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kogti [31]
3 years ago
11

Can someone please help me with this physics question? I'm desperate!

Physics
1 answer:
Lelu [443]3 years ago
3 0

Answer:

a) 2·√10 seconds

b) Linda should be approximately 30.6 meters

c) Jenny's speed at the 100-m mark is approximately 6.325 m/s

Explanation:

The speed with which Linda is running = 8.6 m/s

The point Jenny starts = The 80-m mark

The acceleration of Jenny = 1.0 m/s²

a) The time it takes Jenny to run from the 80-m mark to the 100-m mark, <em>t</em>, is given as follows

Δs = u·t + (1/2)·a·t²

Δs = Distance = 100-m - 80-m = 20-m

u = The initial velocity of Jenny = 0

a = Jenny's acceleration = 1.0 m/s²

∴ 20 = 0×t + (1/2) × 1 × t² = t²/2

20 = t²/2

t = √(20 × 2) = 2·√10

The time it takes Jenny to run from the 80-m mark to the 100-m mark = 2·√10 seconds

b) The distance Linda runs in t = 2·√10 seconds, d = v × t

Given that Linda's velocity, v = 8.6 m/s, we have;

d = 8.0 × 2·√10 = 16·√10

The distance Linda runs in t = 2·√10 seconds = 16·√10 meters ≈ 50.6 meters

Therefore, Linda should be approximately (50.6 - 20) meters = 30.6 meters behind Jenny when Jenny starts running

c) Jenny's speed at the 100 m mark is given as follows;

v = u + a·t

t = 2·√10 seconds, a = 1.0 m/s², u = 0

∴ v = 0×t + 1.0×2·√10 = 2·√10 ≈ 6.325

Jenny's speed at the 100-m mark ≈ 6.325 m/s

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The question is incomplete. Here is the complete question.

To understand the decibel scale. The decibel scale is a logarithmic scale for measuring the sound intensity level. Because the decibel scale is logarithmic, it changes by an additive constant when the intensity when the intensity as measured in W/m² changes by a multiplicative factor. The number of decibels increase by 10 for a factor of 10 increase in intensity. The general formula for the sound intensity level, in decibels, corresponding to intensity I is

\beta=10log(\frac{I}{I_{0}} )dB,

where I_{0} is a reference intensity. for sound waves, I_{0} is taken to be 10^{-12} W/m^{2}. Note that log refers to the logarithm to the base 10.

Part A: What is the sound intensity level β, in decibels, of a sound wave whose intensity is 10 times the reference intensity, i.e. I=10I_{0}? Express the sound intensity numerically to the nearest integer.

Part B: What is the sound intensity level β, in decibels, of a sound wave whose intensity is 100 times the reference intensity, i.e. I=100I_{0}? Express the sound intensity numerically to the nearest integer.

Part C: Calculate the change in decibels (\Delta \beta_{2},\Delta \beta_{4} and \Delta \beta_{8}) corresponding to f = 2, f = 4 and f = 8. Give your answer, separated by commas, to the nearest integer -- this will give an accuracy of 20%, which is good enough for sound.

Answer and Explanation: Using the formula for sound intensity level:

A) I=10I_{0}

\beta=10log(\frac{10I_{0}}{I_{0}} )

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<u>The sound Intensity level with intensity 10x is </u><u>10dB</u>.

B) I=100I_{0}

\beta=10log(\frac{100I_{0}}{I_{0}} )

\beta=10log(100)

β = 20

<u>With intensity 100x, level is </u><u>20dB</u>.

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I=2I_{0}

\beta=10log(\frac{2I_{0}}{I_{0}} )

\beta=10log(2)

β = 3

For \Delta \beta_{4}:

I=4I_{0}

\beta=10log(\frac{4I_{0}}{I_{0}} )

\beta=10log(4)

β = 6

For \Delta \beta_{8}:

I=8I_{0}

\beta=10log(\frac{8I_{0}}{I_{0}} )

β = 9

Change is

\Delta \beta_{2},\Delta \beta_{4}, \Delta \beta_{8} = 3,6,9 dB

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Answer:

total stretch of the double-length spring will be 20 cm

Explanation:

given data

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to find out

the total stretch of the double-length spring will be

solution

we can say here spring constant is

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k is spring constant and m is mass and g is acceleration due to gravity

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kx2 = 2mg   ........................3

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Complete Question

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Answer:

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