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nlexa [21]
4 years ago
15

You throw a 20-n rock vertically into the air from ground level. you observe that when it is a height 16.0 m above the ground, i

t is traveling at a speed of 24.7 m/s upward.
Physics
1 answer:
timama [110]4 years ago
6 0

The problem is missing some parts. But here are the questions and answers.

a.       Use the work-energy theorem to find its speed just as it left the ground 

Initial Kinetic Energy = GPE + KE

1/2 * m * V^2 = m * g * h + 1/2 * m * v^2

V^2 = 2 * 9.8 * 15.6 + 25.6^2

=sqrt 968.96

V (initial speed) ≈ 31.130 m/s

 

b.      Use the work-energy theorem to look for its maximum height. 

H = 968.96 / 2 * 9.8 ≈ 49.44 <span>m</span>

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Water flows from the bottom of a storage tank at a rate of r(t) = 300 − 6t liters per minute, where 0 ≤ t ≤ 50. Find the amount
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r(t) models the water flow rate, so the total amount of water that has flowed out of the tank can be calculated by integrating r(t) with respect to time t on the interval t = [0, 35]min

∫r(t)dt, t = [0, 35]

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1. Stops: Using the information learned in this course, explain three things you will not do when driving.
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The Great Sandini is a 60 kg circus performer who is shotfrom a cannon (actually a spring gun). You don't find many men ofhis ca
d1i1m1o1n [39]

Answer:

V=15.3 m/s

Explanation:

To solve this problem, we have to use the energy conservation theorem:

U_e+K_i+U_{gi}+W_{friction}=K_f+U_{gf}

the elastic potencial energy is given by:

U_e=\frac{1}{2}*k*x^2\\U_e=\frac{1}{2}*1100N/m*(4m)^2\\U_e=8800J

The work is defined as:

W_{friction}=F_f*d*cos(\theta)\\W_{friction}=40N*2.5m*cos(180)\\W_{friction}=-100J

this work is negative because is opposite to the movement.

The gravitational potencial energy at 2.5 m aboves is given by:

U_{gf}=m*g*h\\U_{gf}=60kg*9.8*2.5\\U_{gf}=1470J

the gravitational potential energy at the ground and the kinetic energy at the begining are 0.

8800J+0+0+-100J=\frac{1}{2}*62kg*v^2+1470J\\v=\sqrt{\frac{2(8800J-100J-1470J)}{62kg}}\\v=15.3m/s

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3 years ago
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