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Dmitry [639]
3 years ago
15

Please try to answer... one time my battery died in my truck and a stranger stopped to give me a jump. She didn't know how so le

t me hook the jumper cables up, I forgot which to hook first, positive or negative, in a hurry and at the time didn't think it was a big deal I hooked both negatives up first then hooked the positives to the battery. After a minute my car started.. a month later i randomly ran into this person (we lived close by) she preceded to tell me that something happened to her car battery and it was my fault
Physics
1 answer:
8090 [49]3 years ago
3 0
<span>(1) Connect one red clamp to the positive (+) battery post of the "dead" battery.(2) Connect the other red clamp to the positive (+) post of the good battery.(3) Connect one black-end clamp to the negative (-) post of the good battery.(4) Carefully connect the other black-end clamp to some large metallic part of your car's engine block. Never connect it to the negative (-) post of the "dead" battery because this causes sparking which could ignite battery gases.<span>hope this helps </span></span>
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Assuming that 70 percent of the Earth’s surface
Aneli [31]
We need to find the volume of a spherical shell with a radius of
6.37 million meters and a thickness of 0.95 mile.

The technically correct way to do this is to find the volume of the
outside of the shell, then find the volume of the inside of the shell,
and subtract the inside volume from the outside volume.  That's
the REAL way to do it.

But look.  This 'shell' (the 0.95 mile of water) is only about  1530 meters thick,
on a sphere with a radius of 6.37 million meters.  The depth of the water is like
0.024 percent of the radius !  There's not a whole lot of difference between the
sphere outside the water and the sphere inside it.

So I want to do this problem the easier way ... Let's say that the volume
of the water is going to be

                  (the surface area that it covers on the Earth)
         times
                  (the thickness of the coating of water) .

The area of a sphere is  4 pi Radius² .
That's
                         (4 pi) x (6.37 x 10⁶ m)²

                   =    (4 pi) x (40.58 x 10¹² m²)

We're only interested in 70% of the total surface area.

                   =   (0.7) x (4 pi) x (40.58 x 10¹²) m²

                   =            3.57 x 10¹⁴  square meters of Earth's surface.

The volume of the water covering that area is

               (the area) times (average depth of 0.95 mile) .

We have to change that 0.95 mile to meters.
The question reminds us that                         1 mile = 1609 meters .    
So the volume of the water is

                      (the area) times (0.95 x 1609 meters).

But we're not there yet.  The question isn't asking for the volume.
It's asking for the mass of the water. 
We're ready to get the volume in cubic meters.
We're supposed to know that each cubic meter is 1,000 liters,
   and the mass of 1 liter of water is 1 kilogram.
So each cubic meter of volume is 1,000 kilograms of mass.

Now we're ready to dump all the numbers into the machine and
turn the crank.  The mass of all this water will be

         (the surface area) x (0.95 x 1609 meters) x (1,000 kg/m³)   

  =    (3.57 x 10¹⁴  m²)  x   (1528.6 m)  x  (1,000 kg/m³)

  =            5.457 x 10²⁰ kilograms .

This is my answer, and I'm stickin to it.

But ... just like all the other problems you get in high school, the
answer doesn't matter.  The teacher doesn't need the answer,
and YOU don't need the answer.  The reason you got this problem
for an assignment is to give you practice in HOW TO FIND the
answer ... how to plan what you're going to do with the problem,
and then how to carry it out.

I don't know how much effort you put into this problem, but somewhere
along the way, you chickened out and posted it on Brainly.  So far, the
result of that decision was:  The person who got all the practice was ME.
I got the good stuff, and all YOU got was the answer.

I hope my work is clear enough that you can go through it, and pick up
some of the good stuff for yourself.
3 0
2 years ago
Plz hurry How fast can work be done by a 7 horsepower engine?
den301095 [7]

Answer:

It will go about 45-50 mph.

Explanation:

Depending on what you are hauling.

7 0
2 years ago
Read 2 more answers
What length of tube would be required to produce a second tone under the same experimental conditions? Explain your answer.
Alisiya [41]

There is an indirect relationship between length and frequency. The longer the length the pipe has, the higher frequency it is. The shorter the length the pipe has, the lower frequency it is.

<u>Explanation:</u>

The four properties of the string that affect its frequency are length, diameter, tension, and density. These properties are described below: When the length of a string is changed, it will vibrate with a different frequency. Shorter strings have higher frequency and therefore higher pitch.

The longer the tube is the lower the pitch of the note that it can emit. When a tube is heated it expands and so is longer! As the gas in the tube gets warmer the molecules move faster, that means they can carry the vibrations of the sound wave more rapidly and so the pitch goes up.

7 0
3 years ago
As an ice cube melts, its molecules
fomenos

Answer:

A. absorb heat energy and move farther apart

Explanation:

The answer is A since when a ice cube takes in heat, it changes form and melts down to a liquid : water.

The molecules in water are farther apart since they are moving around alot but can be contained in a container.

3 0
3 years ago
video for A door 1 m wide, of mass 15 kg, is hinged at one side so that it can rotate without friction about a vertical axis. It
natka813 [3]

Answer:

\omega_f = 0.4\ rad/s

Explanation:

given,

width of door dimension  = 1 m

mass of door = 15 Kg

mass of bullet = 10 g = 0.001 Kg

speed of bullet = 400 m/s

I_{total} =I_{door} + I_{bullet}

I_{total} =\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2

a) from conservation of angular momentum  

L_i = L_f  

mv\dfrac{W}{2} = I_{total}\omega_f  

mv\dfrac{W}{2}= (\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2)\omega_f

\omega_f= \dfrac{mv\dfrac{W}{2}}{\dfrac{1}{3}MW^2 + m(\dfrac{W}{2})^2}

\omega_f= \dfrac{\dfrac{mv}{2}}{\dfrac{MW }{3}+(\dfrac{mW}{4})}

\omega_f= \dfrac{\dfrac{0.01\times 400}{2}}{\dfrac{15\times 1 }{3}+(\dfrac{0.01\times 1}{4})}

\omega_f = 0.4\ rad/s

8 0
3 years ago
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