Let V be the volume of the object inside the water surface. Now the buoyant force is equal to the weight of the water displaced.
Hence, Vρg = 2000N
Or, V = 2000/ρg = 0.2041m3.
Thus, the total volume of the object V1is 2* 0.2041 = 0.4082 m3.
We have the relation, F = (ρf - ρb)Vg
Where, ρfis the density of the fluid and ρbis the density of the object or the body and V is the volume of the displaced liquid. Here, F = 2000 N.
Therefore, 2000 = (1000-ρb)9.8*0.2041
Or, ρb = 0.0899 m3.
Now the true weight of the object is given as V1ρbg = 0.4082* 0.0899*9.8 = 0.359 N.
The weight of the displaced water = Vρg. Here, ρ is the density of water displaced = 1000 kg.m3. and g is the acceleration due to gravity = 9.8 m.s–2. Given that the weight of the displaced water is 2000 N.
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A homogeneous solid object floats on water with 80% of its volume below the surface. The same object when placed in a second liquid floats on that liquid with 73% of its volume below the surface. Determine the density of the object.
Many people have imagined that if they were to float the top of a flexible snorkel tube out of the water, they would be able to breathe through it while walking underwater. However, they generally do not consider just how much water pressure opposes the expansion of the chest and the inflation of the lungs. Suppose you can barely breathe while lying on the floor with a 370 N weight on your chest. How far below the surface of the...
105.00 L 5.00 kg 0.00 N Fluid Oil Water Weight of 5.00 kg wood block = 49 N Step 2: Place the wood block in the water and notice that it floats. When the wood block rests on the ground, the downward gravitational force is balanced by an upward normal (“contact") force. When floating, the gravitational force is still there, but the normal (contact") force is not. The force exerted by the fluid is the buoyant force (B). When floating,...
The
mass is 25 grams. Please label and put a box around the
solution.
Purpose: To calculate the apparent weight of a mass when immersed in water. Students will also have the opportunity to learn how to use calipers to accurately measure lenghs. Background: Archimedes discovered that he could find the density of an object by comparing it's weight o its apparent weight when immersed in water (see example 10-8 in he textbook). When mmersed, the buoyancy due to the...
University Physics Name: Solve only 10 problems. M. Ard Spring 2019 test 3 7) An object floats in water with 5/8 of its volume submerged. What is the ratio of the densi of the object to that of water?
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An object that weighs 1000 N floats with 80% of its volume submerged below water. What is the boyance force? What is the volume of the object?
1. (20 Marks) An object has an L-shape and can be treated as 2 beams fixed together at right angle. Beam 1 (B) has a weight wi = 180 N and a length Li-2.00 m and Beam 2 (B) has a weight w2 90.0 N and a length L2- 1.00 m. This object is hinged by its corner O to the ground via a hinge that allows it to rotate clockwise counter clockwise, when possible. B2 rope B2 0 Figure...
This question is in regards of a chemistry experiment dealing
with Archimedes’s principal and obtaining the volume of an object
submerged in water.Thank you!
Why does the difference between the weight of the object in the air and its weight in water equal the weight of the displaced water (think about the forces which are actually working on the submerged object)? (3 points)