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2 years ago

Please answer the following question.

Physics

1 answer:

Dima020 [189]2 years ago

6 0

The oceanic lithosphere is a layer on the Earth's crust that consists of rocks of the mantle.

<h3>What is the lithosphere?</h3>

The lithosphere can be defined as the outer rock layer present on the Earth's surface.

The oceanic lithosphere is composed of different rocks of the mantle such as, among others, gabbro and/or basalt.

The structure of the oceanic lithosphere is similar around the world because it is slender compared to the continental lithosphere and always determined by the supply of magma.

In conclusion, the oceanic lithosphere is a layer on the Earth's crust that consists of rocks of the mantle.

Learn more about the lithosphere here:

brainly.com/question/10508547

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Someone is trying to balance a (110cm) plank with certain forces.

Mashutka [201]

Answer:

a. i. 30 Nm ii. This moment is a clockwise positive moment.

b. i. 15 Nm ii, This moment is a counter-clockwise negative moment.

c. i. The plank will not balance. ii. The plank would tip up.

d. 150 N

Explanation:

a) Calculate the moment of the 60N force (about O), then name its type.

i. Calculate the moment of the 60N force (about O)

Since moment = Force × perpendicular distance from point of moment ,

M = Fd

Since F = 60 N and d = 50 cm = 0.5 m

M = 60 N × 0.5 m = 30 Nm

ii. Then name its type.

This moment is a clockwise positive moment.

b) Calculate the moment of the 30N force (about O), then name its type.

i. Calculate the moment of the 30N force (about O),

Since moment = Force × perpendicular distance from point of moment ,

M' = F'd'

Since F' = 30 N and d' = 50 cm = 0.5 m

M' = 30 N × 0.5 m = 15 Nm

ii. Then name its type.

This moment is a counter-clockwise negative moment.

c) Will the plank balance? If not, which way will it tip?

i. Will the plank balance?

The plank will balance if the net moment on it is zero

So net moment, M' = positive moment - negative moment = M - M' = 30 Nm - 15 Nm = 15 Nm

Since the net moment on the plank is M" = 15 Nm ≠ 0,<u>the plank will not balance.</u>

ii Which way will it tip?

The plank would tip in the direction of the greater moment since the net moment is positive. <u>This moment tilts the plank in a clockwise direction, so the plank would tip up.</u>

d) What extra force would be needed at (B) to balance the plank?

The extra force must balance the net moment,

So M" = F"d" where F" = force and d" = distance of force from O = 10 cm = 0.10 m

F" = M"/d"

= 15 Nm/0.10 m

= 150 N

4 0

3 years ago

What is the weight of a 42 kg object if the object was on the moon?

sveticcg [70]

(1.6 m/s²)(42 Kg)= 80 N

4 0

3 years ago

A rocket has landed on planet x, which has half the radius of earth. An astronaut onboard the rocket weighs twice as much on pla

Nastasia [14]

Answer:

Option (c) u0

Explanation:

The escape velocity has a formula as:

V = √(2gR)

Where V is the escape velocity,

g is the acceleration due to gravity

R is the radius of the earth.

Now, from the question, we were told that the escape velocity for the rocket taking off from earth is u0 i.e

V(earth) = u0

V(earth) = √(2gR)

u0 = √(2gR) => For the earth

Now, let us calculate the escape velocity for the rocket taking off from planet x. This is illustrated below below:

g(planet x) = 2g (earth) => since the weight of the astronaut is twice as much on planet x as on earth

R(planet x) = 1/2 R(earth) => planet x has half the radius of earth

V(planet x) =?

Applying the formula V = √(2gR), the escape velocity on planet x is obtained as follow:

V(planet x) = √(2g(x) x R(x))

V(planet x) = √(2 x 2g x 1/2R)

V(planet x) = √(2 x g x R)

V(planet x) = √(2gR)

The expression obtained for the escape velocity on planet x i.e V(planet x) = √(2gR), is exactly the same as that obtained for the earth i.e V(earth) = √(2gR)

Therefore,

V(planet x) = V(earth) = √(2gR)

But from the question, V(earth) is u0

Therefore,

V(planet x) = V(earth) = √(2gR) = u0

So, the escape velocity on planet x is u0

4 0

3 years ago

What are the characteristics of the radiation emitted by a blackbody? According to Wien's Law, how many times hotter is an objec

jasenka [17]

Answer:

a) What are the characteristics of the radiation emitted by a blackbody?

The total emitted energy per unit of time and per unit of area depends in its temperature (Stefan-Boltzmann law).

The peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase (Wien’s displacement law).

The spectral density energy is related with the temperature and the wavelength (Planck’s law).

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave length of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

Explanation:

A blackbody is an ideal body that absorbs all the thermal radiation that hits its surface, thus becoming an excellent emitter, as these bodies express themselves without light radiation, and therefore they look black.

The radiation of a blackbody depends only on its temperature, thus being independent of its shape, material and internal constitution.

If it is study the behavior of the total energy emitted from a blackbody at different temperatures, it can be seen how as the temperature increases the energy will also increase, this energy emitted by the blackbody is known as spectral radiance and the result of the behavior described previously is Stefan's law:

$E = \sigma T^{4}$ (1)

Where $\sigma$ is the Stefan-Boltzmann constant and T is the temperature.

The Wien’s displacement law establish how the peak of emission of the spectrum will be displace to shorter wavelengths as the temperature increase (inversely proportional):

$\lambda max = \frac{2.898x10^{-3} m. K}{T}$ (2)

Planck’s law relate the temperature with the spectral energy density (shape) of the spectrum:

$E_{\lambda} = {{8 \pi h c}\over{{\lambda}^5}{(e^{({hc}/{\lambda \kappa T})}-1)}}}$ (3)

b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wavelength of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?

It is need it to known the temperature of both objects before doing the comparison. That can be done by means of the Wien’s displacement law.

Equation (2) can be rewrite in terms of T:

$T = \frac{2.898x10^{-3} m. K}{\lambda max}$ (4)

Case for the object with the blackbody emission spectrum peak in the blue:

Before replacing all the values in equation (4), $\lambda max$ (450 nm) will be express in meters:

$450 nm . \frac{1m}{1x10^{9} nm}$ ⇒ $4.5x10^{-7}m$

$T = \frac{2.898x10^{-3} m. K}{4.5x10^{-7}m}$

$T = 6440 K$

Case for the object with the blackbody emission spectrum peak in the red:

Following the same approach above:

$700 nm . \frac{1m}{1x10^{9} nm}$ ⇒ $7x10^{-7}m$

$T = \frac{2.898x10^{-3} m. K}{7x10^{-7}m}$

$T = 4140 K$

Comparison:

$\frac{6440 K}{4140 K} = 1.55$

The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.

4 0

3 years ago

Kai takes a hamburger off of the grill and puts a piece of cheese on it. Explain how and why the cheese melts when Kai puts it o

Yanka [14]

Because the hamburger is still hot from the grill, the cheese melts because of that heat.

3 0

3 years ago

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