How does a shiny , silver and smoothsurface reduce the heat loss of a hot object to its cooler surroundings

Shiny surface are poor radiators of heat

A shiny , silver and smooth surface reduce the heat loss of a hot object to its cooler surroundings because good radiators and good absorbed of heat have dull, black and rough surfaces while on the other hand poor radiators have smooth and shiny surfaces therefore they reduce the heat loss of a hot object.

3

step-by-step explanation:

10 - 2.8 = 7.2

2 2/5 = 2.4

7.2 / 2.4 = 3

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s