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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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8 0
3 years ago
An RL circuit contains a resistor with R = 6800 Ω and an inductor with L = 2300 µH. If the impedance of this circuit is 160,000
Rainbow [258]

| Impedance | = √ [R² +(ωL)²]

R² = 6800² = 4.624 x 10⁷
 
(ωL)² = (2 · π · f · 2.3 · 10⁻³)²

          = 2.0884 x 10⁻⁴  f²

| Z | =  √[ (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²) ]  =  1.6 x 10⁵

     (1.6 x 10⁵)²  =  (4.624 x 10⁷) + (2.0884 x 10⁻⁴ f²)

     (2.56 x 10¹⁰) - (4.624 x 10⁷)  =  2.0884 x 10⁻⁴ f²


Frequency² =   (2.56 x 10¹⁰ - 4.624 x 10⁷)  /  2.0884 x 10⁻⁴

                    =       2.555 x 10¹⁰ / 2.0884 x 10⁻⁴

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7 0
3 years ago
You are holding one end of an elastic cord that is fastened to a wall 3.0 m away. You begin shaking the end of the cord at 2.3 H
Karo-lina-s [1.5K]

Answer:

Time take to fill the standing wave to the entire length of the string is 1.3 sec.

Explanation:

Given :

The length of the one end x= 3m, frequency of the wave f = 2.3 Hz, wavelength of the wave λ = 1 m.

Standing wave is the example of the transverse wave, standing wave doesn't transfer energy in a medium.

We know,

∴ v = fλ

Where v = speed of the standing wave.

also, ∴ v=\frac{x}{t}

where t = time take to fill entire length of the string.

Compare above both equation,

⇒   t = \frac{3}{2.3} sec

     t = 1.3sec

Therefore, the time taken to fill entire length 0f the string is 1.3 sec.

7 0
3 years ago
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