Do you know what you need to do to solve this problem?
Answer:
(2×2×2)×(2×2×2)×(2×2×2)×(2×2×2)×-(2×2×2)×(2×2×2)
Addition will be the last operation
Step 1: We make the assumption that 498 is 100% since it is our output value.
Step 2: We next represent the value we seek with $x$x.
Step 3: From step 1, it follows that $100\%=498$100%=498.
Step 4: In the same vein, $x\%=4$x%=4.
Step 5: This gives us a pair of simple equations:
$100\%=498(1)$100%=498(1).
$x\%=4(2)$x%=4(2).
Step 6: By simply dividing equation 1 by equation 2 and taking note of the fact that both the LHS
(left hand side) of both equations have the same unit (%); we have
$\frac{100\%}{x\%}=\frac{498}{4}$
100%
x%=
498
4
Step 7: Taking the inverse (or reciprocal) of both sides yields
$\frac{x\%}{100\%}=\frac{4}{498}$
x%
100%=
4
498
$\Rightarrow x=0.8\%$⇒x=0.8%
Therefore, $4$4 is $0.8\%$0.8% of $498$498.
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