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aleksklad [387]
3 years ago
9

List the Layers of Earth in order from Thickest to thinnest using the Data table on the Earth’s interior: (Combine numbers for L

ower and Upper Mantle
)1.________________________________________________________________2. ________________________________________________________________3. ________________________________________________________________4. ________________________________________________________________
Physics
2 answers:
wolverine [178]3 years ago
7 0
1. Crust
2. Mantle
3. Outer core
4. Inner core
stealth61 [152]3 years ago
3 0

Answer:

Crust, Mantle, Outer Core, Inner Core

I hope this helped! :))

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If a hiker that weighs 600 newtons climbs a 50 meter hill, how much gravitational potential energy has the hiker gained?
Illusion [34]

The gravitational potential energy U is defined as the product of mass m, the acceleration of gravity g and the height of object h.

U = mgh

We do not have the mass of the hiker. But we know that its W weight is:

W = mg

Where

g = 9.8 m/s^2

So:

m = \frac{W}{g}\\\\m = \frac{600}{g}.

So:

U = (\frac{600}{g})gh\\\\U = (600N)(50m)

U = 30000J

The hiker has gained 30,000 J of energy

6 0
3 years ago
Oil having a density of 930 kg/m^3 floats on
Zolol [24]

Answer:

0.0268 m

Explanation:

Draw a free body diagram of the block.  There are three forces: weight force mg pulling down, buoyancy of the oil B₁ pushing up, and buoyancy of the water B₂ pushing up.

Sum of forces in the y direction:

∑F = ma

B₁ + B₂ − mg = 0

ρ₁V₁g + ρ₂V₂g − mg = 0

ρ₁V₁ + ρ₂V₂ = m

ρ₁V₁ + ρ₂V₂ = ρV

ρ₁Ah₁ + ρ₂Ah₂ = ρAh

ρ₁h₁ + ρ₂h₂ = ρh

(930 kg/m³)h₁ + (1000 kg/m³)h₂ = (968 kg/m³) (4.93 cm)

Since the block is fully submerged, h₁ + h₂ = 4.93 cm.

(930 kg/m³) (4.93 cm − h₂) + (1000 kg/m³)h₂ = (968 kg/m³) (4.93 cm)

h₂ = 2.68 cm

h₂ = 0.0268 m

4 0
3 years ago
The current required to stimulate the heart during ventricular fibrillation is about 110mA (1000mA = 1A). Assuming that a conduc
algol [13]

Answer:

Explanation:

Given

current required I=110 mA

Internal Resistance R=300 \Omega

According to ohm law

Current flows in a conductor is directly Proportional to the voltage applied.

V\propto I

V=IR

V=110\times 10^{-3}\times 300

V=33 V  

5 0
3 years ago
How much force can a 2.5 kg sledge hammer excerpt on a nail if you can swing the hammer at 20 m/s and the hammer contacts the na
AleksAgata [21]

Answer:

625 N

Explanation:

The impulse given to the nail is equal to the change in momentum of the hammer:

I=F \Delta t=m \Delta v

where

F is the force exerted by the hammer

\Delta t=0.08 s is the time of contact

m=2.5 kg is the mass

\Delta v=20 m/s is the change in velocity of the hammer

Substituting the data and re-arranging the equation, we can find the force:

F=\frac{m \Delta v}{\Delta t}=\frac{(2.5 kg)(20 m/s)}{0.08 s}=625 N

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4 years ago
A feeding buffer protects ______ path from delays in ______ ____.
Bad White [126]
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