Answer:
not sure if im right but this is my problem
3b=7
Step-by-step explanation:
Answer:
Step-by-step explanation:
(8x-6)
=2(4x-3)
Answer:
Quantity of heat needed (Q) = 722.753 × 10³
Step-by-step explanation:
According to question,
Mass of water (m) = 40 kg
Change in temperature ( ΔT) = 18°c
specific heat capacity of water = 4200 j kg^-1 k^-1
The specific heat capacity is the amount of heat required to change the temperature of 1 kg of substance to 1 degree celcius or 1 kelvin .
So, Heat (Q) = m×s×ΔT
Or, Q = 40 kg × 4200 × 18
or, Q = 3024 × 10³ joule
Hence, Quantity of heat needed (Q) = 3024 × 10³ joule
In calories 4.184 joule = 1 calorie
So, 3024 × 10³ joule = 722.753 × 10³
Answer: 13095238095/100000000000
Step-by-step explanation: To write 0.13095238095 as a fraction you have to write 0.13095238095 as numerator and put 1 as the denominator. Now you multiply numerator and denominator by 10 as long as you get in numerator the whole number.
0.13095238095 = 0.13095238095/1 = 1.3095238095/10 = 13.095238095/100 = 130.95238095/1000 = 1309.5238095/10000 = 13095.238095/100000 = 130952.38095/1000000 = 1309523.8095/10000000 = 13095238.095/100000000 = 130952380.95/1000000000 = 1309523809.5/10000000000 = 13095238095/100000000000
