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3 years ago

A 1650 kg car accelerates at a rate of 4.0 m/s^2. How much force is the car's engine producing?

Physics

2 answers:

Westkost [7]3 years ago

7 0

F = M A

Force = (mass) x (acceleration)

= (1,650 kg) (4 m/s²) = 6,600 kg-m/s² = <em>6,600 Newtons</em>

Eduardwww [97]3 years ago

3 0

According to Newtons laws of motion F = MA. This means Force = Mass x Acceleration.

So...

4 x 1650 = 66,000
The units for Force is Newtons

So the answer is 66,000 Newtons

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30 points! Please answer at least one and how you found it! :)

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1 Is advantage because with out is. you couldn't carry the piano at all.
2 Efficiency because it is going against you.

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2 years ago

What will be the acceleration of a 40-kilogram object that is pushed with a net force of 80 newtons?

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= 80 N/40 kg
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3 years ago

) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note tha

vaieri [72.5K]

(a) 0.0021 s, 2926.5 rad/s

The frequency of the B note is

$f= 466 Hz$

The time taken to make one complete cycle is equal to the period of the wave, which is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{466 Hz}=0.0021 s$

The angular frequency instead is given by

$\omega = 2\pi f$

And substituting

f = 466 Hz

We find

$\omega = 2\pi (466 Hz)=2926.5 rad/s$

(b) 20 Hz, 125.6 rad/s

In this case, the period of the sound wave is

T = 50.0 ms = 0.050 s

So the frequency is equal to the reciprocal of the period:

$f=\frac{1}{T}=\frac{1}{0.050 s}=20 Hz$

While the angular frequency is given by:

$\omega = 2\pi f = 2 \pi (20 Hz)=125.6 rad/s$

(c) $4.30\cdot 10^{14} Hz, 7.48\cdot 1^{14} Hz, 2.33\cdot 10^{-15} s, 1.34\cdot 10^{-15}s$

The minimum angular frequency of the light wave is

$\omega_1 = 2.7\cdot 10^{15}rad/s$

so the corresponding frequency is

$f=\frac{\omega}{2 \pi}=\frac{2.7\cdot 10^{15}rad/s}{2\pi}=4.30\cdot 10^{14} Hz$

and the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{4.30\cdot 10^{14}Hz}=2.33\cdot 10^{-15}s$

The maximum angular frequency of the light wave is

$\omega_2 = 4.7\cdot 10^{15}rad/s$

so the corresponding frequency is

$f=\frac{\omega}{2 \pi}=\frac{4.7\cdot 10^{15}rad/s}{2\pi}=7.48\cdot 10^{14} Hz$

and the period is the reciprocal of the frequency:

$T=\frac{1}{f}=\frac{1}{7.48\cdot 10^{14}Hz}=1.34\cdot 10^{-15}s$

(d) $2.0\cdot 10^{-7}s, 3.14\cdot 10^{7} rad/s$

In this case, the frequency is

$f=5.0 MHz = 5.0 \cdot 10^6 Hz$

So the period in this case is

$T=\frac{1}{f}=\frac{1}{5.0\cdot 10^6 Hz}=2.0 \cdot 10^{-7} s$

While the angular frequency is given by

$\omega = 2\pi f=2 \pi (5.0\cdot 10^{6}Hz)=3.14\cdot 10^{7} rad/s$

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Explain why engineers leave a gap in a concrete bridge (answer must relate to density)

seraphim [82]

<span>Different materials expand and contract at different rates based on temperature. Just like if you leave a plastic bottle full of water in a freezer it will burst, but if you leave it partially full no problem.....Ok?Expansion joints do the same for bridges. There is a gap to allow for temperature related expansions and contractions. Sometimes you drive over bridges and roadways where this movement is constricted and you might notice a bumpy ride. Engineers can predict the variation of structural length based on span lengths and leave the necessary gaps.....btw, NICE QUESTION:)</span>

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3 years ago