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sukhopar [10]
3 years ago
12

Anna drives north at a speed of 50 km/h for the first hour. Then, she drives north for a second hour but slows down to 30 km/h.

Compare or contrast the first hour of her trip to the second hour of her trip.
Group of answer choices

The distance is greater in the first hour because her speed is faster.

The distance is greater in the second hour because her speed is slower.

The displacement is the same for both hours because she is travelling north the entire time.

The distances are the same for both hours because she is travelling north the entire time.
Physics
2 answers:
8090 [49]3 years ago
8 0

Answer:

The correct answer is A The distance is greater in the first hour because her speed is faster.

Explanation:

During the first hour, Anna is driving at a speed of 50 km/h. During the second hour, she is only driving at a speed of 30 km/h. The faster she goes, the farther she will go.

Hope this helps,

♥<em>A.W.E.</em><u><em>S.W.A.N.</em></u>♥

kkurt [141]3 years ago
3 0

Answer:

The distance is greater in the first hour because her speed is faster.

Explanation:

As we know that the distance of an object is given as

d = speed \times time

here we know that in first hour speed is

v_1 = 50 km/h Towards North

So the distance moved by the car is

d_1 = 50 \times 1

d_1 = 50 km

Now similarly we have speed in next hour as

v_2 = 30 km/h Towards North

so distance will be

d_2 = v_2 t

d_2 = 30 \times 1

d_2 = 30 km

So correct answer will be

The distance is greater in the first hour because her speed is faster.

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MA_775_DIABLO [31]

The linear speed of the ladybug is 4.1 m/s

Explanation:

First of all, we need to find the angular speed of the lady bug. This is given by:

\omega=\frac{2\pi}{T}

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The period of revolution is the time taken by the ladybug to complete one revolution: in this case, since it does 1 revolution every second, the period is 1 second:

T = 1 s

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\omega=\frac{2\pi}{1 s}=6.28 rad/s

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v=\omega r

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v=(6.28)(0.65)=4.1 m/s

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3 years ago
9) If a wave has a speed of 362 m/s and a period of 4.17 s, what is its wavelength?
kakasveta [241]

Answer:

Option B (1.51 m)

Explanation:

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2 years ago
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What component of a longitudinal sound wave is analogous to a trough of a transverse wave?
anzhelika [568]

Explanation:

There are two components of a longitudinal sound wave which are compression and rarefaction. Similarly, there are two components of the transverse wave, the crest, and trough.

The crest of a wave is defined as the part that has a maximum value of displacement while the trough is defined as the part which corresponds to minimum displacement.

While compression is that space where the particles are close together while the rarefaction is that space where the particles are far apart from each other.

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3 years ago
A stone is dropped into a well. The sound of the splash is heard 6.25 s later. What is the depth of the well? (Take the speed of
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Answer:

depth of well is 163.30 m

Explanation:

Given data

speed of sound = 343 m/s

timer = 6.25 s

to find out

depth of well

solution

let us consider depth d

so equation will be

depth = 1/2 ×g ×t²    ..............1

and

depth = velocity of sound × time    .................2

here we have given time 6.25 that is sum of 2 time

when stone reach at bottom that time

another is sound reach us after stone strike on bottom

so time 1 + time 2 = 6.25 s

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1/2 ×9.8 × t1² = 343 × (6.25 - t1 )

t1 = 5.77376 sec

so height = 1/2 ×g ×t²

height = 1/2 ×9.8 × (5.773)²

height = 163.30 m

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