This type of listening response is called back-channel signal. This allows the speaker to know that the listener is attentive or willing to engage a conversation between them. It is shown through short utterances, facial expressions, head nods and others.
In order to answer this exercise you need to use the formulas
S = Vo*t + (1/2)*a*t^2
Vf = Vo + at
The data will be given as
Vf = final velocity = ?
Vo = initial velocity = 1.4 m/s
a = acceleration = 0.20 m/s^2
s = displacement = 100m
And now you do the following:
100 = 1.4t + (1/2)*0.2*t^2
t = 25.388s
and
Vf = 1.4 + 0.2(25.388)
Vf = 6.5 m/s
So the answer you are looking for is 6.5 m/s
Answer:
A) 12.08 m/s
B) 19.39 m/s
Explanation:
A) Down the hill, we will apply Newton’s second law of motion in the downward direction to get:
mg(sinθ) – F_k = ma
Where; F_k is frictional force due to kinetic friction given by the formula;
F_k = (μ_k) × F_n
F_n is normal force given by mgcosθ
Thus;
F_k = μ_k(mg cosθ)
We now have;
mg(sinθ) – μ_k(mg cosθ) = ma
Dividing through by m to get;
g(sinθ) – μ_k(g cosθ) = a
a = 9.8(sin 12.03) - 0.6(9.8 × cos 12.03)
a = -3.71 m/s²
We are told that distance d = 24.0 m and v_o = 18 m/s
Using newton's 3rd equation of motion, we have;
v = √(v_o² + 2ad)
v = √(18² + (2 × -3.71 × 24))
v = 12.08 m/s
B) Now, μ_k = 0.10
Thus;
a = 9.8(sin 12.03) - 0.1(9.8 × cos 12.03)
a = 1.08 m/s²
Using newton's 3rd equation of motion, we have;
v = √(v_o + 2ad)
v = √(18² + (2 × 1.08 × 24))
v = 19.39 m/s
Explanation:
It is given that,
Mass of the ball, m = 1 lb
Length of the string, l = r = 2 ft
Speed of motion, v = 10 ft/s
(a) The net tension in the string when the ball is at the top of the circle is given by :



F = 18 N
(b) The net tension in the string when the ball is at the bottom of the circle is given by :



F = 82 N
(c) Let h is the height where the ball at certain time from the top. So,


Since, 

Hence, this is the required solution.
Explanation:
The structural diversity of carbon-based molecules is determined by following properties:
1. the ability of those bonds to rotate freely,
2.the ability of carbon to form four covalent bonds,
3.the orientation of those bonds in the form of a tetrahedron.