We will convert the 1dm3 in terms of cm3 as follows:
1dm^3 = (10 cm)^3
= 1000 cm^3
The mass of platinum is equal to 900 lb.
Then we will convert the mass in terms of grams as follows:
1 lb = 453.6 g
900 = 900 x 453.6 g
= 408240 g
Then density of platinum is equal to 21.4 g/cm^3
We will calculate the volume of platinum in mass 408240 g as follows:
Volume of platinum = mass of platinum / density of platinum
= 408240 g / 21.4 g/cm^3
= 19076.6 cm^3
The total volume of platinum is 19076.6 cm^3
The volume of platinum in 1 L bar is 1000cm^3
So, to calculate the number of bars we will use the formula as follows;
Number of bars = volume of platinum available / volume of platinum required in 1 L bar
= 19076.6 cm^3 / 1000 cm^3
= 19
So, the number of bars are 19.
Answer:
The maximum height is 2881.2 m.
Explanation:
Given that,
Acceleration = 29.4 m/s²
Time = 7.00 s
We need to calculate the distance
Using equation of motion
Put the value into the formula
We need to calculate the velocity
Using formula of velocity
Put the value into the formula
We need to calculate the height
Using formula of height
Put the value into the formula
We need to calculate the maximum height
Using formula for maximum height
Put the value into the formula
Hence, The maximum height is 2881.2 m.
Answer:
True
Explanation:
Pre-questioning may help a reader focus on information s/he hopes to find in the reading selection.
Answer:
The work done to get you safely away from the test is 2.47 X 10⁴ J.
Explanation:
Given;
length of the rope, L = 70 ft
mass per unit length of the rope, μ = 2 lb/ft
your mass, W = 120 lbs
mass of the 70 ft rope = 2 lb/ft x 70 ft
= 140 lbs.
Total mass to be pulled to the helicopter, M = 120 lbs + 140 lbs
= 260 lbs
The work done is calculated from work-energy theorem as follows;
W = Mgh
where;
g is acceleration due gravity = 32.17 ft/s²
h is height the total mass is raised = length of the rope = 70 ft
W = 260 Lb x 32.17 ft/s² x 70 ft
W = 585494 lb.ft²/s²
1 lb.ft²/s² = 0.0421 J
W = 585494 lb.ft²/s² = 2.47 X 10⁴ J.
Therefore, the work done to get you safely away from the test is 2.47 X 10⁴ J.
