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DochEvi [55]
3 years ago
8

Which of the following is example of convection check all that apply

Physics
1 answer:
sladkih [1.3K]3 years ago
6 0

Answer air circulating in a hot air balloon

Explanation:

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(11%) Problem 5: A submarine is stranded on the bottom of the ocean with its hatch 25 m below the surface. In this problem, assu
V125BC [204]

Answer:

F = 1.24*10^4 N

Explanation:

Given

Depth of the ship, h = 25 m

Density of water, ρ = 1.03*10^3 kg/m³

Diameter of the hatch, d = 0.25 m

Pressure of air, P(air) = 1 atm

Pressure of water =

P(w) = ρgh

P(w) = 1.03*10^3 * 9.8 * 25

P(w) = 2.52*10^5 N/m²

P(net) = P(w) + P(air) - P(air)

P(net) = P(w)

P(net) = 2.52*10^5 N/m²

Remember,

Pressure = Force / Area, so

Force = Area * Pressure

Area = πr² = πd²/4

Area = 3.142 * 0.25²/4

Area = 3.142 * 0.015625

Area = 0.0491 m²

Force = 0.0491 * 2.52*10^5

F = 12373 N

F = 1.24*10^4 N

5 0
3 years ago
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A velocity selector in a mass spectrometer uses a 0.150 T magnetic field. (a) What electric field strength (in volts per meter)
Alekssandra [29.7K]

Answer:

The electric field strength is 6.6\times10^{5}\ V/m

Explanation:

Given that,

Magnetic field = 0.150 T

Speed v= 4.40\times10^{6}\ m/s

We need to calculate the electric field strength

Using formula of velocity

v=\dfrac{E}{B}

E=v\times B

Where, v = speed

B = magnetic field

Put the value into the formula

E=4.40\times10^{6}\times0.150

E=660000\ V/m

E=6.6\times10^{5}\ V/m

Hence, The electric field strength is 6.6\times10^{5}\ V/m

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3 years ago
The simple sugar made during photosynthesis is called _____.
Andre45 [30]

Answer:

glucose

Explanation:

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A long metal cylinder with radius a is supported on an insulating stand on the axis of a long, hollow, metal tube with radius b.
bija089 [108]

a)

i) Potential for r < a: V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

ii) Potential for a < r < b:  V(r)=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r}

iii) Potential for r > b: V(r)=0

b) Potential difference between the two cylinders: V_{ab}=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

c) Electric field between the two cylinders: E=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}

Explanation:

a)

Here we want to calculate the potential for r < a.

Before calculating the potential, we have to keep in mind that the electric field outside an infinite wire or an infinite cylinder uniformly charged is

E=\frac{\lambda}{2\pi \epsilon_0 r}

where

\lambda is the linear charge density

r is the distance from the wire/surface of the cylinder

By integration, we find an expression for the electric potential at a distance of r:

V(r) =\int Edr = \frac{\lambda}{2\pi \epsilon_0} ln(r)

Inside the cylinder, however, the electric field is zero, because the charge contained by the Gaussian surface is zero:

E=0

So the potential where the electric field is zero is constant:

V=const.

iii) We start by evaluating the potential in the region r > b. Here, the net electric field is zero, because the Gaussian surface of radius r here contains a positive charge density +\lambda and an equal negative charge density -\lambda. Therefore, the net charge is zero, so the electric field is zero.

This means that the electric potential is constant, so we can write:

\Delta V= V(r) - V(b) = 0\\\rightarrow V(r)=V(b)

However, we know that the potential at b is zero, so

V(r)=V(b)=0

ii) The electric field in the region a < r < b instead it is given only by the positive charge +\lambda distributed over the surface of the inner cylinder of radius a, therefore it is

E=\frac{\lambda}{2\pi r \epsilon_0}

And so the potential in this region is given by:

V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0}  (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r} (1)

i) Finally, the electric field in the region r < a is zero, because the charge contained in this region is zero (we are inside the surface of the inner cylinder of radius a):

E = 0

This means that the potential in this region remains constant, and it is equal to the potential at the surface of the inner cylinder, so calculated at r = a, which can be calculated by substituting r = a into expression (1):

V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

And so, for r<a,

V(r)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

b)

Here we want to calculate the potential difference between the surface of the inner cylinder and the surface of the outer cylinder.

We have:

- Potential at the surface of the inner cylinder:

V(a)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

- Potential at the surface of the outer cylinder:

V(b)=0

Therefore, the potential difference is simply equal to

V_{ab}=V(a)-V(b)=\frac{\lambda}{2\pi \epsilon_0} ln(\frac{b}{a})

c)

Here we want to find the magnitude of the electric field between the two cylinders.

The expression for the electric potential between the cylinders is

V(r)=\int\limits^b_r {Edr} = \frac{\lambda}{2\pi \epsilon_0}  (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0}  ln\frac{b}{r}

The electric field is just the derivative of the electric potential:

E=-\frac{dV}{dr}

so we can find it by integrating the expression for the electric potential. We find:

E=-\frac{d}{dr}(\frac{\lambda}{2\pi \epsilon_0} (ln(b)-ln(r))=\frac{\lambda}{2\pi \epsilon_0} \frac{1}{r}

So, this is the expression of the electric field between the two cylinders.

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0
3 years ago
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