If 0.484J of heat is added to 0.1372g of water, how much will the temperature increase? 0.263 °C
0.927 °c
0.843 °C
0.460 °C

ΔT = 0.843 °C

Explanation:

Given data:

Heat added = 0.484 J

Mass of water = 0.1372 g

Temperature increase = ΔT = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Specific heat capacity of water is 4.184 j/g.°C

Now we will put the values in formula.

0.484 j = 0.1372 g × 4.184 j/g.°C× ΔT

0.484 j = 0.574j/°C× ΔT

ΔT = 0.484 j / 0.574j/°C

ΔT = 0.843 °C

because of all the things that happen around the us, many people get into arguments which sadly could get physical, but we should actually make a change for that.

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied