A balloon and a mass are attached to a rod that is pivoted at P.
1 answer:
Answer:
The correct option is;
B Move both the balloon and mass 10 cm to the right
Explanation:
Given that the system is in equilibrium, we have;
Force of balloon = ↑
Force of mass = ↓
The direction of the balloon is having an upward motion which gives a clockwise moment or motion to the rod while the direction of the force of the mass weight is downwards, giving the rod an anticlockwise moment
for the rod to rotate clockwise, the moment of the balloon should be larger than that of the rod
At the present equilibrium we have;
× 30 =
× 20
Therefore;
= 1.5×
Moving both balloon and mass 10 cm to the right gives;
The moment of the balloon = × (30 - 10) =
× 20 = 20×
,
The moment of the mass = × (20 - 10) =
× 10
When we substitute = 1.5×
in the moment equation for the mass, we have;
The moment of the mass = × 10 = 1.5×
×10 = 15×
Therefore, the balloon now has a larger momentum than that of the mass and the rod will rotate clockwise.
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Complete Question
The complete question is shown on the first uploaded image
Answer:
The angle between shuttle's velocity and the Earth's field is
Explanation:
From the question we are told that
The length of eire let out is
The emf generated is
The earth magnetic field is
The speed of the shuttle and tether is
The emf generated is mathematically represented as
making the subject of the formula
substituting values
Answer:
<em>The final charge on the 6.0 mF capacitor would be 12 mC</em>
Explanation:
The initial charge on 4 mF capacitor = 4 mf x 50 V = 200 mC
The initial Charge on 6 mF capacitor = 6 mf x 30 V =180 mC
Since the negative ends are joined together the total charge on both capacity would be;
q =
q = 200 - 180
q = 20 mC
In order to find the final charge on the 6.0 mF capacitor we have to find the combined voltage
q = (4 x V) + (6 x V)
20 = 10 V
V = 2 V
For the final charge on 6.0 mF;
q = CV
q = 6.0 mF x 2 V
q = 12 mC
Therefore the final charge on the 6.0 mF capacitor would be 12 mC
If no frictional work is considered, then the energy of the system (the driver at all positions is conserved.
Let
position 1 = initial height of the diver (h₁), together with the initial velocity (v₁).
position 2 = final height of the diver (h₂) and the final velocity (v₂).
The initial PE = mgh₁ and the initial KE = (1/2)mv₁²
where g = acceleration due to gravity,
m = mass of the diver.
Similarly, the final PE and KE are respectively mgh₂ and (1/2)mv₂².
PE in position 1 is converted into KE due to the loss in height from position 1 to position 2.
Therefore
(KE + PE) ₁ = (KE + PE)₂
Evaluate the given answers.
A) The total mechanical energy of the system increases.
FALSE
B) Potential energy can be converted into kinetic energy but not vice versa.
TRUE
C) (KE + PE)beginning = (KE + PE) end.
TRUE
D) All of the above.
FALSE
Let
position 1 = initial height of the diver (h₁), together with the initial velocity (v₁).
position 2 = final height of the diver (h₂) and the final velocity (v₂).
The initial PE = mgh₁ and the initial KE = (1/2)mv₁²
where g = acceleration due to gravity,
m = mass of the diver.
Similarly, the final PE and KE are respectively mgh₂ and (1/2)mv₂².
PE in position 1 is converted into KE due to the loss in height from position 1 to position 2.
Therefore
(KE + PE) ₁ = (KE + PE)₂
Evaluate the given answers.
A) The total mechanical energy of the system increases.
FALSE
B) Potential energy can be converted into kinetic energy but not vice versa.
TRUE
C) (KE + PE)beginning = (KE + PE) end.
TRUE
D) All of the above.
FALSE