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rodikova [14]
3 years ago
14

What type of climate do most islands within Oceania have? A. tundra B. tropical humid C. humid continental D. desert

Physics
2 answers:
neonofarm [45]3 years ago
8 0
Most of the islands have a tropical humid climate.

Answer choice:

B. Tropical humid

Explanation:

Oceania includes more than 10,000 islands. Oceania is generally hot and humid year-round. The islands of this region have no true winter or summer, but many areas experience seasonal changes in winds, ocean currents, and rainfall. The climate usually remains<span> hot and humid throughout the year.
</span>
A humid subtropical climate is a zone of climate characterized by hot and humid summers, and mild winters. The Humid Subtropical climate is found on the east coast of continents between 20° and 40° north and south of the equator, including the Islands of Oceania.
 
Tomtit [17]3 years ago
4 0
B
Most of the islands have a tropical humid climate
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An engine moves a motorboat through water at a constant velocity of 22 meters/second. If the force exerted by the motor on the b
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Answer: Option B: 1.3×10⁵ W

Explanation:

Power = \frac{Work \hspace{1mm} done}{Time}

P=\frac{W}{t}

Work Done, W= F.s

Where s is displacement in the direction of force and F is force.

\Rightarrow P = \frac{F.s}{t} =F \times \frac{s}{t}=F.v

where, v is the velocity.

It is given that, F = 5.75 × 10³N

v = 22 m/s

P = 5.75 × 10³N×22 m/s = 126.5 × 10³ W ≈1.3×10⁵W

Thus, the correct option is B

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3 years ago
An interference pattern is produced by light with a wavelength 580 nm from a distant source incident on two identical parallel s
irakobra [83]

Answer:

Explanation:

1 )

Here

wave length used that is λ = 580 nm

=580 x 10⁻⁹

distance between slit d = .46 mm

= .46 x 10⁻³

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= 580 x 10⁻⁹ / .46 x 10⁻³

= 0.126 x 10⁻² radian

2 )

Angular position of second order interference maxima

2 x  0.126 x 10⁻² radian

= 0.252 x 10⁻² radian

3 )

For intensity distribution the formula is

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For angular position of θ1

δ = .126 x 10⁻² radian

I = I₀ cos².126x 10⁻²/2

= I₀ X .998

For angular position of θ2

I = I₀ cos².126x2x 10⁻²/2

=  I₀ cos².126x 10⁻²

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The force between two moving objects that are touching
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A 7450 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci
bagirrra123 [75]

Answer:

a) 520m

b) 10.30 s

c) 100,95 m/s

Explanation:

a) According the given information, the rocket suddenly stops when it reach the height of 520m, because the engines fail, and then it begins the free fall.

This means the maximum height this rocket reached before falling  was 520 m.

b) As we are dealing with constant acceleration (due gravity) g=9.8 \frac{m}{s^{2}} we can use the following formula:

y=y_{o}+V_{o} t-\frac{gt^{2}}{2}   (1)

Where:

y_{o}=520 m  is the initial height of the rocket (at the exact moment in which it stops due engines fail)

y=0  is the final height of the rocket (when it finally hits the launch pad)

V_{o}=0 is the initial velocity of the rocket (at the exact moment in which it stops the velocity is zero and then it begins to fall)

g=9.8m/s^{2}  is the acceleration due gravity

t is the time it takes to the rocket to hit the launch pad

Clearing t:

0=520 m+0-\frac{9.8m/s^{2} t^{2}}{2}   (2)

t^{2}=\frac{-520 m}{-4.9 m/s^{2}}   (3)

t=\sqrt{106.12 s^{2}   (4)

t=10.30 s   (5)  This is the time

c) Now we need to find the final velocity V_{f} for this rocket, and the following equation will be perfect to find it:

V_{f}=V_{o}-gt  (6)

V_{f}=0-(9.8 m/s^{2})(10.30 s)  (7)

V_{f}=-100.95 m/s  (8) This is the final velocity of the rocket. Note the negative sign indicates its direction is downwards (to the launch pad)

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