Please answer and provide an explanation. Thank you in advance. A bullet of mass
ID: 1307618 • Letter: P
Question
Please answer and provide an explanation. Thank you in advance.
A bullet of mass 12.0g is fired into a ballistic pendulum of mass 2.60kg. The bullet passes through pendulum bob and emerges with a speed of 50.0m/s. The pendulum bob reaches a maximum height of 15.00cm. Write what speed the bullet was fired? The person 's spine and upper body are represented as uniform horizontal beam of weight 400.0N, pivoted at the base of the spine... The erector spinalis muscle is attached at a point that is 0.72L from the base of the spine, where L is the length of the spine. The muscle makes an angle of 12.5 degree with the spine. The person is lifting an object of mass 25.00kg.Explanation / Answer
7.initial kinetic energy of bullet=final kinetic energy of bullet+potential energy of bob
0.5*0.012*(v^2)=0.5*0.012*(50^2) + 2.6*9.8*(15/100)
v=56 m/s
5.a) using Continuity eqn.-
A1*V1=A2*V2
pi*((d/2)^2)*3.45=pi*((d/4)^2)*v
v=13.8 m/s
b) using bernoullis eqn-
P + 0.5*rho*(v1 ^2)+ rho*g*h =constant
P+ (0.5*1000*(3.45^2) + 0 = 120k + 0.5*1000*(13.8^2) + 1000*9.8*25
P=454.26K Pa
8. a)change in potential energy of the system- m*g*h=m*g*lcos65
42.5*9.8*8.25cos65 = 1452.17 Joules
b) energy dissipated by friction=loss in potential energy of the system=1452.17 Joules
c) kinetic energy of the boy=change in the potential energy of the system
(0.5*42.5*v^2)=1452.17
v=8.26 m/s
Q6. a. Balancing the torque-(l is length of horizontal beam)
clockwise torque=anti-clockwise torque
torque by load + torque by mass of horizontal beam = torque by spinal muscles
25*9.8*l +400*l/2 = T*cos(90-12.5)*0.72l
T=2855.55 N
b. since all torques are balanced and the only force left is due to the t*cos12.5 component of the spinal muscles,the force on the base is-
Fx=Tcos12.5
2787.86 N along the horizontal beam.
Q2. a. temp at point A,applying PV=nRT ( R=0.0821 lts-atm / k /mol)
2.6 * 20=2 * 0.0821 *T
T=316.68 K
temp at B-
PV=nRT
2.6*40=2*0.0821*T
T=633.37 K
temp at C-
PV=nRT
1.3*40=2*.0821*T
T=316.68 K
b. Work done in Process 1 = P * delta(V) = 2.6*(40-20) = 52 J
Work done in Process 2 = 1.3*(40-40) = 0 J
Work done in Process 3 = work done in isothermal process= 2.3026* n*R*T*log(V2/V1) = 2.3026*2*0.0821*316.68*log(20/40) = -36.04 J
c. By law of conservation of energy-
Initial System energy(0.5*n*R*T) = Energy used in all process + delta H
0.5*2*0.0821*316.68=Process 1 + Process 2 + Process 3 + delta H
25.99=52 + 0 +(-36.04) + delta H
delta H = 10.03 J