Answer:
<h2>2 N</h2>
Explanation:
The force engine exert on the car can be found by using the formula
w is the workdone
d is the distance
From the question we have
We have the final answer as
<h3>2 N</h3>
Hope this helps you
Force applied = F = 628 N
<span>Acceleration = a m/s² </span>
<span>Newton's 2nd law of motion : F = Ma </span>
<span> a = F/M -------- (1) </span>
<span>New mass of the crate = M1 = 3.8M kg </span>
<span>New acceleration = a1 = F/M1 = F/(3.8 M) ----- (2) </span>
<span>a1/a = {F/(3.8M)}/(F/M) = 1/3.8 = 10/38 = 5/19 ------- Answer</span>
Answer:
The moon moves because the earth rotates around its axis of rotation. Although the moon moves too around its orbit, the speed is too small to notice. In other words, the moon's position changes insignificantly, however, the earth rotates fast enough to notice.
Answer:
A) r = 0.03 m
B) r = 0.0533 m
C) B_max = 0.00003 T
Explanation:
Formula for magnetic field inside the capacitor when it is parallel to the length element is;
B_in = (μ_o•I•r/(2πR²)
Formula for maximum magnetic field is;
B_max = (μ_o•I/(2πR)
Formula for magnetic field outside the capacitor is; B_out = (μ_o•I/(2πr)
A) Magnetic field inside the capacitor is gotten from our first equation above;
B_in = (μ_o•I•r/R²)
Since we want to find the radius at which the magnitude of the induced magnetic field equal to 75% or 0.75 of its maximum value.
Thus;
B_in = 0.75B_max
(μ_o•I•r/(2πR²) = 0.75((μ_o•I/(2πR))
μ_o•I and 2πR will cancel out to give;
r/R = 0.75
r = 0.75R
We are given R = 40 mm = 0.04 m
r = 0.75 × 0.04
r = 0.03 m
B) magnetic field outside the capacitor is; B_out = (μ_o•I/(2πr)
Thus for the magnitude of the induced magnetic field equal to 75% or 0.75 of its maximum value:
B_out = 0.75B_max
(μ_o•I/(2πr) = 0.75((μ_o•I/(2πR))
μ_o•I and 2π will cancel out to give;
1/r = 0.75/R
r = R/0.75
r = 0.04/0.75
r = 0.0533 m
C) B_max = μ_o•I/(2πR)
μ_o is a constant known as vacuum of permeability with a value of 4π × 10^(-7) T.m/A
Thus;
B_max = (4π × 10^(-7) × 6)/(2π × 0.04)
B_max = 0.00003 T
