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Step2247 [10]
3 years ago
15

Earth can be modeled with five layers based on the strengths and physical properties of the materials in them. Put the layers in

order, with the outermost surface of Earth on top. The two layers of the core have been lumped together here.
Physics
1 answer:
andre [41]3 years ago
4 0

Answer:

Crust, Upper mantle, mantle, outer core, inner core

Explanation:

The Earth's layers have been clasified in 5 according to the materials that conform them, theri physical properties, strengths and also their state of matter. We all know how the outer layer of the Earth looks like, but if we start to dig a huge hole we are going to see different types por materials due to a change in pressure, temperature, and other factors. At the very center of the Earth there's what's called "core". The core is liquid and at extremely high temperatures. This is because of the enormous amount of pressure the rest of the Earth is putting it under. So, if we list the different layers of the Earth according to the materials they are made of, from the Earth's surface to the core, the answer is:

1) Crust (surface)

2) Upper Mantle

3) Mantle

4) Outer core

5) Inner core

In some books you may find a layer called Lithosphere. Tis layer consists not only of the crust, but also it contains the transition zone between the upper mantle and the crust.

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a 1.5 kg ball is thrown vertically upward with an initial speed of 15 m/s. if the initial potential energy is taken as zero, fin
trapecia [35]

Answer:

a) E_{p} = 0

E_{k} = 168.7 J

E_{m} = 168.7 J

b) E_{p} = 73.6 J

E_{k} = 95.8 J

E_{m} = 169.4 J

c) E_{p} = 169.2 J

E_{k} = 0

E_{m} = 169.2 J

Explanation:

We have:

m: is the ball's mass = 1.5 kg

v₀: is the initial speed = 15 m/s

g: is the gravity acceleration = 9.81 m/s²

a) In the initial position we have:

h: is the height = 0

The potential energy is given by:

E_{p} = mgh = 0

The kinetic energy is:

E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}*1.5*(15)^{2} = 168.7 J

And the mechanical energies:

E_{m} = E_{p} + E_{k} = 0 + 168.7 J = 168.7 J

b) At 5 m above the initial position we have:

h = 5 m

The potential energy is:

E_{p} = mgh = 1.5*9.81*5 = 73.6 J

Now, to find the kinetic energy we need to calculate the speed at 5 m:

v_{f}^{2} = v_{0}^{2} - 2gh = (15)^{2} - 2*9.81*5 = 126.9

v_{f} = \sqrt{126.9} = 11.3 m/s

E_{k} = \frac{1}{2}mv^{2} = \frac{1}{2}*1.5*(11.3)^{2} = 95.8 J

And the mechanical energies:

E_{m} = E_{p} + E_{k} = 73.6 + 95.8 J = 169.4 J

c) At its maximum height:

v_{f}: is the final speed = 0

h = \frac{v_{0}^{2}}{2g} = \frac{(15)^{2}}{2*9.81} = 11.5 m

Now, the potential, kinetic and mechanical energies are:

E_{p} = mgh = 1.5*9.81*11.5 = 169.2 J

E_{k} = \frac{1}{2}mv^{2} = 0

E_{m} = 169.2 J + 0 = 169.2 J

I hope it helps you!    

7 0
3 years ago
A 6157 N piano is to be pushed up a(n) 2.41 m frictionless plank that makes an angle of 21.9 ◦ with the horizontal. Calculate th
xz_007 [3.2K]

Answer:

Work done, W = 5534.53 J

Explanation:

It is given that,

Force acting on the piano, F = 6157 N

It is pushed up a distance of 2.41 m friction less plank.

Let W is the work done in sliding the piano up the plank at a slow constant rate. It is given by :

W=Fd\ sin\theta

Since, F=F\ sin\theta (in vertical direction)

W=6157\times 2.41\ sin(21.9)

W = 5534.53 J

So, the work done in sliding the piano up the plank is 5534.53 J. Hence, this is the required solution.

3 0
4 years ago
Suppose an oven's radiation wavelength is 0.125 m. a container with 350.00 g of water was placed in the oven, and the temperatur
ivann1987 [24]
The heat (energy) needed to raise the temperature of the water is given by
Q=m C_S (T_f - T_i)=(350.0 g)(4.18 J/gC)(80C-20C)=87780 J

The wavelength of the radiation of the oven is \lambda=0.125 m, so the energy of a single photon of this radiation is
E=h \frac{c}{\lambda}=(6.6 \cdot 10^{-34}J) \frac{3\cdot 10^8 m/s}{0.125 m}=1.6 \cdot 10^{-24} J

So, the number of photons required to heat the water is the total energy absorbed by the water divided by the energy of a single photon:
N= \frac{Q}{E}= \frac{87780 J}{1.6\cdot 10^{-24}J}=  5.5 \cdot 10^{28} photons
8 0
3 years ago
How much force must be applied on a blade of length 4cm and thickness of 0.1mm to exert a pressure of 4000000pa?
Viktor [21]

Answer:

F= 403429 kpa

Explanation:

Pressure is the product of force and area

Mathematically,

P=F*A -------where F is force and A is area.

A= 40 *0.1 = 4mm² -----convert to m²

A= 4e⁻⁶ m²

P= 4000000 pa

F= P/A = 4000000/4e⁻⁶

F= 403428793.493 pa

F= 403429 kpa

7 0
3 years ago
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