D) a car speeding up may i have brainliest hope this help
Using the formula F=ma
500N=50kg (a)
a= 10 m/s^2
I think the question should be the below:
<span>What is the total distance, side to side, that the top of the building moves during such an oscillation?
</span>
Answer is the below:
<span>Acceleration .. a = (-) ω² x </span>
<span>(ω = equivalent ang. vel. = 2π.f) (x = displacement from equilibrium position) </span>
<span>x (max) = a(max) /ω² </span>
<span>x = (0.015 x 9.8m/s²) / (2π.f)² .. .. (0.147) / (2π*0.22)² .. .. ►x(max) = 0.077m .. (7.70cm)</span>
Answer:
W = 8.01 × 10^(-17) [J]
Explanation:
To solve this problem we need to know the electron is a subatomic particle with a negative elementary electrical charge (-1,602 × 10-19 C), The expression to calculate the work is given by:
W = q*V
where:
q = charge = 1,602 × 10^(-19) [C]
V = voltage = 500 [V]
W = work [J]
W = 1,602 × 10^(-19) * 500
W = 8.01 × 10^(-17) [J]
Making a wire thicker has the same effect as making a road wider. It makes it easier for the electron traffic to flow. The resistance decreases, and the current (traffic) increases.
