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3 years ago

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Two cars are traveling along a straight line in the same direction, the lead car at 24.7m/s and the other car at 29.9m/s. at the

moment the cars are 38.1m apart, the lead driver applies the brakes, causing her car to have an acceleration of -1.96m/s2. how long does it take for the lead car to stop? (in s)

1 answer:

sattari [20]3 years ago

7 0

Answer: 12.6 s

We would use the equation of motion:

$v-u=at$

where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

It is given that the lead car has an initial velocity, $u=24.7\hspace{1mm} m/s$ . It decelerates with $a=-1.96 \hspace{1mm} m/s^{2}$ to stop (v=0). we need to calculate the time taken by the lead car to stop.

$\Rightarrow t=\frac{v-u}{a}\\ \Rightarrow t=\frac{0-24.7}{-1.96}=12.6\hspace{1mm} s$

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a 50kg person is standing still in ice skates throws their 2.0kg helmet to the right at 25m/s while on a friction less surface.

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Answer:

Vp = 1 [m/s]

Explanation:

In order to solve this problem, we must use the principle of conservation of the amount of movement. In the same way, analyze the before and after of the actions.

<u>The moment before</u>

The 50kg person is still hold (no movement) with the 2kg helmet

<u>The moment after</u>

The helmet moves at 25[m/s] in one direction, the person moves in the opposite direction, due to the launch of the helmet.

In this way we can apply the principle of conservation of movement, expressing the before and after. To the left we have the before and to the right of the equal sign we have the after.

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As the people sing in church, the sound level everywhere inside is 101 dB . No sound is transmitted through the massive walls, b

katovenus [111]

The sound level from 1 km away is 46.4 dB and the sound energy radiated through the windows and doors in 20 min is 332.6 J

<h3>What do you mean by sound radiates?</h3>

Multiple plate modes participate in the forced vibration of a baffled plate that produces sound. The emitted sound power depends on the activation of each plate mode by the external pressures as well as the radiation characteristics of the individual plate modes and their mutual interaction via their radiated sound, as shown in eq (20) and (24). The net contribution of the mutual interaction between plate modes to the overall sound power may be disregarded if a plate is activated in a manner that results in a plate vibration field that can be defined as reverberant with an equal distribution of energy over all modes.

(Lw) = 10·log (W/Wo) dB

Given:

sound level, $\beta= 101 dB$

Area, A = $22\;m^{2}$

Time, $\triangle t = 20\;min=1200\;s$

Intensity, $I=1\times 10^{-12}\;W/m^{2}$

$r=1\;km=1000\;m$

(a)

We know that, Sound level is,

$\beta=10\times log(\frac{I}{I_{o} } )$

Solving the above equation for sound intensity,

$I=I_{o} \times 10^{\frac{\beta}{10} }$

$I=1 \times 10^{-12} \times 10^{\frac{101}{10} }$

$I=0.0126\;W/m^{2}$

Therefore, The sound energy is,

$E=P\times \triangle t$

Substitute $P=I \times A$ in the above equation,

$E=I \times A \times \triangle t$

$E=0.0126 \times 22 \times 1200$

$E=332.6\;J$

(b)

Half of a sphere area with 1 km radius is equal to Area of the sound at 1 km distance,

$A_{hemisphere} = \frac{1}{2} \times 4 r^{2} \pi$

Substitute the known value in the above equation ,

$A_{hemisphere} = \frac{1}{2} \times 4 (1000)^{2} \pi$

$A_{hemisphere} = 6283185\;m^{2}$

Sound Intensity is,

$I = \frac{P}{A_{hemisphere}}$

Substitute $P=I \times A$ in the above equation,

$I = \frac{I \times A}{A_{hemisphere}}$

Substitute the known value in the above equation,

$I = \frac{0.0126 \times 22}{6283185}$

$I = 4.4 \times 10^{-8}\;W/m^{2}$

Sound level is,

$\beta=10\times log(\frac{I}{I_{o} } )$

Substitute the known value in the above equation,

$\beta=10\times log(\frac{4.4 \times 10^{-8} }{1 \times 10^{-12} } )$

$\beta=46.4\;dB$

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