Answer:
Explanation:
All the displacement will be converted into vector, considering east as x axis and north as y axis.
5.3 km north
D = 5.3 j
8.3 km at 50 degree north of east
D₁= 8.3 cos 50 i + 8.3 sin 50 j.
= 5.33 i + 6.36 j
Let D₂ be the displacement which when added to D₁ gives the required displacement D
D₁ + D₂ = D
5.33 i + 6.36 j + D₂ = 5.3 j
D₂ = 5.3 j - 5.33i - 6.36j
= - 5.33i - 1.06 j
magnitude of D₂
D₂²= 5.33² + 1.06²
D₂ = 5.43 km
Angle θ
Tanθ = 1.06 / 5.33
= 0.1988
θ =11.25 ° south of due west.
The choices are:
a. Normal Force
b. Gravity Force
c. Applied Force
d. Friction Force
e. Tension Force
f. Air Resistance Force
Answer:
The answer is letter e, Tension Force.
Explanation:
Force refers to the "push" and "pull" of an object, provided that the object has mass. This results to acceleration or a change in velocity. There are many types of forces such as <em>Normal Force, Gravity Force, Applied Force, Friction Force, Tension Force and Air Resistance Force.</em>
The situation above is an example of a "tension force." This is considered the force that is being applied to an object by strings or ropes. This is a type force that allows the body to be pulled and not pushed, since ropes are not capable of it. In the situation above, the tension force of the rope is acting on the bag and this allows the bag to be pulled.
Thus, this explains the answer.
Answer:
Explanation:
The acceleration of a circular motion is given by
where 
is the angular velocity and 
is the radius.
Angular velocity is related to the period, T, by
Substitute into the previous formula.
This acceleration does not depend on the linear or angular displacement. Hence, the amount of rotation does not change it.
Coulomb's law:
Force = (<span>8.99×10⁹ N m² / C²<span>) · (charge₁) · (charge₂) / distance²
= (</span></span><span>8.99×10⁹ N m² / C²<span>) (1 x 10⁻⁶ C) (1 x 10⁻⁶ C) / (1.0 m)²
= (8.99×10⁹ x 1×10⁻¹² / 1.0) N
= 8.99×10⁻³ N
= 0.00899 N repelling.
Notice that there's a lot of information in the question that you don't need.
It's only there to distract you, confuse you, and see whether you know
what to ignore.
-- '4.0 kg masses'; don't need it.
Mass has no effect on the electric force between them.
-- 'frictionless table'; don't need it.
Friction has no effect on the force between them,
only on how they move in response to the force.
</span></span>
Answer:
h’ = 1/9 h
Explanation:
This exercise must be solved in parts:
* Let's start by finding the speed of sphere B at the lowest point, let's use the concepts of conservation of energy
starting point. Higher
Em₀ = U = m g h
final point. Lower, just before the crash
Em_f = K = ½ m 
energy is conserved
Em₀ = Em_f
m g h = ½ m v²
v_b = 
* Now let's analyze the collision of the two spheres. We form a system formed by the two spheres, therefore the forces during the collision are internal and the moment is conserved
initial instant. Just before the crash
p₀ = 2m 0 + m v_b
final instant. Right after the crash
p_f = (2m + m) v
the moment is preserved
p₀ = p_f
m v_b = 3m v
v = v_b / 3
v = ⅓
* finally we analyze the movement after the crash. Let's use the conservation of energy to the system formed by the two spheres stuck together
Starting point. Lower
Em₀ = K = ½ 3m v²
Final point. Higher
Em_f = U = (3m) g h'
Em₀ = Em_f
½ 3m v² = 3m g h’
we substitute
h’= 
h’ = 
h’ = 1/9 h
