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2 years ago

A car travels in a straight line for 4.4 h at a constant speed of 81 km/h. What is its acceleration?

Physics

2 answers:

Fed [463]2 years ago

7 0

The speed and length of time don't matter.

Constant speed in a straight line is the DEFINITION of ZERO acceleration.

irga5000 [103]2 years ago

5 0

Answer:

$a = 0m/s^2$

Explanation:

we can use the equation:

$v_f = a(t) + v_i$

And plug in our numbers:

$81km/h = am/s^2(4.4h) + 81hm/h$

$0 = am/s^2(4.4h)$

$0 = am/s^2$

$a = 0m/s^2$

Because the speed was constant, there was no difference between the final and initial velocities, meaning that once you do the math you see there is no acceleration affecting the speed.

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Answer:

Lion Average speed:60 km/h

Explanation:

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Answer:

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3 years ago

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lesantik [10]

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2 years ago

Read 2 more answers

Consider a taut inextensible string. You shake the end of the string with some frequency, causing a wave to travel down the stri

masha68 [24]

Answer:

Part 1:

Option B is correct (It will remain unchanged).

Part 2:

Option C is correct (It will increase by a factor of √ 2)

Part 3:

Option E is correct (It will be half as fast/long.)

Part 4:

Option C is correct (It will increase by a factor of √ 2.)

Explanation:

Formula we are going to use:

V=f*λ

Where:

V is the speed of Sound

f is the frequency of wave

λ is the wavelength.

The speed of wave , tension and linear density have following relation:

$V=\sqrt{F/\rho}$

Where:

V is the speed of Sound (Initial)

F is the tension in string (Initial)

$\rho$ is the linear density of string (Constant)

Terms:

V' is the new speed

f' is the new frequency

λ' is the wavelength

Solution:

Part 1:

From $V=\sqrt{F/\rho}$ :

Speed of Sound is independent of the frequency of shaking so speed well remain unchanged.

Option B is correct (It will remain unchanged)

Part 2:

If F'=2F then

$V=\sqrt{F/\rho}$

$V'=\sqrt{F'/\rho}\\V'=\sqrt{2F/\rho}\\V'= \sqrt{2} * \sqrt{F/\rho}\\V'=\sqrt{2}V$

Option C is correct (It will increase by a factor of √ 2)

Part 3:

Formula we are going to use:

V=f*λ

Given f'=2f,

Even though frequency is doubled we will keep velocities same. V=V' in order to find the changing wavelength.

V'=f'*λ'

f*λ=f'*λ'

f*λ=2f*λ'

Solving above Equation:

λ'=λ/2

Option E is correct (It will be half as fast/long.)

Part 4:

T'=2T means $V'=\sqrt{2}V$ (From Part 1)

f'=f

Now:

V'=f'*λ'

$\sqrt{2}f*\lambda=f'*\lambda '\\\sqrt{2}f*\lambda=f*\lambda '\\ \lambda '=\sqrt{2}*\lambda$

Option C is correct (It will increase by a factor of √ 2.)

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2 years ago

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