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3 years ago

The atmosphere is composed of five?

2 answers:

6 0

The atmosphere is composed of five major layers. These layers are, in order from the surface of the earth, troposphere, stratosphere, mesosphere, ionosphere, and exosphere.

troposphere - lowest layer that provides majority of our weather. contains four-fifths of air on the earth and extends to a maximum height of 11 miles at the equator and lesser height at the poles.

stratosphere - extends to the height of 30 miles from the equator and it's the location of the ozone layer.

mesosphere - extends to the height of 52 miles from the equator and is where meteors generally burn up

ionosphere - extends to the height of 430 miles from the equator and is already considered as part of the outer space. It is where satellites orbit around the world.

exosphere - extends to the height range of 620 miles to 6,214 miles from the Earth's surface and merges with interplanetary space

motikmotik3 years ago

5 0

The atmosphere is made up of 5 major layers.

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According to this equation, F=ma, how much force is needed to accelerate an 82-kg runner at 7.5m/s2?

Murljashka [212]

You multiply the mass by the acceleration 82*7.5=615; that's what I would do

5 0

3 years ago

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Three point charges are placed on the x−y plane: a + 50.0-nC charge at the origin, a −50.0-nC charge on the x axis at 10.0 cm, a

butalik [34]

Answer:

(a) F = 0.00322i - 0.00793j with magnitude |F| = 0.00856N

(b) E = -42846.7 N/C

Explanation:

The diagram attached below explains some parameters.

Parameters given:

Charge Q1 = +50 nC at point (0, 0)

Charge Q2 = -50 nC at point (0.1, 0)

Charge Q3 = +150 nC at point (0.1, 0.08)

* The distances are in meters.

(a) The total electric force on the charge Q3 due to Q1 and Q2 is the vector sum of the forces due to Q1 and Q2. Mathematically,

F = F1 + F2

FORCE DUE TO Q1 i.e. F(Q1, Q3)

We have to find the x and y components.

From the diagram, we can find θ using SOHCAHTOA:

θ = tan⁻¹ (0.08/0.1)

θ = 38.66⁰

The distance between Q1 and Q3 can be found using Pythagoras theorem:

x² = 0.08² + 0.1²

x = 0.128 m

F1 = Fx(Q1, Q3)i + Fy(Q1, Q3)j

F1 = iF(Q1, Q3)cosθ + jF(Q1, Q3)sinθ

F(Q1, Q3) = (k * Q1 * Q3) / r²

k = Coulombs constant

F(Q1, Q3) = (9 * 10⁹ * 50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.128)²

F(Q1, Q3) = 0.00412N

F1 = i0.00412 * cos38.66 + j0. 00412 * sin38.66

F1 = 0.00322i + 0.00257j N

FORCE DUE TO Q2 i.e. F(Q2, Q3)

We have to find the x and y components.

F2 = Fx(Q2, Q3)i + Fy(Q2, Q3)j

F2 = iF(Q2, Q3)cos90 + jF(Q2, Q3)cos0

F(Q2, Q3) = (k * Q2 * Q3) / r²

F(Q2, Q3) = (9 * 10⁹ * -50 * 10⁻⁹ * 150 * 10⁻⁹) /(0.08)²

F(Q2, Q3) = -0.0105N

F2 = -i0.0105 * cos90 - j0.0105 * cos0

F2 = - 0.0105j N

Hence, the total force will be

F = F1 + F2

F = 0.00322i + 0.00257j - 0.0105j

F = 0.00322i - 0.00793j N

The magnitude of this force is:

|F| = √(0.00322² + (-0.00793²)

|F| = 0.00856N

(b) The electric field at charge Q3 is the sum of the electric fields due to Q1 and Q2:

E = E1 + E2

E1, electric field due to Q1 = kQ1/r²

E1 = (9 * 10⁹ * 50 * 10⁻⁹) / (0.128²)

E1 = 27465.8 N/C

E2, electric field due to Q2 = (9 * 10⁹ * -50 * 10⁻⁹) / (0.08²)

E1 = -70312.5N/C

The total electric field:

E = E1 + E2

E = 27465.8 - 70312.5

E = -42846.7 N/C

3 0

3 years ago

If the wave speed stays the same, which of the following decreases as the frequency increases?

dalvyx [7]

Answer:yes

Explanation:

8 0

3 years ago

An amount of work W is done on an object of mass m initially at rest, and as result it winds up moving at speed v. Suppose inste

sesenic [268]

Answer:

Explanation:

Given

W amount of work is done on the system such that it acquires v velocity after operation(initial velocity)

According to work energy theorem work done by all the forces is equal to change in kinetic energy of object

$W=\frac{1}{2}mv^2---1$

where m=mass of object

v=velocity of object

When the object is already have velocity v then the final speed is given by work energy theorem

$W=\frac{1}{2}mv_f^2-\frac{1}{2}mv^2-----2$

From 1 and 2 we get

$\frac{1}{2}mv^2=\frac{1}{2}mv_f^2-\frac{1}{2}mv^2$

$2\times \frac{1}{2}mv^2=\frac{1}{2}mv_f^2$

$v_f^2=2v^2$

$v_f=\sqrt{2}v$

8 0

3 years ago

Match the type of fitness with the correct example.

jeyben [28]

cardiovascular fitness: 3, 4, 7

flexibility: 1, 5

muscular fitness: 2, 6

7 0

3 years ago

Read 2 more answers