Slicing a tomato is still considered a physical change because you are only altering the shape of it by slicing it. It does not change color or make bubble when slicing the tomato. Therefore, it is still a physical change
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Answer:
Q = 7272 Kilojoules.
Explanation:
<u>Given the following data;</u>
Mass = 2.0*101kg = 202kg
Initial temperature, T1 = 10°C
Final temperature, T2 = 90°C
We know that the specific heat capacity of iron = 450J/kg°C
*To find the quantity of heat*
Heat capacity is given by the formula;
Where;
- Q represents the heat capacity or quantity of heat.
- m represents the mass of an object.
- c represents the specific heat capacity of water.
- dt represents the change in temperature.
dt = T2 - T1
dt = 90 - 10
dt = 80°C
Substituting the values into the equation, we have;
Q = 7272KJ or 7272000 Joules.
Answer:
(a) 7 m
(b) 1 m
Explanation:
Given:
The magnitude of displacement vector 'a' is 3 m
The magnitude of displacement vector 'b' is 4 m.
The vector 'c' is the vector sum of vectors 'a' and 'b'.
(a)
Now, when the angle between the vectors is 0°, it means that the vectors are in the same direction. When vectors are in the same direction, then their resultant magnitude is simply the sum of their magnitudes.
So, magnitude of 'c' when 'a' and 'b' are in same direction is given as:

Therefore, the magnitude of vector 'c' is 7 m when angle between 'a' and 'b' is 0°.
(b)
When the angle between the vectors is 180°, it means that the vectors are exactly in the opposite direction. When the vectors are in opposite direction, then their resultant magnitude is the subtraction of their magnitudes.
So, magnitude of 'c' when 'a' and 'b' are in opposite direction is:

Therefore, the magnitude of vector 'c' is 1 m when angle between 'a' and 'b' is 180°.
Answer:
A. To find the mass flow rate.
We use= 220 x 0.355/ 60
= 1.3kg/s
B. Volume flowrate is = mass flowrate / density
But density is 1000kg/m³
= 1.3kg/s/ 1000kg/m³
= 0.0013m³/s
C. Flow speead at 1
= 0.0013m³/s / (2 x 10-2m)²
= 6.5m/s
D.flow speed at 2
0.0013m³/s / (8x 10-2m)²
=1.63m/s
E. Gauge pressure at point 1
= 152+ 1/1000 ( 1.63)²- 6.5² + 1000( 9.8) ( 0-1.35)
= 119kpa