Answer:
by answering your question
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.
I’m imagining imagining imagining an imagination...
Answer:
Solution
Verified by Toppr
Correct option is
C
3 cm
RI=apparent depthreal depth
Substituting, 34=apparentdepth12
Therefore, apparent depth=412×3=9
The height by which it appears to be raised is 12−9=3cm
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SIMILAR QUESTIONS
A coin is placed at the bottom of a glass tumbler and then water is added. It appeared that the depth of the coin has been reduced because
Medium
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A tank is filled with water to a height of 12.5 cm. The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again?
Answer:
The horizontally applied force = 2360 N
Explanation:
<em>Force:</em> Force can be defined as the product of mass and acceleration. the S.I unit of force is Newton (N)
Fh = Fr + ma......... Equation 1
Where Fh = horizontally applied force, Fr = friction force, m = mass of the crate, a = acceleration of the crate.
<em>Given: m = 400 kg, a = 1 m/s²</em>
Fr = 1/2 W, W = mg ⇒W = 400×9.8 = 3920 N
∴Fr = 1/2(3920), Fr = 1960 N
Substituting these values into equation 1
Fh = 1960 + 400×1
Fh = 1960 + 400
Fh = 2360 N
Therefore the horizontally applied force = 2360 N
