A these were large creatures similar to dinosaurs
Answer:
1.53 g
Explanation:
Given,
Initial speed = 30 m/s
Final speed = 0 m/s
Period of stretch, t = 2 s
average deceleration = ?
we know
a = -15 m/s²
Deceleration of the jumper = 15 m/s²
Deceleration in terms of g
Hence, the deceleration of the jumper is equal to 1.53 g
Answer:
Work done = 0.3142 Nm
Explanation:
Mass of Object is 50 g
Circular path of radius is 10 cm ⇒ 0.1 m
Work done = Force × Distance = ?
*Distance moved (circular path) ⇒ Circumference of the circular path
2πr = 2 × 3.142 × 0.1 ⇒ 0.6284 m
*Force that is enough to move a 50 g must be equal or more than its weight.
therefore convert 50 grams to newton = 0.5 N
Recall that; work done is force times distance
∴ 0.5 N × 0.6284 m
Work done = 0.3142 Nm
Answer:
T/√8
Explanation:
From Kepler's law, T² ∝ R³ where T = period of planet and R = radius of planet.
For planet A, period = T and radius = 2R.
For planet B, period = T' and radius = R.
So, T²/R³ = k
So, T²/(2R)³ = T'²/R³
T'² = T²R³/(2R)³
T'² = T²/8
T' = T/√8
So, the number of hours it takes Planet B to complete one revolution around the star is T/√8
Explanation:
(a) Hooke's law:
F = kx
7.50 N = k (0.0300 m)
k = 250 N/m
(b) Angular frequency:
ω = √(k/m)
ω = √((250 N/m) / (0.500 kg))
ω = 22.4 rad/s
Frequency:
f = ω / (2π)
f = 3.56 cycles/s
Period:
T = 1/f
T = 0.281 s
(c) EE = ½ kx²
EE = ½ (250 N/m) (0.0500 m)²
EE = 0.313 J
(d) A = 0.0500 m
(e) vmax = Aω
vmax = (0.0500 m) (22.4 rad/s)
vmax = 1.12 m/s
amax = Aω²
amax = (0.0500 m) (22.4 rad/s)²
amax = 25.0 m/s²
(f) x = A cos(ωt)
x = (0.0500 m) cos(22.4 rad/s × 0.500 s)
x = 0.00919 m
(g) v = dx/dt = -Aω sin(ωt)
v = -(0.0500 m) (22.4 rad/s) sin(22.4 rad/s × 0.500 s)
v = -1.10 m/s
a = dv/dt = -Aω² cos(ωt)
a = -(0.0500 m) (22.4 rad/s)² cos(22.4 rad/s × 0.500 s)
a = -4.59 m/s²
