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stepan [7]
3 years ago
8

Which layer in the atmosphere has the least air

Physics
2 answers:
djverab [1.8K]3 years ago
8 0
Exosphere is my answer



please can i have a brainlier

stich3 [128]3 years ago
7 0
The Exosphere has the least amount of oxygen.

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The cylinder of gravity of cylinder is where
Elodia [21]

Explanation:

In uniform gravity it is the same as the centre of mass. For regular shaped bodies it lies at the centre of the that particular body. Hence for a cylinder centre of gravity lies at the midpoint of the axis of the cylinder.

5 0
4 years ago
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A 860g block of titanium metal has a volume of 200 cm3 and the density of 4.3g/cm^3.
valina [46]

Answer:

4.3 g/ cm^3

Explanation:

Mass / Volume = density . so 2150/500= 4.3 g/cm^3

6 0
3 years ago
A concave mirror has a radius of curvature of 10 cm. Find the location an height of the image if the distance of the object is 8
puteri [66]
R=10
F=R/2
F=10/2=5
F=-5(CONCAVE MIRROR)
U=-8(CONCAVE MIRROR)
HEIGHT OF OBJECT=1.5
V=?
HEIGHT OF IMAGE=?
I/F=1/U+1/V
-I/5=-1/8-1/V
-1/V=-1/5+1/8
-1/V=-8+5/40
-1/V=-3/40
1/V=3/40
V=40/3

HEIGHT OF IMAGE/HEIGHT OF OBJECT =-V/U
HEIGHT OF IMAGE=40/3*1/-8*15/10
                              =-20/8
                              =-2.5
7 0
3 years ago
LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.
Svetradugi [14.3K]

Answer:

i) E = 269 [MJ]    ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

i)

E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]

8 0
3 years ago
What is a neutral atom?
12345 [234]
An atom in which is the number of electrons that surrounds the nucleus
4 0
3 years ago
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