The atmosphere has many roles, including:

Answers

Answer 1

Answer: There are five layers of the atmosphere and these are; troposphere, stratosphere, mesosphere, thermosphere and exosphere. The first layer, troposphere, is where we are able to do most of our activities. This is where we can see the formation of clouds, the production of rain, hail, snow and other weather phenomenon. Also, this layer is where the greatest amount of air pressure because most of the molecules of air are in this area. Like us, air has also mass and the pressure is brought down by the earth’s gravity causing an increase in weight exerted on you as you descend lower into the atmosphere. So, as you enter into the other layers of atmosphere above the troposphere, the air pressure starts to decrease. Below the atmosphere is the hydrosphere. This is where all liquid forms are located. And since the seawater has a greater mass than air, it has the greatest pressure.

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A boy kicks a ball with a mass of 5 kg so that it accelerates at 4 m/s2 in the forward direction. A 40 N B 0 N C 20 N backward D 20 N forward
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If the velocity of a body changes from 13 m/s to 30 m/s while undergoing constant acceleration, what's the average velocity of the body? A. 28 m/s
B. 17 m/s
C. 21.5 m/s
D. 19.5 m/s

Answers

Since the acceleration is constant, the average velocity is simply the average of the initial and final velocities of the body:

$v_(avg) = (v_f+v_i)/(2)=(30 m/s+13 m/s)/(2)=21.5 m/s$

We can proof that the distance covered by the body moving at constant average velocity $v_(avg)$ is equal to the distance covered by the body moving at constant acceleration a:

- body moving at constant velocity $v_(avg)$ : distance is given by

$S=v_(avg)t = (v_f+v_i)/(2)t$

- body moving at constant acceleration $a=(v_f-v_i)/(t)$ : distance is given by

$S=v_i t+ (1)/(2)at^2 = v_i t + (1)/(2)(v_f-v_i)/(t)t^2=(v_i+(1)/(2)(v_f-v_i))t=(v_f+v_i)/(2)t$

The answer is 21.5 m/s

Heather is doing an acid-base reaction. She has 25.00 mL of hydrochloric acid of unknown concentration in a flask. She wants to add a basic solution to the acid drop by drop, until all the acid has reacted. Which tool would be best for adding the basic solution dropwise and for most precisely measuring the volume of solution added?

a graduated cylinder
an eye dropper
a beaker
a buret

Answers

Answer: a buret

A graduated cylinder and a beaker cannot be used. This is because the basic solution needs to be added drop by drop. This is not possible using a graduated cylinder or a beaker because it is difficult to pour drop by drop from these. An eye dropper cannot be used because it does not measure the volume being added.

A buret would be the best choice because it precisely measures the amount being added and we can add the basic solution to the acid drop by drop.

A burette would be the best tool for the required use.
The process described is called a 'titration'

What are examples or centripetal acceleration?

Answers

Centripetal acceleration is necessary, and is present, whenever any object
moves in a closed path, or along any part of a circular path. Some examples
I can think of are:

-- a planet or a comet in orbit around the sun
-- a moon in orbit around a planet
-- the little ball making circles around the spinning roulette wheel
-- a stone tied to the end of a string, when you spin it around your head
-- the end of the spinning string on a weed-whacker
-- the tip of the blade of a fan or a windmill
-- any little bit of rubber on a car tire or bicycle tire, when they're moving
-- a strawberry seed spinning around the inside of the blender jar when you're
making a smoothie
-- a car driving around a curve in the road
-- you on almost any amusement park ride; whenever you're not moving in a
straight line, you're experiencing centripetal acceleration.

centripedal acceleration :

⇒ the moon in orbit around a planet (or the sun).
⇒ The Earth (or an other planet) in orbit around the sun.

A 25 N object is 3 meters up. How much potential energy does it have.

Answers

We have: Potential Energy, U = m×g×h
U = F×h
Here, F = 25 N
h = 3 m

Substitute their values into the expression,
U = 25 × 3
U = 75 Nm
U = 75 J

So, You Final answer would be 75 Joules.

Hope this helps!

What is the potential energy of a 1-kilogram ball is thrown into the air with an initial velocity of 30m/sec?

Answers

the ball thrown from height=0
potental energy(PE) =0
kinetic energy(KE) = 0.5mv^2 = 0.5(1)(30^2) = 450

at the highest point the ball does not moved, v=0
potential energy at its maximum
kinetic energy = 0

Energy is conserved then
total energy before = after
PE1 + KE1 = PE2 + KE2
0 + 450 = PE2 + 0

conservation of energy is fun fact

mass=1kg
g=10m/s^2 (assuming)
u=30m/s
height(h)=u^2/2g
=900/20=45m
P.E.=mgh
=1×10×45=450 joules

A train is accelerating at a rate of 2 km/hr/s. If its initial velocity is 20 km/hr, what is its velocity after 30 seconds?

Answers

"2 km/hr/s" means that in each second, its engines can increase its speed by 2 km/hr.

If it keeps doing that for 30 seconds, its speed has increased by 60 km/hr.

On top of the initial speed of 20 km/hr, that's 80 km/hr at the end of the 30 seconds.

This whole discussion is of speed, not velocity. Surely, in high school physics,
you've learned the difference by now. There's no information in the question that
says anything about the train's direction, and it was wrong to mention velocity in
the question. This whole thing could have been taking place on a curved section
of track. If that were the case, it would have taken a team of ace engineers, cranking
their Curtas, to describe what was happening to the velocity. Better to just stick with
speed.

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