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2 years ago

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Two people are talking at a distance of 3.0 m from where you are and you measure the sound intensity as 1.1 × 10-7 W/m2. Another

student is 4.0 m away from the talkers. What sound intensity does the other student measure? Assume that the sound spreads out uniformly and undergoes no significant reflections or absorption

1 answer:

ioda2 years ago

8 0

Answer:

$6.1875\times 10^{-8}$

Explanation:

Assuming uniform spread of sound with no significant reflections or absorption. We know that sound intensity varies $I=\frac {k}{r^{2}}$ where r is the distance

Since intensity is given then when at 3 m

$1.1\times 10^{-7}= \frac {k}{3^{2}}$

$k=3^{2}\times 1.1\times 10^{-7}= 9.9\times 10^{-7}$

Since we have the constant then at 4m

Intensity, $I= \frac {9.9\times 10^{-7}}{4^{2}}=6.1875\times 10^{-8}$

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Sally finds herself stranded on a frozen pond so slippery that she can't stand up or walk on it. To save herself, she throws one

8_murik_8 [283]

Answer:

a) 2.5 m/s. (In the opposite direction to the direction in which she threw the boot).

b) The centre of mass is still at the starting point for both bodies.

c) It'll take Sally 12 s to reach the shore which is 30 m from her starting point.

Explanation:

Linear momentum is conserved.

(mass of boot) × (velocity of boot) + (mass of sally) × (velocity of Sally) = 0

5×30 + 60 × v = 0

v = (-150/60) = -2.5 m/s. (Minus inicates that motion is in the opposite direction to the direction in which she threw the boot).

b) At time t = 10 s,

Sally has travelled 25 m and the boot has travelled 300 m.

Taking the starting point for both bodies as the origin, and Sally's direction as the positive direction.

Centre of mass = [(60)(25) + (5)(-300)]/(60+5)

= 0 m.

The centre of mass is still at the starting point for both bodies.

c) The shore is 30 m away.

Speed = (Distance)/(time)

Time = (Distance)/(speed) = (30/2.5)

Time = 12 s

Hope this Helps!!!

7 0

2 years ago

Read 2 more answers

A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wal

Studentka2010 [4]

(a) 18.9 m/s

The motion of the stone consists of two independent motions:

- A horizontal motion at constant speed

- A vertical motion with constant acceleration ( $g=9.8 m/s^2$ ) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

$u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s$

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

$v=18.9 m/s$

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

$v_y^2 - u_y^2 = 2ah$ (1)

where

$u_y = 16.4 m/s$ is the initial vertical velocity

$v_y$ is the vertical velocity at height h

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (negative because it is downward)

At the top of the parabolic path, $v_y = 0$ , so we can use the equation to find the maximum height

$h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m$

So, at half of the maximum height,

$h = \frac{13.7}{2}=6.9 m$

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

$v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s$

And so, the speed of the stone at half of the maximum height is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s$

(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

$v_1 = 18.9 m/s$

while the speed at half of the maximum height (part b) is

$v_2 = 22.2 m/s$

So the difference is

$\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s$

And so, in percentage,

$\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%$

So, the stone in part (b) is moving 17.4% faster than in part (a).

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2 years ago

4. In a closed system consisting of a cannon and a cannonball, the kinetic energy of a cannon is 72,000 J. If the cannonball is

FromTheMoon [43]

Answer:

D an B

Explanation:

3 0

2 years ago

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1) My 14V car battery could be used to charge my laptop, but I need to use an inverter to first convert it to a standard 120V. T

LenaWriter [7]

Answer:

1) Charge chord resistance is 75 Ω

2) Charge chord resistance is 6.33 Ω

Explanation:

1) To answer the question, we note that the the formula voltage is found as follows;

V = IR

Therefore,

$R = \frac{V}{I} = \frac{120}{1.6} = 75 \, \Omega$

2) Where the voltage, V = 19.5 V and the current, I = 3.33 A, we have;

Initial resistance R₁ = 19.5 V/(3.33 A) = 5.86 Ω

However, to reduce the current to 1.6 A, we have;

$R_T = \frac{19.5}{1.6} = 12.1875 \ \Omega$

Therefore, where the resistance is found by the sum of the total resistance we have;

$R_T$ = R₁ + Charge chord resistance

∴ 12.1875 = 5.86 + Charge chord resistance

Hence, charge chord resistance = 6.33 Ω

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2 years ago

The resistivity of a silver wire with a radius of 5.04 × 10–4 m is 1.59 × 10–8 ω · m. if the length of the wire is 3.00 m, what

Alexxx [7]

<span>5.98 x 10^-2 ohms. Resistance is defined as: R = rl/A where R = resistance in ohms r = resistivity (given as 1.59x10^-8) l = length of wire. A = Cross sectional area of wire. So plugging into the formula, the known values, including the area of a circle being pi*r^2, gives: R = 1.59x10^-8 * 3.00 / (pi * (5.04 x 10^-4)^2) R = (4.77 x 10^-8) / (pi * 2.54016 x 10 ^-7) R = (4.77 x 10^-8) / (7.98015 x 10^-7) R = 5.98 x 10^-2 ohms So that wire has a resistance of 5.98 x 10^-2 ohms.</span>

4 0

2 years ago