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pishuonlain [190]
2 years ago
6

The deepest point of the pacific ocean is 11,033 m, in the mariana trench. what is the gauge pressure in the water at that point

? the density of seawater is 1025 kg/m3.
Physics
1 answer:
Liula [17]2 years ago
4 0

Given that,

Depth of seawater, h = 11,033 m

Density seawater, p (rho) = 1025 kg/m³

Gauge Pressure , P = ??

Since, we know that:

Pressure, P = pgh

Pressure = 1025 * 9.81 * 11033

Pressure = 1109395723.3 N/m²

or

Pressure = 1.1 x 10∧8 Pascal

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A very long uniform line of charge has charge per unit length λ1 = 4.80 μC/m and lies along the x-axis. A second long uniform li
Elodia [21]

Answer:

a) E=228391.8 N/C

b) E=-59345.91N/C

Explanation:

You can use Gauss law to find the net electric field produced by both line of charges.

\int \vec{E_1}\cdot d\vec{r}=\frac{\lambda_1}{\epsilon_o}\\\\E_1(2\pi r)=\frac{\lambda_1}{\epsilon_o}\\\\E_1=\frac{\lambda_1}{2\pi \epsilon_o r_1}\\\\\int \vec{E_2}\cdot d\vec{r}=\frac{\lambda_2}{\epsilon_o}\\\\E_2=\frac{\lambda_2}{2\pi \epsilon_o r_2}

Where E1 and E2 are the electric field generated at a distance of r1 and r2 respectively from the line of charges.

The net electric field at point r will be:

E=E_1+E_2=\frac{1}{2\pi \epsilon_o}(\frac{\lambda_1}{r_1}+\frac{\lambda_2}{r_2})

a) for y=0.200m, r1=0.200m and r2=0.200m:

E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.200m}-\frac{2.26*10^{-6}C}{0.200m}}]=228391.8N/C

b) for y=0.600m, r1=0.600m, r2=0.200m:

E=\frac{1}{2\pi(8.85*10^{-12}C^2/Nm^2)}[\frac{4.80*10^{-6}C}{0.600m}-\frac{2.26*10^{-6}C}{0.200m}}]=-59345.91N/C

5 0
2 years ago
As shown in the figure below, a bullet is fired at and passes through a piece of target paper suspended by a massless string. Th
NikAS [45]

Answer:

M = 0.730*m

V = 0.663*v

Explanation:

Data Given:

v_{bullet, initial} = v\\v_{bullet, final} = 0.516*v\\v_{paper, initial} = 0\\v_{paper, final} = V\\mass_{bullet} = m\\mass_{paper} = M\\Loss Ek = 0.413 Ek

Conservation of Momentum:

P_{initial} = P_{final}\\m*v_{i} = m*0.516v_{i} + M*V\\0.484m*v_{i} = M*V .... Eq1

Energy Balance:

\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2}  ....... Eq 2

Substitute Eq 2 into Eq 1

0.484*m*v_{i} = \frac{0.320744*m*v^2_{i} }{V^2} *V  \\0.484 = 0.320744*\frac{v_{i} }{V} \\\\V = 0.663*v_{i}

Using Eq 1

0.484m*v_{i} = M* 0.663v_{i}\\\\M = 0.730*m

7 0
2 years ago
Why must the height of the meniscus in the graduated cylinder match the height of the water in the tub when measuring volume?
galben [10]

Answer:

See explanation

Explanation:

First, in order for you to understand, remember the basic concept of meniscus in graduated cylinder.

<em>"The meniscus is the curve seen at the top of a liquid in response to its container. The meniscus can be either concave or convex, depending on the surface tension of the liquid and its adhesion to the wall of the container".</em>

Now, according to this definition, and for water, the reading of the volume must be donde at the bottom of the curve of the meniscus. This is because the water  gives a concave curve.

If you read it and matches the height of water, you are getting two results:

One, get an accurate value or volume, because it's been done at eye level.

The second fact is that when you do the reading this way, The total pressure is made equal to the atmospheric pressure by adjusting the height of the cylinder until the water level is equal.

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2 years ago
What sentence best supports the statement that hormones are involved in the regulation of homeostasis? A. The hormone cortisol s
juin [17]

B. The hormone erythropoeitin increases the production of red blood cells when oxygen levels are low.

3 0
2 years ago
Now suppose the initial velocity of the train is 4 m/s and the hill is 4 meters tall. If the train has a mass of 30000 kg, what
hoa [83]

Answer:

<h2>187,500N/m</h2>

Explanation:

From the question, the kinectic energy of the train will be equal to the energy stored in the spring.

Kinetic energy = 1/2 mv² and energy stored in a spring E = 1/2 ke².

Equating both we will have;

1/2 mv² = 1/2ke²

mv² = ke²

m is the mass of the train

v is the velocity of then train

k is the spring constant

e is the extension caused by the spring.

Given m = 30000kg, v = 4 m/s, e = 4 - 2.4 = 1.6m

Substituting this values into the formula will give;

30000*4² =  k*1.6²

k = \frac{30,000*16}{1.6^2}\\ \\k = \frac{480,000}{2.56}\\ \\k = 187,500Nm^{-1}

The value of the spring constant is 187,500N/m

7 0
2 years ago
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