A fuel oil tank is an upright cylinder, buried so that its circular top is 14 feet beneath ground level. the tank has a radius o
1 answer:
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Answer:
Explanation:
Electrostatic Forces
The force exerted between two point charges and
separated a distance d is given by Coulomb's formula
The forces are attractive if the charges have different signs and repulsive if they have equal signs.
The problem described in the question locates three point charges in a straight line. The charges have the values shown below
The distance between and
is
The distance between and
is
We must find the value of such that
Applying Coulomb's formula for is
Now for
If the total force on is zero, both forces must be equal. Note that being q2 negative, the force on q3 is to the right. The force exerted by q1 must go to the left, thus q1 must be positive. Equating the forces we have:
Simplfying and solving for
11)
12)
13) 50.0 m/s
14) 41.6 m/s
Explanation:
11)
The potential energy of an object is the energy possessed by the object due to its position relative to the ground. It is given by
where
m is the mass of the object
g is the acceleration due to gravity
h is the height relative to the ground
Here in this problem, when the train is at the top, we have:
m = 8325 kg (mass of the train + riders)
(acceleration due to gravity)
h = 127 m (height of the train at the top)
Substituting,
12)
According to the law of conservation of energy, the total mechanical energy of the train must be conserved (in absence of friction). So we can write:
where
is the kinetic energy at the top
is the potential energy at the top
is the kinetic energy at the bottom
is the potential energy at the bottom
The kinetic energy is the energy due to motion; since the train is at rest at the top, we have
Also, at the bottom the height is zero, so the potential energy is zero
Therefore, we find:
13)
The kinetic energy of an object is the energy of the object due to its motion. Mathematically, it is given by
where
m is the mass of the object
v is the speed of the object
From question 12), we know that the kinetic energy of the train at the bottom is
We also know that the mass is
m = 8325 kg
Therefore, we can calculate the speed of the train at the bottom:
14)
At the top of the second hill, the total mechanical energy of the train is still conserved.
Therefore, we can write again:
where
is the kinetic energy at the top of the 1st hill
is the potential energy at the top of the 1st hill
is the kinetic energy at the top of the 2nd hill
is the potential energy at the top of the 2nd hill
From the previous questions, we know that
and
The height of the second hill is
h = 39 m
So we can also find the potential energy at the second hill:
So, the kinetic energy at the second hill is
And so, the speed is
Answer:
0.25m/s
Explanation:
Given parameters
m₁ = 5kg
v₁ = 1.0m/s
m₂ = 15kg
v₂ = 0m/s
Unknown:
velocity after collision = ?
Solution:
Momentum before collision and after collision will be the same. For inelastic collision;
m₁v₁ + m₂v₂ = v(m₁ + m₂)
Insert parameters and solve for v;
5 x 1 + 15 x 0 = v (5 + 15 )
5 = 20v
v = = 0.25m/s