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3 years ago

Starting over, so here's some points! Take em ~Jayden

Physics

2 answers:

Serhud [2]3 years ago

8 0

Answer:

Thank you again!!!! :D

Nat2105 [25]3 years ago

7 0

Answer:

tysm!

Explanation:

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In moving out of a dormitory at the end of the semester, a student does 1.20 x 104 J of work. In the process, his internal energ

Zielflug [23.3K]

Answer:

(a) W=1.20×10⁴J

(b) U= -5.46×10⁴J

Explanation:

Given that student does 1.20×10⁴J work

(a) W=1.20×10⁴J

Work done by student,so positive sign

During the process, his internal energy decreases by 5.46×10⁴J.

(b) U= -5.46×10⁴J.

As the Energy decreases therefore negative sign

For (c) Q

We know the formula

$Q=W+U\\Q=1.2*10^{4}+(-5.46*10^{4} )\\ Q=-4.26*10^{4}J$

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3 years ago

Mrs. Rohaley gives you a piece of iron and asks you to determine the physical and chemical properties of the metal. She wants yo

Fofino [41]

Answer:The answer is B

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3 years ago

An engineer is designing a runway. She knows that a plane, starting at rest, needs to reach a speed of 180mph at take-off. If th

kvv77 [185]

Answer:

The plane would need to travel at least $8,\!580\; {\rm ft}$ ( $8.58 \times 10^{3}\; {\rm ft}$ .)

The $10,\!000\; {\rm ft}$ runway should be sufficient.

Explanation:

Convert unit of the the take-off velocity of this plane to $\rm ft\cdot s^{-1}$ :

$\begin{aligned}v &= 180\; {\rm mph} \\ &= 180\; {\rm mi \cdot hrs^{-1}} \times \frac{1\; {\rm hrs}}{3600\; {\rm s}} \times \frac{5280\; {\rm ft}}{1\; {\rm mi}} \\ &= 264\; {\rm ft \cdot s^{-1}}\end{aligned}$ .

Initial velocity of the plane: $u = 0\; {\rm ft \cdot s^{-1}}$ .

Take-off velocity of the plane $v =264\; {\rm ft\cdot s^{-1}}$ .

Let $x$ denote the distance that the plane travelled along the runway. Since acceleration is constant but unknown, make use of the SUVAT equation $x = ((u + v) / 2) \, t$ .

Notice that this equation does not require the value of acceleration. Rather, this equation make use of the fact that the distance travelled (under constant acceleration) is equal to duration $t$ times average velocity $(u + v) / 2$ .

The distance that the plane need to cover would be:

$\begin{aligned}x &= \left(\frac{u + v}{2}\right)\, t \\ &= \frac{0\; {\rm ft \cdot s^{-1}} + 264\; {\rm ft \cdot s^{-1}}}{2} \times 65.0\; {\rm s} \\ &= 8.58\times 10^{3}\; {\rm ft}\end{aligned}$ .

4 0

3 years ago