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sladkih [1.3K]
3 years ago
6

A scuba tank, when fully submerged, displaces 15.7 l of seawater. the tank itself has a mass of 13.7 kg and, when "full," contai

ns 3.00 kg of air. the density of seawater is 1025 kg/m3. assume that only its weight and the buoyant force act on the tank.
Physics
1 answer:
julia-pushkina [17]3 years ago
7 0
The problem is asking for the buoyant force.
The first step is to know the weights of the tank and air. Tank = 13.7 * 9.8 = 134.26 N Air = 3 * 9.8 = 29.4 N Total weight of the tank and air is 163.66 N 
To know the buoyant force, we must convert 15.7 liters to cubic meters. 
1 liter = 1000 ml = 1000 cm^3 1 m = 100 cm 1 m^3= 1,000,000 cm^3 One liter = 1000 divided by 1,000,000 = 0.001 m^3 V = 15 * 0.001 = 0.015 m^3 
Then compute for the buoyant force:Buoyant force = 0.015 * 1025 * 9.8 = 150.675 N
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Two cars travel westward along a straight highway, one at a constant velocity of 85 km/h, and the other at a constant velocity o
spayn [35]

To solve letter a:

d1 = 85t1 = 16 km, 
85t1 = 16, 
t1 = 16 / 85 = 0.1882 h = 11.29 min. 

d2 = 115t2 = 16 km, 
115t2 = 16, 
t2 = 16 / 115 = 0.139 h = 8.35 min. 

t1 - t2 = 11.29 - 8.35 = 2.94 min. 
Car #2 arrives 2.94 minutes sooner.

To solve letter b:

15 min = 1/4 h = 0.25 h. 
d1 = d2, 
115t = 85(t + 0.25), 
115t = 85t + 21.25, 
115t - 85t = 21.25, 
30t = 21.25, 
t = 21.25 / 30 = 0.71 h, 

d = 115 * 0.71 = 81.65 km.

8 0
3 years ago
4. Look at the graph above. What information is on the graph and what does it mean? Please help!!!
Aliun [14]

Answer:

CO2 increases in the winter

Explanation:

6 0
3 years ago
A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne
kati45 [8]

Answer:

The average emf that will be induced in the wire loop during the extraction process is 37.9 V

Explanation:

The average emf induced can be calculated from the formula

Emf = -N\frac{\Delta \phi}{\Delta t}

Where N is the number of turns

\Delta \phi is the change in magnetic  flux

\Delta t is the time interval

The change in magnetic flux is given by

\Delta \phi = \phi _{f} - \phi _{i}

Where \phi _{f} is the final magnetic flux

and \phi _{i} is the initial magnetic flux

Magnetic flux is given by the formula

\phi = BAcos(\theta)

Where B is the magnetic field

A is the area

and \theta is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is, \theta = 0^{o}

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

A = \pi r^{2}

∴ A = \pi (0.15)^{2}

A = 0.0225\pi

∴\phi_{i}  = (1.5)(0.0225\pi)cos(0^{o} )

\phi_{i}  = (1.5)(0.0225\pi)

( NOTE: cos (0^{o}) = 1 )

\phi_{i}  = 0.03375\pi Wb

For \phi_{f}

The field pointed upwards, that is \theta = 90^{o}. Since cos (90^{o}) = 0

Then

\phi_{f} = 0

Hence,

\Delta \phi = 0- 0.03375\pi

\Delta \phi = - 0.03375\pi

From the question

\Delta t = 2.8 ms = 2.8 \times 10^{-3} s

Here, N = 1

Hence,

Emf = -N\frac{\Delta \phi}{\Delta t} becomes

Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }

Emf = 37.9 V

Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.

5 0
3 years ago
In a mixture what happens when the ingredients intermingled
Zinaida [17]
The ingredients do not react with or chemically bond to each other.
7 0
3 years ago
Read 2 more answers
What is the gauge pressure in Pascals inside a honey droplet of a 0.1 cm diameter? Assume that air is surrounding this droplet a
Vikki [24]

Answer:

The gauge pressure in Pascals inside a honey droplet is 416 Pa

Explanation:

Given;

diameter of the honey droplet, D = 0.1 cm

radius of the honey droplet, R = 0.05 cm = 0.0005 m

surface tension of honey, γ = 0.052 N/m

Apply Laplace's law for a spherical membrane with two surfaces

Gauge pressure =  P₁ - P₀ = 2 (2γ / r)

Where;

P₀ is the atmospheric pressure

Gauge pressure = 4γ / r

Gauge pressure = 4 (0.052) / (0.0005)

Gauge pressure = 416 Pa

Therefore, the gauge pressure in Pascals inside a honey droplet is 416 Pa

5 0
3 years ago
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