Answer:
Explanation:
a )
initial velocity u = 45 m/s
acceleration a = - 5 m/s²
final velocity v = 0
v = u - at
0 = 45 - 5 t
t = 9 s
b )
s = ut - 1/2 at²
= 45 x 9 - .5 x5x 9²
405 - 202.5
202.5 m
2 )
a )
s = ut + 1/2 a t²
u = 0
s = 1/2 at²
= .5 x 9.54 x 6.5²
= 201.5 m
b )
v = u + at
= 0 + 9.54 x 6.5
= 62.01 m / s
3
a )
acceleration = (v - u) / t
= (34 - 42) / 2.4
= - 3.33 m /s²
b )
v² = u² - 2 a s
34² = 42² - 2 x 3.33² s
s = 27.41 m
c )
Average velocity
Total displacement / time
= 27.41 / 2.4
= 11.42 m /s
4 )
a )
v = u + at
v = 0 + 3 x 4
= 12 m /s
b )
s = ut + 1/2 a t²
= o + .5 x 3 x 4²
= 24 m
Answer:
(a) 81.54 N
(b) 570.75 J
(c) - 570.75 J
(d) 0 J, 0 J
(e) 0 J
Explanation:
mass of crate, m = 32 kg
distance, s = 7 m
coefficient of friction = 0.26
(a) As it is moving with constant velocity so the force applied is equal to the friction force.
F = 0.26 x m x g = 0.26 x 32 x 9.8 = 81.54 N
(b) The work done on the crate
W = F x s = 81.54 x 7 = 570.75 J
(c) Work done by the friction
W' = - W = - 570.75 J
(d) Work done by the normal force
W'' = m g cos 90 = 0 J
Work done by the gravity
Wg = m g cos 90 = 0 J
(e) The total work done is
Wnet = W + W' + W'' + Wg = 570.75 - 570.75 + 0 = 0 J
The linear speed of the ladybug is 4.1 m/s
Explanation:
First of all, we need to find the angular speed of the lady bug. This is given by:
where
T is the period of revolution
The period of revolution is the time taken by the ladybug to complete one revolution: in this case, since it does 1 revolution every second, the period is 1 second:
T = 1 s
Therefore, the angular speed is
Now we can find the linear speed of the ladybug, which is given by
where:
is the angular speed
r = 65.0 cm = 0.65 m is the distance of the ladybug from the axis of rotation
Substituting, we find
Learn more about angular speed:
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The answer is wind forces and Earth’s rotation
