Answer:
The block will not move.
Explanation:
We'll begin by calculating the frictional force. This can be obtained as follow:
Coefficient of friction (µ) = 0.6
Mass of block (m) = 3 Kg
Acceleration due to gravity (g) = 10 m/s²
Normal reaction (R) = mg = 3 × 10 = 30 N
Frictional force (Fբ) =?
Fբ = µR
Fբ = 0.6 × 30
Fբ = 18 N
From the calculations made above, the frictional force of the block is 18 N. Since the frictional force (i.e 18 N) is bigger than the force applied (i.e 14 N), the block will not move.
Answer:
Keq = 2k₃
Explanation:
We can solve this exercise using Newton's second one
F = m a
Where F is the eleatic force of the spring F = - k x
Since we have two springs, they are parallel or they are stretched the same distance by the object and the response force Fe is the same for the spring age due to having the same displacement
F + F = m a
k₃ x + k₃ x = m a
a = 2k₃ x / m
To find the effective force constant, suppose we change this spring to what creates the cuddly displacement
Keq = 2k₃
Light rays change direction when they hit a mirror. The phenomenon is known as reflection. Light rays travels in a straight light. They strike the surface of the mirror at a particular angle called incident angle. It is the angle between the ray and normal at the point of contact. The rays leaves the surface making the same angle with the normal called reflection angle but in different direction.
Answer:
Rita and Katrina both followed similar paths into the Gulf.
Explanation:
Answer:

Explanation:
= Cambio en la longitud de la cuerda = 0.25 cm
T = tensión en cuerda
A = Área de la cadena = 
d = Diámetro de la cuerda = 0.2 cm
L = Longitud original de la cuerda = 1.6 m
El cambio de longitud de una cuerda viene dado por

La tensión en la cuerda es
.