Cookies would not be an effective pre-exercise energizing snack, as their high sugar content would be absorbed quickly by the body, not sustaining the athlete for the duration of their event.
Radiated away as electromagnetic radiation.
Answer:
a. Concave mirror, radius of curvature = 16 cm b. magnification = 2
Explanation:
a. Since the image is upright and larger than the object, we need a concave mirror.
We know image height, h'/object height, h = -image distance, d'/object distance, d
h'/h = -d'/d
Using the real is positive convention,
h'= + 5.4 cm, h = + 2.7 cm and d = + 12 cm.
So, + 5.4 cm/+ 2.7 cm = -d'/+ 12 cm
2 = -d'/12
d' = -2 × 12 cm
= -24 cm
Using the mirror formula 1/d + 1/d' = 2/r where r = radius of curvature of the mirror
1/+12 + 1/- 24 = 2/r
1/12(1 - 1/2) = 2/r
1/12(1/2) = 2/r
1/24 = 2/r
r/2 = 24
r = 2 × 24
r = 48 cm
b.
magnification = image height, h'/object height,h = + 5.4 cm/+ 2.7 cm = 2
Answer:
Explanation:
Given the equation modelled by the height of the train given as:
s(t) = 18t²-2t³ for for 0 ≤ t ≤ 9
a) Velocity is the rate of change of displacement.
Velocity = dS(t)/dt
V = dS(t)/dt = 36t - 6t² miles
Velocity at t = 3hrs is determiner by substituting t = 3 into the velocity function.
V = 36(3) -6(3)²
V= 108 - 72
Velocity = 36mi/hr
b) for Velocity at time = 7hrs
V(7) = 36(7) - 6(7)²
V(7) = 252 - 294
V(7) = -42mi/hr
The velocity at t = 7hrs is -42mi/hr
c) Acceleration is the rate of change of velocity.
a(t) = dV(t)/dt
Given v(t) = 36t - 6t²
a(t) = 36 - 12t
Acceleration at t=1 is given as:
a(1) = 36 -12(1)
a(1) = 24mi/hr²
