The large truck can rest at a stable equilibrium as compare to the small,from the definition of inertia, Inertia is the tendency of an object to remain at rest or in motion.

6 0

3 years ago

A 0.54 kg particle has a speed of 5.0 m/s at point A and kinetic energy of 7.5 J at point B. What is

Wewaii [24]

Answers

(a) 6.75 Joules.

(b) 5.27 m/s

Explanation

Kinetic energy is the energy possessed by a body in motion.

(a) its kinetic energy at A?

K.E = 1/2 mv²

= 1/2 × 0.54 × 5²

= 6.75 Joules.

(b) its speed at point B?

K.E = 1/2 mv²

7.5 = 1/2 × 0.54 × V²

V² = 7.5 ÷ 0.27

= 27.77778

V = √27.77778

= 5.27 m/s

Work done = 7.5 - 6.75

= 0.75 Joules

4 0

3 years ago

A railroad car of mass 3.45 ✕ 104 kg moving at 2.60 m/s collides and couples with two coupled railroad cars, each of the same ma

Alex

Answer:

a) 1.67 m/s

b) 23kJ

Explanation:

We need to apply the linear momentum conservation formula, that states:

$m1*v_{o1}+m2*v_{o2}=m1*v_{f1}+m2*v_{f2}$

in this case:

$3.45*10^4kg*2.60m/s+2*3.45*10^4kg*1.20m/s=3*m1*v_{f}\\v_f=1.67m/s$

the initital kinetic energy is:

$K_i=\frac{1}{2}*3.45*10^4kg*(2.60m/s)^2+2(\frac{1}{2}*3.45*10^4kg*(1.20m/s)^2\\K_i=167kJ$

and the final:

$K_f=3*\frac{1}{2}*3.45*10^4kg*(1.67m/s)^2\\K_f=144kJ$

The energy lost is given by:

$E_l=|K_f-K_i|\\E_l=23kJ$

8 0

3 years ago