The distance covered by the object is 42.4 m
Explanation:
The motion of the object is a uniformly accelerated motion (at constant acceleration), therefore we can use the following suvat equation:
where
v is the final velocity
u is the initial velocity
a is the acceleration
s is the distance covered
For the object in this problem, we have:
u = 0 (it starts from rest)
v = 24 m/s (final velocity)
Solving for s, we find the distance travelled by the object:
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A) it has a net force because we know that it’s going down in a direction, so there is no force being cancelled out.
Answer:
897
Explanation:
Speed of the car, v = 126 km/h, converting to m/s, we have v = 35 m/s and
Radius of the curve, R = 150 mm = 0.15 m
The centripetal acceleration a(c) is given by the formula = v² / R so that
a(c) = 35² / 0.15
a(c) = 1225 / 0.15
a(c) = 8167 m/s²
The force that causes the acceleration is frictional force = µ m g, where
µ = coefficient of friction
m = the mass of the car and
g = acceleration due to gravity, 9.81
From Newton's law:
µ m g = m a(c) , so that
µ = a(c) / g
µ = 8167 / 9.81
µ = 897
Therefore, the coefficient of static friction must be as big as 897
Answer:
C the baseball began at rest and rolls at a rate of 14.7 m/s after 1.5 seconds
Answer:
(a) the average acceleration of the bus while braking is 8.333 m/s
(b) if the bus took twice as long to stop, the acceleration will be half of the value obtained in part a. [¹/₂ (8.333 m/s) = 4.16 s]
Explanation:
Given;
initial velocity of the bus, v = 25 m/s
time of the motion, t = 3 s
(a) the average acceleration of the bus while braking
a = dv/dt
where;
a is the bus acceleration
dv is change in velocity
dt is change in time
a = 25 / 3
a = 8.333 m/s
(b) If the bus took twice as long to stop, the duration = 2 x 3s
a = 25 / (2 x 3s)
a = ¹/₂ x (25 / 3)
a = ¹/₂ (8.333 m/s) = 4.16 s
Thus, if the bus took twice as long to stop, the acceleration will be half of the value obtained in part a.
