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lozanna [386]
3 years ago
6

Is the distance traveled during a specific unit of

Physics
1 answer:
Anton [14]3 years ago
4 0

Answer:

velocity

Explanation:

I think i dont really know

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Find the average velocity of the object from point B to C.
ExtremeBDS [4]
The average velocity of the object in moving from point-B to point-C is

                 (the straight-line distance and direction from point-B to point-C)
divided by
                 (the time the object takes to make the trip) .
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How many kilocalories (Calories) does the snack bar contain?
laiz [17]

Answer:

See the explanation below.

Explanation:

Attached is a picture of a chocolate bar that can be easily found on the market. In this picture, we see the table of contents with the amount of energy that depends on the mass in grams of the chocolate bar.

We see that for the bar size of 100G, the energy value is 483 [kcal].

We can see that the energy depends on the size of the chocolate bar.

7 0
3 years ago
Gloria is Latin for what?
max2010maxim [7]

Answer:

Gloria is the anglicized form of the Latin feminine given name gloriae (Latin pronunciation: ['gloːria]), meaning immortal glory; glory, fame, renown, praise, honor.

Explanation:

mark me as brainlist

5 0
3 years ago
Read 2 more answers
PLEASE HELP ME!!! the equations are around the question itself.
mixas84 [53]
I’m so so sorry I wish I could help.... My best answer is no because they have different shapes but I am not sure
6 0
4 years ago
Kiran drove from City A to City B, a distance of 228 mi. She increased her speed by 12 mi/h for the 400-mi trip from City B to C
Degger [83]

Answer:

From city A to city B her speed was 38 mi/h

Explanation:

The traveled distance can be calculated using this equation:

From city A to city B

228 mi = v · t₁

Where:

v = velocity

t₁ = time it took Kiran to travel the 228 mi from city A to city B

From city B to city C

400 mi = (v + 12 mi/h) · t₂

We also know that the entire trip took 14 h, then:

t₁ + t₂ = 14 h

So, we have a system of three equations with three unknwons:

228 mi = v · t₁

400 mi = (v + 12 mi/h) · t₂

t₁ + t₂ = 14 h

Let´s solve the third equation for t₁:

t₁ = 14 h - t₂

Now let´s replace t₁ in the first equation and solve it for t₂

228 mi = v · t₁

228 mi = v · (14 h - t₂)

228 mi/v - 14 h =  - t₂

t₂ = 14 h - 228 mi/v

Now let´s replace t₂ in the second equation:

400 mi = (v + 12 mi/h) · t₂

400 mi = (v + 12 mi/h) · (14 h - 228 mi/v)

400 mi = 14 h · v - 228 mi + 168 mi - 2736 mi²/(v · h)

400 mi = 14 h · v - 60 mi - 2736 mi²/(v · h)

460 mi = 14 h · v - 2736 mi²/(v · h)

Multiplicate by v both sides of the equation:

460 mi · v = 14 h · v² - 2736 mi²/h

0 = 14 h · v² - 460 mi · v - 2737 mi²/h

Solving the quadratic equation:

v = 38 mi/h

(The other solution of the equation is negative, and therefore discarded)

From city A to city B her speed was 38 mi/h

4 0
3 years ago
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