Storm water picks up pollutants off of these types of surfaces because they are not able to absorb rainfall
1 answer:
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a) The work done is 920 J
b) The work done is 5200 J
Explanation:
a)
In this first part of the problem, the crate is moved horizontally at constant speed.
The work required in this case is given by
where
F is the magnitude of the force applied
d is the displacement of the crate
is the angle between the direction of the force and of the displacement
Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so
F = 230 N
The displacement is
d = 4.0 m
And the angle is , since the force is applied horizontally. Therefore, the work done is
b)
In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore
F = W = 1300 N
The displacement this time is again
d = 4.0 m
And the angle is , since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is
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Answer:
a) The angular acceleration of the beam is 0.5 rad/s²CW (direction clockwise due the tangential acceleration is positive)
b) The acceleration of point A is 3.25 m/s²
The acceleration of point E is 0.75 m/s²
Explanation:
a) The relative acceleration of B with respect to D is equal:
Where
aB = absolute acceleration of point B = 2.5 j (m/s²)
aD = absolute acceleration of point D = 1.5 j (m/s²)
(aB/D)n = relative acceleration of point B respect to D (normal direction BD) = 0, no angular velocity of the beam
(aB/D)t = relative acceleration of point B respect to D (tangential direction BD)
We have that
(aB/D)t = BDα
Where α = acceleration of the beam
BDα = 1 m/s²
Where
BD = 2
b) The acceleration of point A is:
(aA/D)t = ADαj
The acceleration of point E is:
(aE/D)t = -EDαj
Answer:
a) 3.37 x
b) 6.42kg/
Explanation:
a) Firstly we would calculate the volume of the metal using it`s weight in air and water , after finding the weight we would find the density .
Weight of metal in air = 50N = mg implies the mass of metal is 5kg.
Now the difference of weight of the metal in air and water = upthrust acting on it = volume (metal) p (liquid) g = V (1000)(10) = 14N. So volume of metal piece = 14 x
. So density of metal = mass of metal / volume of metal = 5 / 14 x
= 3.37 x
b) Water exerts a buoyant force to the metal which is 50−36 = 14N, which equals the weight of water displaced. The mass of water displaced is 14/10 = 1.4kg Since the density of water is 1kg/L, the volume displaced is 1.4L. Hence, we end up with 3.57kg/l. Moreover, the unknown liquid exerts a buoyant force of 9N. So the density of this liquid is 6.42kg/