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2 years ago

What are the layers of the Earth's atmosphere?

2 answers:

4 0

Layers of the Earth's Atmosphere: Troposphere, Stratosphere, Mesosphere, Thermosphere, and Exosphere - Windows to the Universe.

borishaifa [10]2 years ago

3 0

Layers of the Earth's Atmosphere: Troposphere, Stratosphere, Mesosphere, Thermosphere, and Exosphere - Windows to the Universe.
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A parachutist falls 50.0 m without friction. When the parachute opens, he slows down at a rate of 67 m/s*2. If he reaches the gr

KIM [24]

Answer:

3.49 seconds

3.75 seconds

-43200 ft/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$s=ut+\frac{1}{2}at^2\\\Rightarrow 50=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{50\times 2}{9.81}}\\\Rightarrow t=3.19\ s$

Time the parachutist falls without friction is 3.19 seconds

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 50+0^2}\\\Rightarrow v=31.32\ m/s$

Speed of the parachutist when he opens the parachute 31.32 m/s. Now, this will be considered as the initial velocity

$v=u+at\\\Rightarrow 11=31.32+9.81t\\\Rightarrow t=\frac{11-31.32}{-67}=0.3\ s$

So, time the parachutist stayed in the air was 3.19+0.3 = 3.49 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=0t+\frac{1}{2}\times a\times t^2\\\Rightarrow \frac{s}{2}=\frac{1}{2}at^2$

$s=ut+\frac{1}{2}at^2\\\Rightarrow \frac{s}{2}=u1.1+\frac{1}{2}\times a\times 1.1^2$

Now the initial velocity of the last half height will be the final velocity of the first half height.

$v=u+at\\\Rightarrow v=at$

Since the height are equal

$\frac{1}{2}at^2=u1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow \frac{1}{2}at^2=at1.1+\frac{1}{2}\times a\times 1.1^2\\\Rightarrow 0.5t^2-1.1t-0.605=0\\\Rightarrow 500t^2-1100t-605=0$

$t=\frac{11\left(1+\sqrt{2}\right)}{10},\:t=\frac{11\left(1-\sqrt{2}\right)}{10}\\\Rightarrow t=2.65, -0.45$

Time taken to fall the first half is 2.65 seconds

Total time taken to fall is 2.65+1.1 = 3.75 seconds.

When an object is thrown with a velocity upwards then the velocity of the object at the point to where it was thrown becomes equal to the initial velocity.

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-240^2}{2\times \frac{8}{12}}\\\Rightarrow a=-43200\ ft/s^2$

Magnitude of acceleration is -43200 ft/s²

5 0

2 years ago

If I were a geologist working in South America along one of those plate boundaries, what might I see if I were looking at sedime

faust18 [17]

Answer:

you'll see rocks that have been there for years

4 0

2 years ago

Is (2, 2) a solution of y < 4x-6 Help

vesna_86 [32]

Answer:

B) No.

Explanation:

Okay,so,

this is equation is y=mx +b

mx represents the slope

and b represents the y-intercept

in order to figure this out you need to plot the y-intercept first

that makes its (0,-6) because the 6 is negative in the equation

4x is also equal to 4/1 since we dont know what x is

we have to do rise over run for this

you go up 4 spots on the y intercept from -6 because 4 is positive

then you go to the right 1 time because 1 is positive.

this leaves you at (1,-2)

so, (2,2) is NOT a solution

7 0

2 years ago

A 3.00 kg mass is traveling at an initial speed of 25.0 m/s. What is the

Nitella [24]

Answer:

The magnitude of the force required to bring the mass to rest is 15 N.

Explanation:

Given;

mass, m = 3 .00 kg

initial speed of the mass, u = 25 m/s

distance traveled by the mass, d = 62.5 m

The acceleration of the mass is given as;

v² = u² + 2ad

at the maximum distance of 62.5 m, the final velocity of the mass = 0

0 = u² + 2ad

-2ad = u²

-a = u²/2d

-a = (25)² / (2 x 62.5)

-a = 5

a = -5 m/s²

the magnitude of the acceleration = 5 m/s²

Apply Newton's second law of motion;

F = ma

F = 3 x 5

F = 15 N

Therefore, the magnitude of the force required to bring the mass to rest is 15 N.

4 0

2 years ago

Identify the following terms:

anastassius [24]

Answer:

atom -
the smallest particle of a chemical element that can exist.

atomic mass-
the quantity of matter contained in an atom of an element

atomic weight -
ratio of the average mass of a chemical element's atoms to some standard

protons-
stable subatomic particle that has a positive charge equal in magnitude to a unit of electron charge and a rest mass of 1.67262 × 10−27 kg

electrons-
a stable subatomic particle with a charge of negative electricity, found in all atoms and acting as the primary carrier of electricity in solids

neutrons-
a subatomic particle of about the same mass as a proton but without an electric charge, present in all atomic nuclei except those of ordinary hydrogen.

energy levels-
one of the stable states of constant energy that may be assumed by a physical system
[used especially of the quantum states of electrons in atoms and of nuclei. — called also energy state.]

Covalent bonds

the interatomic linkage that results from the sharing of an electron pair between two atoms.

ionic bonds
type of linkage formed from the electrostatic attraction between oppositely charged ions in a chemical compound.

Valence electrons
a single electron or one of two or more electrons in the outer shell of an atom that is responsible for the chemical properties of the atom.

Lewis Dot Diagram
A way of representing atoms or molecules by showing electrons as dots surrounding the element symbol. One bond is represented as two electrons.

4 0

1 year ago