please answer the all following question.Thank you ,please show step to the answ

Question

please answer the all following question.Thank you ,please show step to the answer.

answer is listed with the question with out the steps.

1) In the figure, a uniform ladder 12 meters long rests against a vertical frictionless wall. The ladder weighs 400 N and makes an angle of 79° with the floor. A man weighing 790 N climbs slowly up the ladder. When he has climbed to a point that is 7.8 m from the base of the ladder, the ladder starts to slip. What is the coefficient of static friction between the floor and the ladder? Ans: 0.12

2)In the figure, a 10.0-m long bar is attached by a frictionless hinge to a wall and held horizontal by a rope that makes an angle of 53° with the bar. The bar is uniform and weighs 39.9 N. How far from the hinge should a 10.0-kg mass be suspended for the tension T in the rope to be 125 N? Ans: 8.15 m from the hinge

3) A plate falls vertically to the floor and breaks up into three pieces, which slide along the floor. Immediately after the impact, a 200-g piece moves along the x-axis with a speed of 2.00 m/s and a 235-g piece moves along the y-axis with a speed of 1.50 m/s. The third piece has a mass of 100 g. In what direction does the third piece move? Ans: 221.4° from the x-axis.

4) A 320-g air track cart is traveling at 1.25 m/s and a 270-g cart traveling in the opposite direction at 1.33 m/s. What is the speed of the center of mass of the two carts? Ans: 0.0693 m/s

5) A piece of thin uniform wire of mass m and length 3b is bent into an equilateral triangle. Find the moment of inertia of the wire triangle about an axis perpendicular to the plane of the triangle and passing through one of its vertices. Ans: mb2

6) In the figure, two blocks, of masses 2.00 kg and 3.00 kg, are connected by a light string that passes over a frictionless pulley of moment of inertia 0.00400 kg · m2 and radius 5.00 cm. The coefficient of friction for the tabletop is 0.300. The blocks are released from rest. Using energy methods, find the speed of the upper block just as it has moved 0.600 m. Ans: 1.40 m/s

7) In the figure, a very light rope is wrapped around a wheel of radius R = 2.0 meters and does not slip. The wheel is mounted with frictionless bearings on an axle through its center. A block of mass 14 kg is suspended from the end of the rope. When the system is released from rest it is observed that the block descends 10 meters in 2.0 seconds. What is the moment of inertia of the wheel? Ans: 54 kg · m2

8) A turntable has a radius of 0.80 m and a moment of inertia of 2.0 kg · m2. The turntable is rotating with an angular velocity of 1.5 rad/s about a vertical axis though its center on frictionless bearings. A very small 0.40- kg ball is projected horizontally toward the turntable axis with a velocity of 3.0 m/s. The ball is caught by a very small and very light cup-shaped mechanism on the rim of the turntable (see figure). What is the angular velocity of the turntable just after the ball is caught? Ans: 1.3 rad/s

9) An 8.0-g bullet is shot into a 4.0-kg block, at rest on a frictionless horizontal surface (see the figure). The bullet remains lodged in the block. The block moves into an ideal massless spring and compresses it by 8.7 cm. The spring constant of the spring is 2400 N/m. The initial velocity of the bullet is closest to? Ans: 1100 m/s.

10) A solid sphere is rolling without slipping along a horizontal surface with a speed of 4.50 m/s when it starts up a ramp that makes an angle of 25.0° with the horizontal. What is the speed of the sphere after it has rolled 3.00 m up the ramp? Ans: 1.58 m/s

Explanation / Answer

1)

T A = - 6 * cos 79* 400 – 7.8*cos 79*790 + 12 F N2 = 0

- 457.94 – 1175.76 + 12 F N2 = 0

F N2 = 136.142 N

FN2 = cos 11 * 136.142 N = 133.64 N

F y = FN1 – F2 – Fm = 0

FN1 = 790 + 400 = 1190 N

F x = FN2 – Ffx = 0

FN2 = Ffx = * F N1

=> = FN2 / F N1

=> = 133.64 / 1190

=> = 0.112 <--- ans.

2)

L*Tsin - xw1 – (L/2)*w2 = 0 or

x = L*[Tsin - (1/2)*w2 ] / w1

x = 10*[125sin53 - (1/2)*39.9 ] / 10*9.81

x = 10*[125sin53 - (1/2)*39.9 ] / 10*9.81

x = 8.14 m <--- ans.

3) Py = 0.235*1.5 = 0.3525

Px = 0.200*2.0 = 0.4

Direction , = 180 + tan^-1 (Py / Px)

Direction , = 180 + tan^-1 (0.3525 / 0.4)

= 221.38° <--- ans.

4) vcm = [(320 g)(1.25 m/s) + (270 g)(-1.33 m/s)] / (320 + 270 g)
vcm = 0.0693 m/s <--- ans.

6) The net force on the pulley is then
Fnet = T1 - T2 = 2g - 2a - 3a - 3*0.3g
Fnet = 1.1g - 5a
and the torque on the pulley is then
= RF = I* = I*a/R with = a/R ( = angular acceleration, I = moment of inertia, R = 0.05 m)
(1.1g - 5a)*0.05 = 0.004*a/0.05
(1.1g - 5a)*0.05^2 = 0.004a
1.1*9.81*0.05^2 - 0.05^2*5a = 0.004a
a = 1.635 m/s^2
since we know a now, we can find v with
v = (2as)
v = (2*1.635*0.6)
v = 1.4007 m/s <--- ans.

7) First find the acceleration of the system.
s = a×t²/2
(10 m) = a×(2 s)²/2
a = 5 m/s²

Now find the tension in the rope.
m×g - T = m×a
(14 kg)×(9.8 m/s²) - T = (14 kg)×(5 m/s²)
T = 67.2 N

And finally, find the moment of inertia.
T×r = I×a/r
(67.2 N)×(2 m) = I×(5 m/s²)/(2 m)
I = 53.76 kgm² <--- ans.

8)Angular momentum I*w is conserved

Initial I*w = 2.0 * 1.5 kg.m^2/s = 3.0 kg.m^2/s = Final I*w

The moment of inertia of a point mass m rotating at radius r is m*r^2
= 0.4kg * 0.8^2 m^2
= 0.256 kg.m^2

So the total moment of inertia just after the ball is caught is 2.256 kg.m^2

So the new angular velocity is 3.0 / 2.256 rad/s = 1.33 rad/s <--- ans.

9) Energy is NOT conserved in the bullet/block collision.

It is a totally inelastic collision. Momentum IS conserved, however:
1) Vbu*m = Vbl*(m+M)
1a) Vbu = Vbl*(M+m)/m

Energy IS conserved when the block hits the spring:
2) ½(M+m)*Vbl² = ½k*x² Vbl = x[k/(M+m)] = 2.129 m/s

Inserting this value into (1a),

Vbu = 1066.6 m/s <--- ans.

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