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3 years ago

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A spherically mirrored ball is slowly lowered at New Years Eve as midnight approaches. The ball has a diameter of 8.0 ft. Assume

you are standing directly beneath it and looking up at the ball. When your reflection is half your size then the mirror is _______ ft above you.

1 answer:

Archy [21]3 years ago

4 0

Answer:

The distance between mirror and you is 2 ft.

Explanation:

diameter, d = 8 ft

radius of curvature, R = 4 ft

magnification, m = 0.5

focal length, f = R/2 = 4/2 = 2 ft

let the distance of object is u and the distance of image is v.

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\\\\\frac{1}{2}=\frac{1}{v}+\frac{1}{u}\\\\v = \frac {2 u}{u - 2}$

Use the formula of magnification

$m = \frac{v}{u}\\\\0.5 =\frac { u}{u - 2}\\ \\u - 2 = 2 u \\\\u = -2 ft$

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It is most accurate to say that body mass index (BMI) provides information about an individual's height-weight ratio. The correct answer is B.

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Two streams merge to form a river. One stream has a width of 8.9 m, depth of 3.8 m, and current speed of 2.3 m/s. The other stre

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<h2>Depth of river is 4.48 m</h2>

Explanation:

Discharge = Area x Velocity

Discharge of river = Discharge of stream 1 + Discharge of stream 2

$A_rv_r=A_1v_1+A_2v_2$

Area of stream 1 = w₁ x d₁ = 8.9 x 3.8 = 33.82 m²

Area of stream 2 = w₂ x d₂ = 6.5 x 3.9 = 25.35 m²

Velocity of stream 1 = 2.3 m/s

Velocity of stream 2 = 2.6 m/s

Velocity of river = 3 m/s

$\texttt{Area of river = }w_r\times d_r=10.7\times d_r$

Substituting

$A_rv_r=A_1v_1+A_2v_2\\\\10.7\times d_r\times 3=33.82\times 2.3+25.35\times 2.6\\\\d_r=4.48m$

Depth of river = 4.48 m

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Which of the following statements is NOT true about absorbents used to soak up different leaks and spills? A) They may be pickin

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The correct option is D.
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A four-wheel-drive vehicle is transporting an injured hiker to the hospital from a point that is 30 km from the nearest point on

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Answer:

$D=200\ km$

Explanation:

distance on terrain, $d_t=30\ km$

distance on the road, $d_r=70\ km$

speed on terrain, $v_t=30\ km.hr^{-1}$

speed on road, $v_r=130\ km.hr^{-1}$

<u>time taken on the terrain,</u>

$t_t=\frac{d_t}{v_t}$

$t_t=\frac{30}{30}$

$t_t=1\ hr$

<u>time taken to cover the distance on the road:</u>

$t_r=\frac{d_r}{v_r}$

$t_r=\frac{70}{130}$

$t_r=\frac{7}{13}\ hr$

<u>Now the distance covered on terrain in the total time:</u>

$D= v_r\times (t_r+t_t)$

$D= 130\times (\frac{7}{13}+1)$

$D=130\times \frac{20}{13}\$

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<em>is the distance the vehicle must target on the road to minimize the time taken in going off the road.</em>

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3 years ago

You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene

uranmaximum [27]

Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is $150\,^{\circ}C$ . According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

$\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T)$ (1)

Where:

$\rho_{k}$ - Density of kerosene, measured in kilograms per cubic meter.

$V_{k}$ - Volume of kerosene, measured in cubic meters.

$c_{k}$ , $c_{t}$ - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

$T_{k,o}$ , $T_{t,o}$ - Initial temperatures of kerosene and tin, measured in degrees Celsius.

$T$ - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

$V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}$

If we know that $m_{t} = 1.83\,kg$ , $c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}$ , $T_{t,o} = 88\,^{\circ}C$ , $T_{k,o} = 24.0\,^{\circ}C$ , $T = 57\,^{\circ}C$ , $c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C}$ and $\rho_{k} = 820\,\frac{kg}{m^{3}}$ , then the volume of the liquid needed to accomplish this task without boiling is:

$V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}$

$V_{k} = 2.273\times 10^{-4}\,m^{3}$

$V_{k} = 0.273\,L$

0.273 liters are needed to accomplish this task without boiling.

3 0

3 years ago