The frequency decreases, and you will hear a lower tone
Answer:
put the car on fire
Explanation:
if you put it on fire you would have a lot of light now
Answer:
L= 2 mH
Explanation:
Given that
Frequency , f= 10 kHz
Maximum current ,I = 0.1 A
Maximum energy stored ,E= 1 x 10⁻⁵ J
The maximum energy stored in the inductor is given as follows

Where ,L= Inductance
I=Current
E=Energy
Now by putting the values in the above equation


L=0.002 H
L= 2 mH
We know that frequency f is given as

C=Capacitance , f=frequency ,L=Inductance
Now by putting the values






Therefore the inductance and capacitance will be 2 mH and 1.26 x 10⁻⁷ F respectively.
Answer:
11 m/s
Explanation:
Draw a free body diagram. There are two forces acting on the car:
Weigh force mg pulling down
Normal force N pushing perpendicular to the incline
Sum the forces in the +y direction:
∑F = ma
N cos θ − mg = 0
N = mg / cos θ
Sum the forces in the radial (+x) direction:
∑F = ma
N sin θ = m v² / r
Substitute and solve for v:
(mg / cos θ) sin θ = m v² / r
g tan θ = v² / r
v = √(gr tan θ)
Plug in values:
v = √(9.8 m/s² × 48 m × tan 15°)
v = 11.2 m/s
Rounded to 2 significant figures, the maximum speed is 11 m/s.