Answer:
Friction, normal force, and weight
Explanation:
If the book slows down, it means that there must be friction acting in the opposite direction of the direction the book is moving in.
Weight is caused by the gravitational pull of the Earth on the book, and normal force is the table pushing the book up because the book is pushing down on the table (3rd law.)
Note that weight and normal force is not the 3rd law action-reaction pair. The pair is the force of the book on the table and the force of the table on the book.
Answer:
1 m/s
Explanation:
Impulse = Change in momentum
Force × Time = Mass(Final velocity) - Mass(Initial Velocity)
(1.0)(1.0) = (1.0)(Final Velocity) - (1.0)(0)
Final velocity = <u>1 m/s</u>
Answer:
(a) 2.85 m
(b) 16.5 m
(c) 21.7 m
(d) 22.7 m
Explanation:
Given:
v₀ₓ = 19 cos 71° m/s
v₀ᵧ = 19 sin 71° m/s
aₓ = 0 m/s²
aᵧ = -9.8 m/s²
(a) Find Δy when t = 3.5 s.
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (19 sin 71° m/s) (3.5 s) + ½ (-9.8 m/s²) (3.5 s)²
Δy = 2.85 m
(b) Find Δy when vᵧ = 0 m/s.
vᵧ² = v₀ᵧ² + 2 aᵧ Δy
(0 m/s)² = (19 sin 71° m/s)² + 2 (-9.8 m/s²) Δy
Δy = 16.5 m
(c) Find Δx when t = 3.5 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.5 s) + ½ (0 m/s²) (3.5 s)²
Δx = 21.7 m
(d) Find Δx when Δy = 0 m.
First, find t when Δy = 0 m.
Δy = v₀ᵧ t + ½ aᵧ t²
(0 m) = (19 sin 71° m/s) t + ½ (-9.8 m/s²) t²
0 = t (18.0 − 4.9 t)
t = 3.67
Next, find Δx when t = 3.67 s.
Δx = v₀ₓ t + ½ aₓ t²
Δx = (19 cos 71° m/s) (3.67 s) + ½ (0 m/s²) (3.67 s)²
Δx = 22.7 m
<span>Slowing an
object down is not a means of accelerating it. It actually decelerates the
motion of an object. Speeding it up, changing its direction and applying
balanced forces accelerate an object. In order for an object to accelerate, a force
must be applied. It follows Newton’s second law of motion where it states that
a body at rest remains at rest unless a force is acted upon it. When you move
an object, you are exerting a force onto it. By exerting a force on the object,
you are actually displacing it from its initial position. You cannot apply
force to the object without altering its position. Keep in mind that when you
exert work, you are exerting energy too. </span>
