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3 years ago

Why does the sea floor spread

Physics

2 answers:

Naily [24]3 years ago

7 0

Because of the volcano's under it makes it open

Brut [27]3 years ago

4 0

Volcano under us hope it helps

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You and a friend each carry identical boxes from the first floor of a building to a room located on the second floor, farther do

melamori03 [73]

Answer:

Work done in both the cases will be same

Explanation:

As we know that the work done against gravity is given as

$W = F_g .d$

here we know that gravitational force is a conservative force and the work done against gravitational force is independent of the path

So here the work done by person to move the object between two different heights will be independent of the path they choose

So for the first person and second person will be same in both the cases because the height through which the boxes are transferred will be same in both the cases

8 0

3 years ago

A spherically-spreading EM wave comes from a 104.0 W source. At a distance of 9.6 m, what is the intensity of the wave?

Nookie1986 [14]

Answer:

Approximately 0.0898 W/m².

Explanation:

The intensity of light measures the power that the light delivers per unit area.

The source in this question delivers a constant power of $\rm 104.0\; W$ . If the source here is a point source, that $\rm 104.0\; W$ of power will be spread out evenly over a spherical surface that is centered at the point source. In this case, the radius of the surface will be 9.6 meters.

The surface area of a sphere of radius $r$ is equal to $4\pi r^{2}$ . For the imaginary 9.6-meter sphere here, the surface area will be:

$\rm 4\pi \times (9.6\; m)^{2} \approx 1158.12\; m^{2}$ .

That $\rm 104.0\; W$ power is spread out evenly over this 9.6-meter sphere. The power delivered per unit area will be:

$\displaystyle\rm \frac{104.0\; W}{1158.12\; m^{2}}\approx 0.0898\; W\cdot m^{-2}$ .

8 0

3 years ago

If you were capable of converting mass to energy with 100%, efficiency, how much mass would you need to produce 3.5x10^12 Joules

Alexeev081 [22]

Answer:

a) 3.9 x 10⁻⁵ kg

Explanation:

The amount of mass required to produce the energy can be given by Einstein's formula:

$E = mc^2\\\\m = \frac{E}{c^2}$

where,

m = mass required = ?

E = Energy produced = 3.5 x 10¹² J

c = speed of light = 3 x 10⁸ m/s

Therefore,

$m = \frac{3.5\ x\ 10^{12}\ J}{(3\ x\ 10^8\ m/s)^2} \\\\m = 3.9\ x\ 10^{-5}\ kg$

Hence, the correct option is:

a) 3.9 x 10⁻⁵ kg

7 0

2 years ago

Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

Part B 

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

7 0

3 years ago

Water boiling Which one is shown? (Look at pic)

Mkey [24]

Answer:

conduction.

Explanation:

Hoped I helped! Im Eve btw have a great day and consider marking this brainliest if you do thank you in advanced!

4 0

3 years ago