Answer:
D-Driving the car faster down the road.
Answer:
potential difference V= 300 volts
Explanation:
Given:
d= 2.0 cm = 0.02m
E = 15 kN/C = 15 × 10³ N/C
For a uniform field between two plates, the Electric Filed Intensity (E) is proportional to the potential difference (V) and inversely proportional to distance between the plates.
E= V/d
⇒ V= E×d = 15 × 10³ N/C × 0.02 m = 300 volts (∴1 Nm/C = 1 J/C= 1 volts)
Answer:
lowest frequency = 535.93 Hz
distance between adjacent anti nodes is 4.25 cm
Explanation:
given data
length L = 32 cm = 0.32 m
to find out
frequency and distance between adjacent anti nodes
solution
we consider here speed of sound through air at room temperature 20 degree is approximately v = 343 m/s
so
lowest frequency will be =
..............1
put here value in equation 1
lowest frequency will be =
lowest frequency = 535.93 Hz
and
we have given highest frequency f = 4000Hz
so
wavelength =
..............2
put here value
wavelength =
wavelength = 0.08575 m
so distance =
..............3
distance =
distance = 0.0425 m
so distance between adjacent anti nodes is 4.25 cm
D. Budgeting time, avoiding stress, and prioritizing.