Answer:
1) as far as I remember
Let's take 20 as vf (final velocity) and 11 as (initial velocity) and 4 as time
So we would use this formula a=vf-vi/t
So 20-11/4
Asnwer 2.25
Answer:
5 .07 s .
Explanation:
The child will move on a circle of radius r
r = 1.5 m
Let the velocity of rotation = v
radial acceleration = v² / r
v² / r = 2.3
v² = 2.3 r = 2.3 x 1.5
= 3.45
v = 1.857 m /s
Time of revolution = 2π r / v
= 2 x 3.14 x 1.5 / 1.857
= 5 .07 s .
Your answer is c holding a brick doesn't contain movement, but energy to grip on it.
hoped it helped!!!
Answer:
(a) 8 m/s
(b) 5 s
Explanation:
(a)
Using,
V² = U²+2gh ......................... Equation 1
Where V = final velocity, U = Initial velocity, g = acceleration due to gravity on the surface of the moon, h = height reached.
Given: V = 0 m/s ( At it's maximum height), g = -1.6 m/s² ( as its moves against gravity), h = 20 m.
Substitute into equation 1
0 = U²+[2×20×(-1.6)]
-U² = - 64
U² = 64
U = √64
U = 8 m/s.
(b)
V = U +gt.................... Equation 2
Where t = time to reach the maximum height.
Given: V = 0 m/s ( At the maximum height), g = -1.6 m/s² ( Moving against gravity), U = 8 m/s.
Substitute into equation 2
0 = 8+(-1.6t)
-8 = -1.6t
-1.6t = -8
t = -8/-1.6
t = 5 s.
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
