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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Answer: The angle between the wire segment and the magnetic field 66.42°
Explanation:
Please see the attachment below
<u>Answer:</u>
Maximum height reached = 35.15 meter.
<u>Explanation:</u>
Projectile motion has two types of motion Horizontal and Vertical motion.
Vertical motion:
We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.
Considering upward vertical motion of projectile.
In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g and final velocity = 0 m/s.
0 = u sin θ - gt
t = u sin θ/g
Total time for vertical motion is two times time taken for upward vertical motion of projectile.
So total travel time of projectile = 2u sin θ/g
Horizontal motion:
We have equation of motion , , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 and time taken = 2u sin θ /g
So range of projectile,
Vertical motion (Maximum height reached, H) :
We have equation of motion, , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.
Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H
In the give problem we have R = 301.5 m, θ = 25° we need to find H.
So
Now we have
So maximum height reached = 35.15 meter.