From rest a balloon begins to rise. the Balloon experiences a lift due to a buoyant force of 8000 N. 1. Total massof the balloon is 600kg. Find Acceleration. 2. The balloon rises a distance of 10m and loses a mass of 50 kg after a weight is thrown off. Find the acceleration of the balloon after this weight loss 3. Find the initial velocity of the mass in the vertical direction at the timeof the jump.

Elements Of Electromagnetics
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From rest a balloon begins to rise. the Balloon experiences a lift due to a buoyant force of 8000 N.
1. Total massof the balloon is 600kg. Find Acceleration.
2. The balloon rises a distance of 10m and loses a mass of 50 kg after a weight is thrown off. Find the
acceleration of the balloon after this weight loss
3. Find the initial velocity of the mass in the vertical direction at the timeof the jump.
4. After falling 10 m, find the mass' final velocity as it comesinto contact with the ground.
5.What is the total time the mass takes to hit the ground?
6. Find the heightof the balloon when the mass hits the ground
7. The mass lands 3 m from the baloon, find the horizontal velocity required to land 3m from the balloon.
8. Since the balloon exeperiences equal forces when the mass is thrown off, what is the horizontal velocity the
baloon experiences during this act and how far does the balloon travel when the mass comes into contact
with the ground?
Transcribed Image Text:From rest a balloon begins to rise. the Balloon experiences a lift due to a buoyant force of 8000 N. 1. Total massof the balloon is 600kg. Find Acceleration. 2. The balloon rises a distance of 10m and loses a mass of 50 kg after a weight is thrown off. Find the acceleration of the balloon after this weight loss 3. Find the initial velocity of the mass in the vertical direction at the timeof the jump. 4. After falling 10 m, find the mass' final velocity as it comesinto contact with the ground. 5.What is the total time the mass takes to hit the ground? 6. Find the heightof the balloon when the mass hits the ground 7. The mass lands 3 m from the baloon, find the horizontal velocity required to land 3m from the balloon. 8. Since the balloon exeperiences equal forces when the mass is thrown off, what is the horizontal velocity the baloon experiences during this act and how far does the balloon travel when the mass comes into contact with the ground?
Step 1
Since you have posted a question with multiple sub-parts, we will solve the first three sub-parts for you.
To get the remaining sub-part solved please repost the complete question and mention the sub-parts to
be solved.
In the given question,
The balloon begins to rise from rest (u) = 0
The balloon experiences a buoyant force of (F) = 8000 N
Mass of the balloon is (m) = 600 kg
Step 2
(a) The balloon experiences a buoyant force (upper force) of 8000 N.
Newton's second law of motion states that the force on the object is equal to the product of mass and its
acceleration.
Thus,
F=8000=ma=8000=a=8000m=8000800-13.33m/s2
which is the required acceleration of the balloon.
Step 3
2.
The ball rises to a distance of 10m (s) = 10m
Now, the initial velocity is (u) = 0m/s
Acceleration (a) = 13.35 m/s²
The balloon loses a mass of 50kg and now the mass of the balloon becomes 600kg-50kg = 550kg
Using Newton's second law of motion, we get
F=ma⇒a= Fm=3000N550 kg-15.54m/s2
which is the acceleration of the balloon after weight loss.
Step 4
3.
Let the velocity of the balloon at the end of 10m is v.
Using Newton's first equation of motion we get
v2=u2+2as=v2 = 2x13.33x10-266.67 v 18.33m/s
Now, when the mass of 50kg jumps from the balloon, it will have the same velocity as the velocity of the
balloon at that point of time.
So, the initial velocity of the mass of 50kg will be 18.33m/s.
Transcribed Image Text:Step 1 Since you have posted a question with multiple sub-parts, we will solve the first three sub-parts for you. To get the remaining sub-part solved please repost the complete question and mention the sub-parts to be solved. In the given question, The balloon begins to rise from rest (u) = 0 The balloon experiences a buoyant force of (F) = 8000 N Mass of the balloon is (m) = 600 kg Step 2 (a) The balloon experiences a buoyant force (upper force) of 8000 N. Newton's second law of motion states that the force on the object is equal to the product of mass and its acceleration. Thus, F=8000=ma=8000=a=8000m=8000800-13.33m/s2 which is the required acceleration of the balloon. Step 3 2. The ball rises to a distance of 10m (s) = 10m Now, the initial velocity is (u) = 0m/s Acceleration (a) = 13.35 m/s² The balloon loses a mass of 50kg and now the mass of the balloon becomes 600kg-50kg = 550kg Using Newton's second law of motion, we get F=ma⇒a= Fm=3000N550 kg-15.54m/s2 which is the acceleration of the balloon after weight loss. Step 4 3. Let the velocity of the balloon at the end of 10m is v. Using Newton's first equation of motion we get v2=u2+2as=v2 = 2x13.33x10-266.67 v 18.33m/s Now, when the mass of 50kg jumps from the balloon, it will have the same velocity as the velocity of the balloon at that point of time. So, the initial velocity of the mass of 50kg will be 18.33m/s.
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