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3 years ago

A car has momentum, p=10880 kg m/s, east. If the car is traveling east at 16 m/s, its mass must be [m] kg.

Physics

2 answers:

GuDViN [60]3 years ago

6 0

<h3><u>Given</u> :</h3>

Momentum of car = 10880 kg m/s

Velocity of car = 16 m/s

We. have to find mass of the car $.$

<h3><u>Solution</u> :</h3>

★ Momentum of body is measured as the <u>product of mass and its velocity</u>.

It is a vector quantity having both <u>magnitude</u> as well as <u>direction</u>.

SI unit : kg m/s or N s

Formula : p = m × v

p denotes momentum
m denotes mass
v denotes velocity

★ <u>Calculation</u> :

⭆ p = m × v

⭆ 10880 = m × 16

⭆ m = 10880/16

⭆ m = 680 kg

∴ <u>Mass of the car = 680 kg</u>

<h3>Hope It Helps!</h3>

MaRussiya [10]3 years ago

5 0

Answer:

<h3>The answer is 67.5 kg</h3>

Explanation:

To find the mass of an object given it's momentum and velocity we use the formula

$m = \frac{p}{v} \\$

where

p is the momentum

v is the velocity

From the question

p = 1080 kgm/s

v = 16 m/s

We have

$m = \frac{1080}{16} = \frac{135}{2} \\$

We have the final answer as

Hope this helps you

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(a) 2 Hz

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$f_n = n f_1$

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$f_3 = 3 \cdot f_1 = 3 \cdot 440 Hz =1320 Hz$

while the frequency of the second harmonic of the E ( $f_1 = 659 Hz$ ) is

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(b) 2 Hz

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$f_B = |f'-f|$

In this case, when the strings are properly tuned, we have

- Frequency of the 3rd harmonic of A-note: 1320 Hz

- Frequency of the 2nd harmonic of E-note: 1318 Hz

So, the beat frequency should be (if the strings are properly tuned)

$f_B = |1320 Hz - 1318 Hz|=2 Hz$

(c) 1324 Hz

The fundamental frequency on a string is proportional to the square root of the tension in the string:

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this means that by tightening the string (increasing the tension), will increase the fundamental frequency also*, and therefore will increase also the frequency of the 2nd harmonic.

In this situation, the beat frequency is 4 Hz (four beats per second):

$f_B = 4 Hz$

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$f_B = |f_3-f_2|$

and $f_3 = 1320 Hz$ , we have two possible solutions for $f_2$ :

$f_2 = f_3 - f_B = 1320 Hz - 4 Hz = 1316 Hz\\f_2 = f_3 + f_B = 1320 Hz + 4 Hz = 1324 Hz$

However, we said that increasing the tension will increase also the frequency of the harmonics (*), therefore the correct frequency in this case will be

1324 Hz

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