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3 years ago

Please help!! Thanks!!

Physics

1 answer:

Igoryamba3 years ago

4 0

B. 200 N to the left.

(The OTHER 100N to the left is cancelled out by the 100N to the right.)

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A commuter backs her car out of her garage with an acceleration of 1.40 m/s^2. (a) How long does it take her to reach a speed of

jonny [76]

Answer:

a) It takes her 1.43 s to reach a speed of 2.00 m/s.

b) Her deceleration is - 2.50 m/s²

Explanation:

The equation of velocity for an object that moves in a straight line with constant acceleration is as follows:

v = v0 + a · t

Where:

v = velocty.

v0 = initial velocity.

a = acceleration.

t = time.

a) Using the equation of velocity, let´s consider that the car moves in the positive direction. Then:

v = v0 + a · t

2.00 m/s = 0 m/s + 1.40 m/s² · t

t = 2.00 m/s / 1.40 m/s²

t = 1.43 s

It takes her 1.43 s to reach a speed of 2.00 m/s

b) Let´s use again the equation of velocity, knowing that at t = 0.800 s the velocity is 0 m/s:

v = v0 + a · t

0 = 2.00 m/s + a · 0.800 s

-2.00 m/s / 0.800 s = a

a = -2.50 m/s²

Her deceleration is - 2.50 m/s²

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___________ used math to show that the planets move in ellipses, not perfect circles, around the Sun.

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A tuning fork of frequency 254 Hz and an open orang pipe of slightly lower frequency are at 15oC. When

katovenus [111]

The temperature of the air in the open orang pipe has been altered by 18.73° C

The frequency of an open orang pipe is estimated by using the formula:

$\mathbf{f = \dfrac{v}{2L}}$

Then, the combination of the frequency of the tuning fork and the open orang pipe is:

$\mathbf{254 - \dfrac{v}{2L} }$

These combinations of frequency produce 4 beats per sound.

i.e.

$\mathbf{254 - \dfrac{v}{2L} =4}$

$\mathbf{ \dfrac{v}{2L} = 254-4 }$

$\mathbf{ \dfrac{v}{2L} = 250 ----(1)}$

When it is altered, the beats first diminish and increase again by 4.

i.e.

$\mathbf{ \dfrac{v'}{2L} = 254+4 }$

$\mathbf{ \dfrac{v'}{2L} = 258 --- (2) }$

If we equate both equations (1) and (2) together, we have:

$\mathbf{\dfrac{v'}{v}= \dfrac{258}{250}}$

However, from our previous knowledge, we understand that the velocity of an object varies directly proportional to the square root of its temperature.

Hence;

when the temperature of the pipe = unknown ???
the temperature of the open orang pipe = 15

∴

$\implies \mathbf{\sqrt{\Big(\dfrac{273 + T}{273 + 15}\Big)}= \dfrac{258}{250}}$

By squaring both sides, we have:

$\implies \mathbf{\Big(\dfrac{273 + T}{273 + 15}\Big)}= \Big (\dfrac{258}{250}\Big )^2}$

$\implies \mathbf{\Big(\dfrac{273 + T}{273 + 15}\Big)= \Big (\dfrac{66564}{62500}\Big )}$

$\implies \mathbf{\Big(\dfrac{273 + T}{288}\Big)= \Big (1.065024\Big )}$

$\implies \mathbf{273 +T =306.726912 }$

T = 306.726912 - 273

T ≅ 33.73 ° C

∴

The change in temperature ΔT = 33.73° C - 15° C

The change in temperature ΔT = 18.73° C

Learn more about wave frequency here:

brainly.com/question/14316711?referrer=searchResults

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3 years ago