If you dropped a rock 4.9 m in 1 s how far would it fall in 3 s? express your answer in meters to three significant figures
1 answer:
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a) Density of the liquid:
b) Weight of the liquid: 1372 N
Explanation:
Translation of the text:
<em>"A full tank with liquid weighs 175kg. Which is 5 times the mass of the empty vessel. Knowing that the inside volume of the vessel is 0.17kl, calculate: </em>
<em>a) the density of the liquid; </em>
<em>b) the weight of the liquid."</em>
a)
We know that the full tank with liquid has a total mass of M = 175 kg. We can write the total mass as
(1)
where
is the mass of the liquid
is the mass of the vessel
We also know that the total mass M is 5 times the mass of the empty vessel, so we have:
which is the mass of the empty vessel.
Therefore, we can find the mass of the liquid only using (1):
The density of the liquid is given by
where
m = 140 kg (mass of the liquid)
V = 0.170 kL = 170 L = (volume of the liquid, which is equal to the volume of the vessel)
So we get
b)
The weight of a body is given by
where
m is its mass
g is the acceleration due to gravity
For the liquid in this problem, we have
m = 140 kg (mass)
(acceleration due to gravity)
Therefore, its weight is
Learn more about density:
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Refer to the diagram shown below.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.