Explanation:
mass of copper block = 0.3 kg at 77 K or -196 °C
Let mass of water taken be m ( maximum) , in this case we will have mixture of water and copper block at 0°C finally.
Heat gained by copper block
= msΔt
.3 x 196 x 385 J
heat lost by water to lower its temp fro 50°C to 0°C
= m x 4186 x 50
Heat lost = heat gained
m x 4186 x 50 = .3 x 196 x 385
m = 108.16 gm
Let mass of water taken be m ( minimum ) , in this case we will have mixture of ice and copper block at 0°C finally.
heat lost by water to lower its temp fro 50°C to 0°C
= m x 4186 x 50
heat lost by water to lower its temp from 0°C to 0°C ice
= m x latent heat of fusion
m x 3.35 x 10⁵
Total heat lost
= m x 3.35 x 10⁵ + m x 4186 x 50
= 544300 m
Heat gained by copper block
= msΔt
.3 x 196 x 385 J
Heat lost = heat gained
544300 m = .3 x 196 x 385 J
m = 41.6 gm
Answer:
Explanation:
the higher up on the Main Sequence the star is, the shorter it lives. The smaller main sequence stars have low luminosities and low surface temperatures. These stars take a very long time to fuse hydrogen into helium and will therefore live a very long life.
Answer:
d_t = 3.05km
v_a = 4.3km/h
Explanation:
42mins*(2/3) = 28mins
42mins-28mins = 14mins
d = v*t
d_1 = (4km/h)*(1h/60mins)*(28mins)
d_1 = 1.87km
d_2 = (5km/h)*(1h/60mins)*(14mins)
d_2 = 1.17km
d_t = d_1+d_2
d_t = 1.87km+1.17km
d_t = 3.05km
v_a = (v_1+v_2)/2
v_a = [(2*4km/h)+5km/h)]/3
v_a = 4.3km/h
Carbon2 hydrogen hydroxide oxygen V carbon dioxide 4 H2O three
In order to answer this, you can use this formula :
F = m.a
which means
a = F/m
a = 18N / 2kg
a = 9 N/kg or 9m/s2
hope this helps