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lana66690 [7]
3 years ago
5

Calculate: A model airplane has a mass of 3.2 kg. Its propeller pulls it forward with a force of 7.0 N. What is the airplane’s a

cceleration?
Physics
1 answer:
vitfil [10]3 years ago
5 0
The airplanes acceleration is 2.1875
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The maximum wavelength that an electromagnetic wave can have and still eject electrons from a metal surface is 459 nm. What is t
spin [16.1K]

Answer:

2.7067 eV

Explanation:

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

c = Speed of light = 3\times 10^8\ m/s

\lambda_0 = Threshold wavelength = 459 nm

Work function is given by

W_0=\frac{hc}{\lambda_0}\\\Rightarrow W_0=\frac{6.626\times 10^{-34}\times 3\times 10^8}{459\times 10^{-9}}\\\Rightarrow W_0=4.33072\times 10^{-19}\ J

Converting to eV

1\ J=\frac{1}{1.6\times 10^{-19}}\ eV

4.33072\times 10^{-19}\ J=4.33072\times 10^{-19}\times \frac{1}{1.6\times 10^{-19}}\ eV=2.7067\ eV

The work function W0 of this metal is 2.7067 eV

4 0
3 years ago
Help<br>pls give me an honest answer ​
Thepotemich [5.8K]

Answer:

0.0025H

Explanation:

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3 years ago
How does an inclined plane change distance and how does it change direction
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 Objects are known to go down because of a unbalanced force
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A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 13.0 cm , giving it a ch
Leokris [45]

a) Electric field inside the paint layer: zero

b) Electric field just outside the paint layer: -3.62\cdot 10^7 N/C

c) Electric field 8.00 cm outside the paint layer: -7.27\cdot 10^7 N/C

Explanation:

a)

We can find the electric field inside the paint layer by applying Gauss Law: the total flux of the electric field through a gaussian surface is equal to the charge contained within the surface divided by the vacuum permittivity, mathematically:

\int EdS = \frac{q}{\epsilon_0}

where

E is the electric field

dS is the element of surface

q is the charge within the gaussian surface

\epsilon_0 = 8.85\cdot 10^{-12}F/m is the vacuum permittivity

Here we want to find the electric field just inside the paint layer, so we take a sphere of radius r as Gaussian surface, where

R = 6.5 cm = 0.065 m is the radius of the plastic sphere (half the diameter)

By taking the sphere of radius r, we note that the net charge inside this sphere is zero, therefore

q=0

So we have

\int E dS=0

which means that the electric field inside the paint layer is zero.

b)

Now we want to find the electric field just outside the paint layer: therefore, we take a Gaussian sphere of radius

r=R=0.065 m

The area of the surface is

A=4\pi R^2

And since the electric field is perpendicular to the surface at any point, Gauss Law becomes

E\cdot 4\pi R^2 = \frac{q}{\epsilon_0}

The charge included within the sphere in this case is the charge on the paint layer, therefore

q=-17.0\mu C=-17.0\cdot 10^{-6}C

So, the electric field is:

E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.065)^2}=-3.62\cdot 10^7 N/C

where the negative sign means the direction of the field is inward, since the charge is negative.

c)

Here we want to calculate the electric field 8.00 cm outside the surface of the paint layer.

Therefore, we have to take a Gaussian sphere of radius:

r=8.00 cm + R = 8.00 + 6.50 = 14.5 cm = 0.145 m

Gauss theorem this time becomes

E\cdot 4\pi r^2 = \frac{q}{\epsilon_0}

And the charge included within the sphere is again the charge on the paint layer,

q=-17.0\mu C=-17.0\cdot 10^{-6}C

Therefore, the electric field is

E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{-17.0\cdot 10^{-6}}{4\pi(8.85\cdot 10^{-12})(0.145)^2}=-7.27\cdot 10^7 N/C

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

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3 years ago
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