/Q₂ = R/r
Where;
Q₁ = The new charge on the on the larger sphere
Q₂ = The new charge on the on the smaller sphere
= 20 × 10⁻⁶ C = Q₁ + Q₂
∴ Q₁ = 20 × 10⁻⁶ C - Q₂ = 20 μC - Q₂
∴ (20 μC - Q₂)/Q₂ = 0.06/(0.03) = 2
20 μC - Q₂ = 2·Q₂
20 μC = 3·Q₂
Q₂ = 20 μC/3
The charge on the smaller sphere, Q₂ = 20 μC/3 = 6. μC
(c) Q₁ = 20 μC - Q₂ = 20 μC - 20 μC/3 = 40 μC/3
The charge on the smaller sphere, Q₁ = 40 μC/3 = 13. μC.
Answer:
The additional trials needed is 48 trials
Explanation:
Given;
initial number of trials, n = 16 trials
the standard deviation, σ = 0.24 s
initial standard error, ε = 0.06 s
The standard error is given by;
To reduce the standard error to 0.03 s, let the additional number of trials = x
Therefore, the additional trials needed is 48 trials.