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3 years ago

An echo is heard from a cliff 3.71 s after a rifle is fired. How many feet away is the cliff

Physics

1 answer:

LenKa [72]3 years ago

3 0

The cliff is 2042 ft away.

We know that the speed of sound in air is directly proportional to the absolute temperature.

First convert the Fahrenheit temperature to Celsius;

°C = 5/9(44.5 - 32)

°C = 6.9 °C

Applying the formula;

V1/V2 = √T1/T2

Where; V1 = velocity of sound in air at 0°C

V2 = Velocity of sound in air at 6.9 °C

1087/V2 = √273/279.9

V2= 1101 ft/s

Given that; V = 2s/t

Where s is the distance of the cliff

t is the time taken

1101 ft/s = 2s/3.71 s

s = 1101 ft/s × 3.71 s/2

s = 2042 ft

Learn more:brainly.com/question/15381147

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Explanation:

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$\Delta T=32-18=14^{\circ}$

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$Q_1=129.01 W$

Heat Transfer due to convection is given by

$Q_2=hA(\Delta T)$

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When an object in simple harmonic motion is at its maximum displacement, its____________ is also at a maximum.

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~~~NEED HELP ASAP~~~ Please solve each section and show all work for each section.

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Explanation:

Forces on Block A:

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass $m_A$ as

$x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)$

$y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)$

Substituting (2) into (1), we get

$\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)$

where $f_N= \mu_kN$ , the frictional force on $m_A.$ Set this aside for now and let's look at the forces on $m_B$

Forces on Block B:

Let the x-axis be (+) up along the inclined plane. We can write the forces on $m_B$ as

$x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)$

$y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)$

From (5), we can solve for N as

$N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)$

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by

$T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)$

Substituting (7) into (4) we get

$m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba$

Collecting similar terms together, we get

$(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag$

$a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)$

Putting in the numbers, we find that $a = 1.4\:\text{m/s}$ . To find the tension T, put the value for the acceleration into (7) and we'll get $T = 21.3\:\text{N}$ . To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get $N = 50.9\:\text{N}$

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