The 1.5kg ball is launched straight upward with an initial velocity of 7m/s. What is the maximum height it will reach?
1 answer:
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The deceleration experienced by the gymnast is the 9 times of the acceleration due to gravity.
Now from Newton`s first law, the net force on gymnast,
Here, W is the weight of the gymnast and a is the acceleration experienced by the gymnast ( acceleration due to gravity)
Therefore,
OR
Given and
Substituting these values in above formula and calculate the force exerted by the gymnast,
a. The free-body diagram for the glass when it is at the top of the circle is attached below.
b. The equation for the net force on the glass at the top of the circle in terms of w, Fn, m, v, and r is mg x g + N - mg x Vtop² /R =0
c. The glass will fall out of the bucket if the normal force between the glass and bucket equals zero. The speed with which she spin the bucket to prevent this from happening is 3.83 m/s.
d. The string will break if the tension on it is more than 100 N. The range of speeds can prevent the string from breaking is 3.83< Vtop<4.99 m/s
<h3>What is Net force?</h3>When two or more forces are acting on the system of objects, then the to attain equilibrium, net force must be zero.
Given, Angela has a bucket of mass 2 kg tied to a string. She places a drinking glass of mass 0.5 kg in the bucket. She spins the bucket in a vertical circle of radius 1.5 m. She must swing the bucket to keep the glass from falling out.
a. The free body diagram of the bucket and glass is attached below.
b. Bucket will undergo centrifugal force
Fb = mVtop² /R
From the equilibrium of forces, we have
For bucket,
T +mb xg - N = mb x Vtop² /R..............(1)
For glass,
mg x g + N = mg x Vtop² /R..............(2)
Thus, this is the net force equation on the glass.
c. On adding both the equations. we have
T + (mb + mg) xg = (mb + mg) Vtop² /R
Substituting the values, T = 0 and from the question, we get
0 + (2+0.5) 9.81 = (2+0.5)(Vtop²/0.5)
Vtop = 3.83 m/s
Thus, the speed of spin to prevent glass from falling out is 3.83 m/s
d. The string will break if the tension on it is more than 100 N
100 + (2+0.5) 9.81 = (2+0.5)(Vtop²/0.5)
Vtop = 4.99 m/s
Thus, the range of velocity is 3.83< Vtop<4.99 m/s
Learn more about net force.
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Answer:
215955.06 m/s^2
Explanation:
length of barrel, s = 0.89 m
initial velocity of the bullet, u = 0 m/s
Final velocity of the bullet, v = 620 m/s
Let a be the acceleration of the bullet in the barrel
Use third equation of motion, we get
a = 215955.06 m/s^2
Thus, the acceleration of the bullet inside the barrel is 215955.06 m/s^2.