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allsm [11]
2 years ago
6

Is the meter, centimeter, or millimeter more useful for measuring the length and width of your classroom? why?

Physics
1 answer:
Verizon [17]2 years ago
8 0
A meter because centimeters and millimeters are way smaller than meters
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1. An object with a mass of m is thrown straight up near the surface of the earth. While the object is going up, the net force o
Vaselesa [24]

Answer:

C: equal to mg

Explanation:

in free-fall, gravity is always the net force on an object

5 0
3 years ago
What is the highest degrees above the horizon the moon ever gets during the year in the Yakima Valley ?
Ivahew [28]

The trickiest part of this problem was making sure where the Yakima Valley is.
OK so it's generally around the city of the same name in Washington State.

Just for a place to work with, I picked the Yakima Valley Junior College, at the
corner of W Nob Hill Blvd and S16th Ave in Yakima.  The latitude in the middle
of that intersection is 46.585° North.  <u>That's</u> the number we need.

Here's how I would do it:

-- The altitude of the due-south point on the celestial equator is always
(90° - latitude), no matter what the date or time of day.

-- The highest above the celestial equator that the ecliptic ever gets
is about 23.5°. 

-- The mean inclination of the moon's orbit to the ecliptic is 5.14°, so
that's the highest above the ecliptic that the moon can ever appear
in the sky.

This sets the limit of the highest in the sky that the moon can ever appear.

90° - 46.585° + 23.5° + 5.14° = 72.1° above the horizon .

That doesn't happen regularly.  It would depend on everything coming
together at the same time ... the moon happens to be at the point in its
orbit that's 5.14° above ==> (the point on the ecliptic that's 23.5° above
the celestial equator).

Depending on the time of year, that can be any time of the day or night.

The most striking combination is at midnight, within a day or two of the
Winter solstice, when the moon happens to be full.

In general, the Full Moon closest to the Winter solstice is going to be
the moon highest in the sky.  Then it's going to be somewhere near
67° above the horizon at midnight.


5 0
2 years ago
Sam is observing the velocity of a car at different times. After three hours, the velocity of the car is 53 km/h. After six hour
Serhud [2]
For the answer to the question above, first find out the gradient. 

<span>m = rise/run </span>
<span>=(y2-y1)/(x2-x1) </span>

<span>the x's and y's are the points given: "After three hours, the velocity of the car is 53 km/h. After six hours, the velocity of the car is 62 km/h" </span>
<span>(x1,y1) = (3,53) </span>
<span>(x2,y2) = (6,62) </span>

<span>sub values back into the equation </span>
<span>m = (62-53)/(6-3) </span>
<span>m = 9/3 </span>
<span>m = 3 </span>

<span>now we use a point-slope form to find the the standard form </span>
<span>y-y1 = m(x-x1) </span>
<span>where x1 and y1 are any set of point given </span>
<span>y-53 = 3(x-3) </span>
<span>y-53 = 3x - 9 </span>
<span>y = 3x - 9 + 53 </span>
<span>y = 3x + 44 </span>

<span>y is the velocity of the car, x is the time.
</span>I hope this helps.
4 0
3 years ago
How does the ratio of tin to copper affect the properties of the alloy bronze?
Pani-rosa [81]

Answer:

Stronger and harder than either of the pure metals

Explanation:

7 0
2 years ago
A group of bike riders took a 4 hour trip. During the first 3 hours, they traveled a total of 50 km, but
zlopas [31]

Answer:

15km/hr

Explanation:

The average speed for the entire trip is the sum of the total distance traveled divided by the total time of the trip.

Total time  = 4hr

distance for the first 3hrs  = 50km

distance for the last 1hr  = 10km

  Total distance  = 50km + 10km  = 60km

Now;

  Average speed  = \frac{total distance }{time taken}  

 Insert the parameters and solve;

 Average speed  = \frac{60km}{4hr}   = 15km/hr

7 0
3 years ago
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