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2 years ago

(10 pt)

Physics

1 answer:

Lubov Fominskaja [6]2 years ago

8 0

Answer:

Explanation:

Wetlands: These are areas or placed where water covers the soil or the area or is present either at or near the surface of the soil all year or for varying periods of time during the year,it may also include the growing season.

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In your opinion, is "avoiding" an effective conflict management strategy? Why or why not?

soldi70 [24.7K]

In my personal opinion, avoiding in not an effective conflict management strategy. When you are avoiding a conflict, you are really just pushing it to the side and it will need to be dealt with sooner or later. Avoiding a conflict is also ammeter thing to do as well, you should take responsibility and handle any situation.

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3 years ago

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N 4. Which of the following can cause a short circuit?

user100 [1]

Answer:

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Explanation:

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2 years ago

A simple generator is used to generate a peak output voltage of 25.0 V . The square armature consists of windings that are 7.0 c

4vir4ik [10]

Answer:

<h3>The 28 loops wound on the square armature</h3>

Explanation:

Peak output voltage $\epsilon _{peak} = 25$ V

Area of square armature $A = (7 \times 10^{-2} )^{2} = 49 \times 10^{-4}$

Magnetic field $B = 0.490$ T

Angular frequency $\omega = 2\pi f = 2 \pi \times 60 = 120\pi$

According to the law of electromagnetic induction,

$\epsilon _{peak} = NBA \omega$

Where $N =$ number of loops of wire.

$N = \frac{25}{49 \times 10^{-4} \times 0.49 \times 120\pi }$

$N = 27.6$ ≅ 28

Thus, 28 loops of wire should be wound on the square armature.

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3 years ago

A rocket moves upward, starting from rest with an acceleration of +30.0 m/s2 for 5.00 s. It runs out of fuel at the end of this

olchik [2.2K]

Answer:

The distance covered by the rocket after fuel ran out is $3442.04 m$

Explanation:

Given that the rocket moves with an acceleration $a=30m/s^2$

time $t=5 s$

Since the rocket starts from rest initial velocity $u=0 s$

The distance it travelled within this time is given by $s=ut+ \frac{1}{2} at^2$ $=0 \times 5+ \frac{1}{2} (30\times25)=375 m$

Velocity at this point is given by $v=u+at$

$v=0+30\times5=150m/s$

Given that at this height it runs out of fuel but travels further. Here final velocity $v=0$ (maximum height), initial velocity $u=150 m/s$ and time to zero velocity $t=\frac{v}{g} = \frac{150}{9.8} =15.3 s.$

Thus it travels $15.3 seconds$ more after fuel running out. The distance covered during this period is given

$s= ut+\frac{1}{2} gt^2=150 \times 15.3+1/2 \times9.8 \times 15.3^2=3442.04 m$

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2 years ago

Calculate the wavelength λ1 for gamma rays of frequency f1 = 7.20×1021 hz .

bagirrra123 [75]

The relationship between wavelength $\lambda$ , frequency f and speed of light c for an electromagnetic wave is
$\lambda= \frac{c}{f}$
Using the data of the problem, we find
$\lambda= \frac{3\cdot 10^8 m/s }{7.20 \cdot 10^{21} Hz}=4.17 \cdot 10^{-14} m$

5 0

3 years ago