Andy has been consuming numbers of calories and recently added and additional 100 calories in the form of a daily snack. Althoug
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
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(a) -2451 N
We can start by calculating the acceleration of the car. We have:
is the initial velocity
v = 0 is the final velocity of the car
d = 125 m is the stopping distance
So we can use the following equation
To find the acceleration of the car, a:
Now we can use Newton's second Law:
F = ma
where m = 1100 kg to find the force exerted on the car in order to stop it; we find:
and the negative sign means the force is in the opposite direction to the motion of the car.
(b)
We can use again the equation
To find the acceleration of the car. This time we have
is the initial velocity
v = 0 is the final velocity of the car
d = 2.0 m is the stopping distance
Substituting and solving for a,
So now we can find the force exerted on the car by using again Newton's second law:
As we can see, the force is much stronger than the force exerted in part a).