Answer:
1. The final velocity of the truck is 15 m/s
2. The distance travelled by the truck is 37.5 m
Explanation:
1. Determination of the final velocity
Initial velocity (u) = 0 m/s
Acceleration (a) = 3 m/s²
Time (t) = 5 s
Final velocity (v) =?
The final velocity of the truck can be obtained as follow:
v = u + at
v = 0 + (3 × 5)
v = 0 + 15
v = 15 m/s
Therefore, the final velocity of the truck is 15 m/s
2. Determination of the distance travelled
Initial velocity (u) = 0 m/s
Acceleration (a) = 3 m/s²
Time (t) = 5 s
Distance (s) =?
The distance travelled by the truck can be obtained as follow:
s = ut + ½at²
s = (0 × 5) + (½ × 3 × 5²)
s = 0 + (½ × 3 × 25)
s = 0 + 37.5
s = 37.5 m
Therefore, the distance travelled by the truck is 37.5 m
Answer:
force for start moving is 7.49 N
force for moving constant velocity 2.25 N
Explanation:
given data
mass = 7.65 kg
kinetic coefficient of friction = 0.030
static coefficient of friction = 0.10
solution
we get here first weight of block of ice that is
weight of block of ice = mass × g
weight of block of ice = 7.65 × 9.8 = 74.97 N
so here Ff = Fa
so for force for start moving is
Fa = weight × static coefficient of friction
Fa = 74.97 × 0.10
Fa = 7.49 N
and
force for moving constant velocity is
Fa = weight × kinetic coefficient of friction
Fa = 74.97 × 0.030
Fa = 2.25 N
Malleable basically means bendable and shapeable. notebook paper is the easiest to bend and shape
Answer:
speed is 4/3 × g × sin β
acceleration is g sin θ
Explanation:
Given data
mass = M
radius = R
distance = s
angle = β
to find out
final speed and the acceleration
solution
we know that cylinder is rolling on inclined plane
so from conservation of mechanical energy
energy at top = energy at bottom
energy at top will be
= m ×g×h
= m ×g×s sinβ ................1
and
energy at bottom will be
= 1/2 ×m × v² + 1/2 × z × ω²
= 1/2 ×m × v² + 1/2 × 1/2 ×m× r² × (v/r)²
= 3/4 × v² .................2
equating now equation 1 and 2
m ×g×s sinβ = 3/4 × v²
so v = 4/3 × g × sin β
so speed is 4/3 × g × sin β
and acceleration is
= mg sin θ ............3
and
acceleration = ma ............4
now equation equation 3 and 4
ma = mg sin θ
so acceleration is g sin θ
