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3 years ago

Can someone please help me?

Physics

1 answer:

Luba_88 [7]3 years ago

5 0

Answer:

1525 meters above ground

Explanation:

So to do this you will need to write this in slope intercept form or $y=mx+b$ . So 650 would be the b, 175 would be the m, and the x would be 5 so the equation would be $y=175(5)+650$ so if you solve or simplify the equation you will get 1525 meters above the ground and that would be our final answer.

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Which of the following is a problem that some people blame on technology?

seropon [69]

The correct answer is B, widespread pollution. If you look closely, you can see that the other answers are not problems at all, but benefits! :)

7 0

4 years ago

Read 2 more answers

It takes Serina 0.68 hours to drive to school. Her route is 34 km long. What is Serina's average speed on her drive to school?

shusha [124]

Speed = distance / time

Speed = 34 km / 0.68 hrs

Speed = 50 km/hr

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3 years ago

Three deer, A, B, and C, are grazing in a field. Deer B is located 62m from deer A at an angle of 51° north of West. Deer C is l

Anna71 [15]

Set deer A's position to be the origin. Let $c$ be the distance from deer A to deer C. We're given that deer B is 95 m away from deer C, which means the length of the vector $B-C$ is 95 (or $C-B$ ). Then

$|B-C|^2=(B-C)\cdot(B-C)=B\cdot B-2B\cdot C+C\cdot C=|B|^2-2B\cdot C+|C|^2$

$|B-C|^2=|B|^2-2|B||C|\cos(180-77-51)^\circ+c^2$

$95^2=62^2-2(62)(c)\cos52^\circ+c^2$

$c^2-124\cos52^\circ c-5181=0\implies c=120\,\mathrm m$

8 0

4 years ago

As a science fair project, you want to launch an 950 g model rocket straight up and hit a horizontally moving target as it passe

Aleksandr-060686 [28]

Answer:

As a science fair project, you want to launch an 950g model rocket straight up and hit a horizontally moving target as it passes 33.0m above the launch point. The rocket engine provides a constant thrust of 20.0N . The target is approaching at a speed of 18.0m/s . At what horizontal distance between the target and the rocket should you launch?

= 43.56m

Explanation:

acceleration =

(20 - (0.95 * 9.8) )/ (0.95)

= 10.68 / 0.95

= 11.24 m/s²

we use

s = ut + (1/2) at²

Given that

s= 40

u =0

s = 0 * t + (1/2) (11.24)t²

t = √(66/1.24)

t = √5.87

t = 2.42sec

hence

Horizontal distance = 18 * 2.42

= 43.56m

5 0

3 years ago

A newly discovered element has two isotopes. One has an atomic weight of 120.9038 amu with 57.25% abundance. The other has an at

Katarina [22]

Answer : The atomic weight of the element is, 121.75 amu

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Mass of isotope 1 = 120.9038 amu

Percentage abundance of isotope 1 = 57.25 %

Fractional abundance of isotope 1 = 0.5725

Mass of isotope 2 = 122.8831 amu

Percentage abundance of isotope 2 = 100 - 57.25 = 42.75 %

Fractional abundance of isotope 2 = 0.4275

Now put all the given values in above formula, we get:

$\text{Average atomic mass of element}=\sum[(120.9038\times 0.5725)+(122.8831\times 0.4275)]$

$\text{Average atomic mass of element}=121.75amu$

Therefore, the atomic weight of the element is, 121.75 amu

6 0

3 years ago