The first law is about force or push and pull
A mechanical wave<span> is a </span>wave<span> that is an oscillation of </span>matter<span>, and therefore transfers energy through a </span>medium.[1]<span> While waves can move over long distances, the movement of the </span>medium of transmission<span>—the material—is limited. Therefore, oscillating material does not move far from its initial equilibrium position. Mechanical waves transport energy. This energy propagates in the same direction as the wave. Any kind of wave (mechanical or electromagnetic) has a certain energy. Mechanical waves can be produced only in media which possess elasticity and inertia.</span>
Answer:
See the explanation below
Explanation:
Density is defined as the relationship between mass and volume, i.e. the following equation can be used:
density = m/v
where:
density [kg/m^3]
m = mass [kg]
v = volume [m^3]
If we change the volume of a body by reducing its size, its mass will also decrease proportionally with a density as seen in the equation.
m = density*v
To understand this concept more clearly, let's use the following example:
We know that the density of water is equal to 1000 [kg/m^3], that is, 1 cubic meter of water contains 1000 kilograms of water, using the equation.
1000 = m /1
m = 1000*1 = 1000 [kg]
Now if we have 500 kilograms of water, that would pass with the volume so that the density remains constant.
1000 = 500/v
v = 500/1000
v = 0.5 [m^3]
We can see that the volume of water has halved. Since the mass of water was reduced by half. That is, the relationship between mass and volume is proportional to the density of the material or substance.
Answer:
a = 3,0 m/s²
Explanation:
En este ejercicio se pide calcular la aceleracion del cuerpo, usemos las ecuaciones de cinematica en una dimensión.
v= v₀ + a t
como el corredor parte del reposo si velocidad inicial es cero
v = at
a = v/t
calculemos
a = 12 /4,0
a = 3,0 m/s²
Answer:
a)5.88J
b)-5.88J
c)0.78m
d)0.24m
Explanation:
a) W by the block on spring is given by
W= 
kx² = (530)(0.149)² = 5.88 J
b) Workdone by the spring = - Workdone by the block = -5.88J
c) Taking x = 0 at the contact point we have U top = U bottom
So, mg
= kx² - mgx
And, = ( kx² - mgx
)/(mg) = 
]/(0.645x9.8)
= 0.78m
d) Now, if the initial initial height of block is 3
= 3 x 0.78 = 2.34m
then, kx² - mgx - mg =0
(530)x² - [(0.645)(9.8)x] - [(0.645)(9.8)(2.34) = 0
265x² - 6.321x - 14.8 = 0
a=265
b=-6.321
c=-14.8
By using quadratic eq. formula, we'll have the roots
x= 0.24 or x=-0.225
Considering only positive root:
x= 0.24m (maximum compression of the spring)
