Answer:
paki sagot naman pls para my isagot ako sa module ko ty
To solve this problem, we begin by first calculating the area of the front lawn. The length and width of the lawn was given and the area of a rectangle is given by the formula: Area = length x width. Thus, the area of the front lawn can be obtained by multiplying 18 ft by 20 ft, wherein we get 360 ft^2 as the area.
Second, the problem indicated that each square foot of lawn accumulates 1450 new snow flakes per minute. This can be translated into the expression 1450 snow flakes/ (minute·ft^2). In this way, we can convert it to units of mass (kg). Afterwards, we simply need to multiply it to the area of the lawn and convert minute to hour. The following expression is then used:
1450 snow flakes/ (minute·ft^2) x 1.90 mg/snow flake x 1 g/1000 mg x 1kg/1000 g x 360 ft^2 x 60 minutes/hour = 59.508 kg snow flake/hour
It is then calculated that 59.508 kg of snow flake accumulates in the lawn every hour.
Liquids and gases. Hope that helps!
The volume of cylinder is 100 mL, inner diameter is 3.2 cm thus, radius will be:
Volume of layer formed=
...... (1)
Volume of layer is also equal to sum of volume of gasoline and water.
Density of gasoline is 
and mass is 34 g thus, volume of gasoline will be:
Now, density of water is 
and mass is 34 g thus, volume of water will be:
Adding both the volumes, volume of layer will be:
Putting the values in equation (1) to solve for height of the layer,
Thus, height of layer is 10.25 cm
Answer:
0.51 cal/g.°C
Explanation:
Step 1: Given data
- Added energy in the form of heat (Q): 14 cal
- Mass of the liquid (m): 12 g
- Initial temperature: 10.4 °C
- Final temperature: 12.7 °C
Step 2: Calculate the temperature change
ΔT = 12.7 °C - 10.4 °C = 2.3 °C
Step 3: Calculate the specific heat of the liquid (c)
We will use the following expression.
c = Q / m × ΔT
c = 14 cal / 12 g × 2.3 °C = 0.51 cal/g.°C
