The two objects with electrical charges interact, which affect the strength of that interaction <span>amount of charge. The answer is letter A. The rest of the choices do not answer the question above.</span>
Explanation:
An electrified comb is charged comb ( let say by running it through the hair) and when it is brought in the proximity of pieces of paper, the pieces tend to cling to it. This happens because the charged comb induces an opposite charge in the paper pieces and as opposite charges attract each other, the pieces are clinged.
Answer:
Capacitance is 0.572×10⁻¹⁰ Farad
Explanation:
Radius = R₁ = 6.25 cm = 6.25×10⁻² m
Radius = R₂ = 15 cm = 15×10⁻² m
Dielectric constant = k = 4.8
Electric constant = ε₀ = 8.854×10⁻¹² F/m
ε/ε₀=k
ε=kε₀
∴ Capacitance is 0.572×10⁻¹⁰ Farad
Answer:
The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
Explanation:
Given;
coefficient of kinetic friction, μ = 0.84
speed of the automobile, u = 29.0 m/s
To determine the the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;
v² = u² + 2ax
where;
v is the final velocity
u is the initial velocity
a is the acceleration
x is the shortest distance
First we determine a;
From Newton's second law of motion
∑F = ma
F is the kinetic friction that opposes the motion of the car
-Fk = ma
but, -Fk = -μN
-μN = ma
-μmg = ma
-μg = a
- 0.8 x 9.8 = a
-7.84 m/s² = a
Now, substitute in the value of a in the equation above
v² = u² + 2ax
when the automobile stops, the final velocity, v = 0
0 = 29² + 2(-7.84)x
0 = 841 - 15.68x
15.68x = 841
x = 841 / 15.68
x = 53.64 m
Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m
As per FBD while its accelerating upwards
we can say that
here normal force is given as
now mass is given as
now we will have
Now while accelerating downwards we can say by FBD
again plug in all values
