1.A raft is constructed of wood having a density of 508.0 kg/m 3 . The surface a
ID: 1596011 • Letter: 1
Question
1.A raft is constructed of wood having a density of 508.0 kg/m3. The surface area of the bottom of the raft is 5.5 m2, and the volume of the raft is 0.59 m3. When the raft is placed in fresh water having density 1.0 103 kg/m3, how deep is the bottom of the raft below water level?
2.A natural-gas pipeline with a diameter of 0.215 m delivers 1.40 m3 of gas per second. What is the flow speed of the gas?
3.A light spring with a spring constant of 88.0 N/m rests vertically on the table, as shown in (a) below. A 2.05 g balloon is filled with helium (0°C and 1 atm pressure) to a volume of 4.88 m3 and connected to the spring, causing the spring to stretch, as shown in (b) below. How much does the spring stretch when the system is in equilibrium. (The density of helium is 0.179 kg/m3. The magnitude of the spring force equals kx.)
m
5.The aorta in an average adult has a cross-sectional area of 2.0 cm2.
(a) Calculate the flow rate (in grams per second) of blood ( = 1.0 g/cm3) in the aorta if the flow speed is 42 cm/s.
(b) Assume that the aorta branches to form a large number of capillaries with a combined cross-sectional area of 3.2 103 cm2. What is the flow speed in the capillaries?
6.A thin, rigid, spherical shell with a mass of 3.96 kg and diameter of 0.208 m is filled with helium (adding negligible mass) at 0°C and 1 atm pressure. It is then released from rest on the bottom of a pool of water that is 4.30 m deep.
(a) Determine the upward acceleration of the shell.
(b) How long will it take for the top of the shell to reach the surface? Disregard frictional effects.
7.A light spring with a spring constant of 15.4 N/m rests vertically on the bottom of a large beaker of water, as shown in (a) below. A 4.64 10-3 kg block of wood with a density of 606.0 kg/m3 is connected to the spring, and the mass-spring system is allowed to come to static equilibrium, as shown in (b) below. How much does the spring stretch?
Explanation / Answer
1.)mass of the raft = density * volume
= 508.0 * 0.59
=299.72kg
mass of water displaced = 299.72 kg
volume of water displaced = mass/density
= 299.72/1000
= 0.29972 m^3
depth of the bottom of the raft below water level =0.29972/5.5
= 0.05449 m
2).
Volume flow rate=flow speed*area if cross-section
Flow speed of the gas=1.60/(*(0.217/2)2) = 43.262 m/sec
=376.3 kg
mass of water displaced = 376.3 kg
volume of water displaced = mass/density
= 376.3/1000
= 0.3763 m^3
depth of the bottom of the raft below water level =0.3763/6.3
= 0.0597 m
6)
For this answer, I am assuming that the upward acceleration of 4.81 m/s^2 is correct.
To find the time it takes for the top of the ball to reach the water's surface, we can use the following equation:
For this equation, the initial position, y0, is the position of the top of the ball. The pool is 3 m deep, and the ball has a diameter of 0.225 m. Using this information, we can deduce that the top of the ball is at a position of y0 = -2.775 m.
The initial velocity of the ball is 0 m/s since the ball is at rest. If the top of the ball is at the water's surface, then y(t) = 0 m. If y(t) and v0 are both zero, the equation above becomes.
Solving for t, we get:
Plugging in the acceleration that you've provided and the initial position that I've provided, find t.
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