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adell [148]
3 years ago
14

Help please!

Physics
1 answer:
Marina CMI [18]3 years ago
6 0

It's the old-but-important concept:  "Speed" depends on who's measuring it and how THEY're moving.  Different observers may very well observe different speeds, and they're all correct.  There's no such thing as "the REAL speed".

When you were a little kid, did you ever get on a moving escalator or walkway, and when you got to the middle, you turned around and walked the <em>opposite</em> way, so that somebody watching you from the outside would see you not moving at all ?

Say you're on a school bus that's driving 10 mph along the street pavement, and you get up out of your seat and run forward up the aisle at 10 mph.  Somebody outside the bus sees you passing them at 20 mph !

If instead, you run toward the BACK of the bus at 10 mph, somebody outside the bus sees you bobbing up and down but not moving forward or backward at all.

All this  problem is saying is:  The bus is driving along at 15 m/s.  A passenger on the bus puts his little baseball down on the floor on its little feet, and it runs backwards down the aisle, toward the back of the bus, at 15 m/s.  Somebody is standing outside looking into the bus as it passes by and the little baseball is scurrying toward the back of the bus.  How will HE describe the motion of the baseball ?    

Have you got it now ?

"Relative to the Earth" just  means how fast an object is passing the stores and telephone poles and people standing still ... things that are attached to the Earth.  "Relative to the bus" would mean how fast an object is passing by people sitting on the bus.

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Which of the following best describes the type of bond formed when sodium reacts with chlorine?
SVEN [57.7K]

Answer: ionic bond

Explanation: This type of chemical bond is called an ionic bond because the bond formed between two ions of opposite charge.

6 0
3 years ago
While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)
ArbitrLikvidat [17]

Answer:

a) See attached picture, b) We know the initial velocity = 0, initial position=0, time=12.0s, acceleration=2.40m/s^{2}, c) the car travels 172.8m in those 12 seconds, d) The car's final velocity is 28.8m/s

Explanation:

a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

b) From the information given by the problem I know:

initial velocity =0

acceleration = 2.40m/s^{2}

time = 12.0 s

initial position = 0

c)

unknown:

displacement.

in order to choose the appropriate equation, I must take the knowns and the unknown and look for a formula I can use to solve for the unknown. I know the initial velocity, initial position, time, acceleration and I want to find out the displacement. The formula that contains all this data is the following:

x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}

Once I got the equation I need to find the displacement, I can plug the known values in, like this:

x=0+0(12s)+\frac{1}{2}(2.40\frac{m}{s^{2}} )(12s)^{2}

after cancelling the pertinent units, I get that  my answer will be given in meters. So I get:

x=\frac{1}{2} (2.40\frac{m}{s^{2}} )(12s)^{2}

which solves to:

x=172.8m

So the displacement of the car in 12 seconds is 172.8m, which makes sense taking into account that it will be accelerating for 12 seconds and each second its velocity will increase by 2.4m/s.

d) So, like the previous part of the problem, I know the initial position of the car, the time it travels, the initial velocity and its acceleration. Now I also know what its final position is, so we have more than enough information to find this answer out.

I need to find the final velocity, so I need to use an equation that will use some or all of the known data and the unknown. In order to solve this problem, I can use the following equation:

a=\frac{V_{f}-V_{0} }{t}

Next, since I need to find the final velocity, I can solve the equation just for that, I can start by multiplying both sides by t so I get:

at=V_{f}-V_{0}

and finally I can add V_{0} to both sides so I get:

V_{f}=at+V_{0}

and now I can proceed and substitute the known values:

V_{f}=at+V_{0}

V_{f}=(2.40\frac{m}{s^{2}}} (12s)+0

which solves to:

V_{f}=28.8m/s

8 0
3 years ago
Read 2 more answers
Why does Doug need to adjust his workouts when they seem too easy?
Aleonysh [2.5K]
Doug needs to change the <em>exercises </em>he is doing. Your body begins to become accustomed to what you do, and if you do the same thing every single time - you will stop making progress, and they'll simply become easy.

However, if you walk into the gym and your body is expecting you to just go over and bench press some, go rock some curls, head over to the squat rack then go home - but instead you start out with some dips, then go do lunges, then attack your triceps instead of biceps, your body will be shocked! 

When you change it up, your body will react. The goal is to be comfortable being uncomfortable. Also, one of the most overlooked things in working out is the intensity. If you bring it everything you got and go to the last gut wretching rep, you'll see results. 

However, if you just leisurely do things with 4 minute rests in between - you won't see true progress.

Hopefully, this helps! =)
6 0
4 years ago
Assume you push a 20 kg child in a little red wagon with a force of 100 N. However, the dirt and rocks below the wheels
GenaCL600 [577]

Answer:

a

Explanation:

4 0
3 years ago
Compare the kinetic energy of a 20,000-kg truck moving at 110 km/h with that of an 80.0-kg astronaut in orbit moving at 27,500 k
laiz [17]
Kinetic energy is the energy that is possessed by an object that is moving. It is calculated by one-half the product of the mass and the square of the velocity of the object. We calculate as follows:<span>

For the truck,
KE = mv^2 / 2
KE = 20000 kg (110 km/h (1 h/3600 s)(1000 m / 1 km))^2 / 2
KE = 611111.11 J

For the astronaut,
KE = 80.0 kg (27500 km/h</span>(1 h/3600 s)(1000 m / 1 km)<span>)^2/2
KE = 611111.11 J

The kinetic energy possessed by the two bodies are the same.

</span>
3 0
3 years ago
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