The car travels a distance <em>d</em> from rest with acceleration <em>a</em> after time <em>t</em> of
<em>d</em> = 1/2 <em>a</em> <em>t</em>²
It covers 69 m with 2.8 m/s² acceleration, so that
69 m = 1/2 (2.8 m/s²) <em>t</em>²
<em>t</em>² = 2 (69 m) / (2.8 m/s²)
<em>t</em> ≈ 7.02 s
where we take the positive square root because we're talking about time *after* the car begins accelerating.
Answer:
a = 12 [m/s²]
Explanation:
To solve this problem we must use Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.
ΣF = m*a
where:
ΣF = sum of forces acting on a body [N] (units of Newtons)
m = mass = 0.5 [kg]
a = acceleration [m/s²]
Let's take the direction of positive forces to the right and negative forces directed to the left
2 + 8 - 4 = 0.5*a
6 = 0.5*a
a = 12 [m/s²]
If an electron, a proton, and a deuteron move in a magnetic field with the same momentum perpendicularly, the ratio of the radii of their circular paths will be:
<h3>How is the ratio of the perpendicular parts obtained?</h3>
To obtain the ratio of the perpendicular parts, one begins bdy noting that the mass of the proton = 1m, the mass of deuteron = 2m, and the mass of the alpha particle = 4m.
The ratio of the radii of the parts can be obtained by finding the root of the masses and dividing this by the charge. When the coefficients are substituted into the formula, we will have:
r = √m/e : √2m/e : √4m/2e
When resolved, the resulting ratios will be:
1: √2 : 1
Learn more about the radii of their circular paths here:
brainly.com/question/16816166
#SPJ4
Complete question:
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)
Answer:
The peak emf generated by the coil is 15.721 kV
Explanation:
Given;
Radius of coil, r = 0.250 m
Number of turns, N = 500-turn
time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s
magnetic field strength, B = 0.425 T
Induced peak emf = NABω
where;
A is the area of the coil
A = πr²
ω is angular velocity
ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s
Induced peak emf = NABω
= 500 x (π x 0.25²) x 0.425 x 376.738
= 15721.16 V
= 15.721 kV
Therefore, the peak emf generated by the coil is 15.721 kV
