By definition we have to:
Applied force: It is the external force that acts directly on a body.
Therefore, we can say that if you have an object and push it towards yourself, you are exerting an external force on the object.
This external force was not acting on the object previously, therefore, it is a force that you are applying at that moment.
Answer:
you exert an Applied Force on an object when you pull it towards you
A. Applied Force
Answer:
<em>The final velocity is 20 m/s.</em>
Explanation:
<u>Constant Acceleration Motion</u>
It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.
Being a the constant acceleration, vo the initial speed, and t the time, the final speed can be calculated as follows:

The provided data is: vo=10 m/s,
, t=2 s. The final velocity is:


The final velocity is 20 m/s.
Well you need to have lots of heat
Answer:
a) 0.05s
b) 4000N
Explanation:
a)When car is stopped its final velocity become zero
U- 10 m/s
V- 0 m/s
S - 0.25 m
t -?
S = (v+u)*t/2
0.25 =(10+0)*t/2
t = 0.05s
b) If we happened to calculate the avarage force we have to consider about acceleration
V= 0
U = 10
t = 0.05 s
a =?
V = U + at
0 = 10 -a * 0.05
a = 200 m/s2
F = m *a
= 20 * 200
= 4000N
Answer:
a) F = 64.30 N, b) θ = 121.4º
Explanation:
Forces are vector quantities so one of the best methods to add them is to decompose each force and add the components
let's use trigonometry
Force F1
sin 170 = F_{1y} / F₁
cos 170 = F₁ₓ / F₁
F_{1y} = F₁ sin 170
F₁ₓ = F₁ cos 170
F_{1y} = 100 sin 170 = 17.36 N
F₁ₓ = 100 cos 170 = -98.48 N
Force F2
sin 30 = F_{2y} / F₂
cos 30 = F₂ₓ / F₂
F_{2y} = F₂ sin 30
F₂ₓ = F₂ cos 30
F_{2y} = 75 sin 30 = 37.5 N
F₂ₓ = 75 cos 30 = 64.95 N
the resultant force is
X axis
Fₓ = F₁ₓ + F₂ₓ
Fₓ = -98.48 +64.95
Fₓ = -33.53 N
Y axis
F_y = F_{1y} + F_{2y}
F_y = 17.36 + 37.5
F_y = 54.86 N
a) the magnitude of the resultant vector
let's use Pythagoras' theorem
F = Ra Fx ^ 2 + Fy²
F = Ra 33.53² + 54.86²
F = 64.30 N
b) the direction of the resultant
let's use trigonometry
tan θ’= F_y / Fₓ
θ'= 
θ'= tan⁻¹ (54.86 / (33.53)
θ’= 58.6º
this angle is in the second quadrant
The angle measured from the positive side of the x-axis is
θ = 180 -θ'
θ = 180- 58.6
θ = 121.4º