Answer:
The required specific heat is 196.94 joule per kg per °C
Step-by-step explanation:
Given as :
The heat generated = Q = 85.87 J
Mass of substance (m)= 34.8 gram = 0.0348 kg
Change in temperature = T2 - T1 = 34.29°C - 21.76°C = 12.53°C
Let the specific heat = S
Now we know that
Heat = Mass × specific heat × change in temperature
Or, Q = msΔt
Or, 85.87 = (0.0348 kg ) × S × 12.53°C
Or , 85.87 = 0.4360 × S
Or, S = 
∴ S = 196.94 joule per kg per °C
Hence the required specific heat is 196.94 joule per kg per °C Answer
Answer:
A
Step-by-step explanation:
Find the volume of both shapes.
Cylinder - V = πr²h = π·2²·4 = 50.26548
Sphere - V = 4/3πr³ = 4/3π·2³ = 33.51032
Find the difference between the two.
50.26548 - 33.51032 = 16.75516, or 16.75
-hope it helps
Answer:
m∠EGF = 65° and m∠CGF = 115°
Step-by-step explanation:
Given;
∠EFG = 50°
EF = FG
Solution,
In ΔEFG m∠EFG = 50° and EF = FG.
Since triangle is an isosceles triangle hence their base angles are always equal.
∴
Let the measure of ∠EGF be x.
∴ 
Now by angle Sum property which states "The sum of all the angles of a triangle is 180°."
m∠EFG + m∠FEG + ∠EGF = 180
Hence
m∠EGF = 65°
Also 'The sum of angles that are formed on a straight line is equal to 180°."
m∠EGF + m∠CGF = 180°
65° + m∠CGF = 180°
m∠CGF = 180° - 65° = 115°
Hence m∠EGF = 65° m∠CGF = 115°
Answer:
Yes, r and h can be changed to produce same volume by;
Increasing "r" and decreasing "h" or by decreasing "r" and increasing "h".
Step-by-step explanation:
What the question is simply asking is if we can get same volume of a particular cylinder if we change the radius and height.
Now, volume of a cylinder is;
V = πr²h
Now, for the volume to remain the same, if we increase "r", it means we have to decrease "h", likewise, if we decrease "r", we now have to increase "h".
Fourty six thousand and fifty six. In expanded form it'll look like this:
40,000
6,000
0
50
6
