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3 years ago

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You are in your car driving on a highway at 23 m/s when you glance in the passenger-side mirror (a convex mirror with radius of

curvature 150 cm) and notice a truck approaching. If the image of the truck is approaching the vertex of the mirror at a speed of 2.0 m/s when the truck is 2.0 m away, what is the speed of the truck relative to the highway?

1 answer:

Irina-Kira [14]3 years ago

8 0

Answer:

v = 26. 88 m/s +23 m/s

Explanation:

u = 23 m/s, r = 150 cm, u₁ = 2.0 m/s, s =2.0 m

$\frac{1}{s} +\frac{1}{s'} = \frac{2}{R}$

$\frac{1}{2.0 m} +\frac{1}{s'} = \frac{2}{1.50 m}$

Solve s'

$\frac{1}{s'} = \frac{2}{1.50 m} - \frac{1}{2.0 m}$

$\frac{1}{s'} = 1.833 m$

$s' = - 0.545 m$

To determine the speed of the trick to the highway

$\frac{ds}{dt}= \frac{s^2* \frac{ds}{dt}}{s' ^2} =\frac{2.0 ^2m * 2.0 m/s}{0.545^2m}$

$\frac{ds}{dt} = 26.88 m/s$

Now to determine the velocity highway is going to be

v = ds/dt + u

v = 26. 88 m/s +23 m/s

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At one point in the circuit, there is an LED and a resistor in parallel with one another. If you measure the voltage drop across

kifflom [539]

Answer:

C. The voltage drop across the resistor is 2.1V and nothing about the current through the resistor.

Explanation:

When connected in parallel, voltage across the resistances are the same. So if 2.1V was dropped across the LED then 2.1V was also dropped across the resistor. However, this tells us nothing about the current through the resistor. We can find the current across the resistor if we know the resistance of the resistor, but that's about it.

If it were a series connection, then the current would have been the same, but the voltage drop were another story.

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3 years ago

The temperature of a smelting furnace is found to be 2000 ℃. Find the

IRISSAK [1]

Answer:

a) The equivalent temperature of 2000 ºC is 3632 ºF.

b) The equivalent temperature of 2000 ºC is 4091.67 R.

c) The equivalent temperature of 2000 ºC is 2273.15 K.

d) The equivalent temperature of 2000 ºC is 1600 ºRe.

Explanation:

a) The equivalent temperature on the Fahrenheit scale is defined by the following formula:

$T_{F} = \frac{9}{5}\cdot T_{C}+32$ (Eq. 1)

Where:

$T_{C}$ - Temperature, measured in degrees Celsius.

$T_{F}$ - Temperature, measured in degrees Fahrenheit.

If we know that $T_{C} = 2000\,^{\circ}C$ , then the temperature is:

$T_{F} = \frac{9}{5}\cdot (2000\,^{\circ}C)+32\,^{\circ}F$

$T_{F} = 3632\,^{\circ}F$

The equivalent temperature of 2000 ºC is 3632 ºF.

b) From result in a) we determine the equivalent temperature on the Rankine scale by using the following formula:

$T_{R} = T_{F}+459.67$ (Eq. 2)

Where:

$T_{F}$ - Temperature, measured in degrees Fahrenheit.

$T_{R}$ - Temperature, measured in Rankine.

If we know that $T_{F} = 3632\,^{\circ}F$ , then the temperature is:

$T_{R} = 3632\,^{\circ}F+459.67\,R$

$T_{R} = 4091.67\,R$

The equivalent temperature of 2000 ºC is 4091.67 R.

c) The equivalent temperature on the absolute scale is calculated by using this expression:

$T_{K} = T_{C}+273.15$ (Eq. 3)

Where:

$T_{C}$ - Temperature, measured in degrees Celsius.

$T_{K}$ - Temperature, measured in Kelvin.

If we know that $T_{C} = 2000\,^{\circ}C$ , then the temperature is:

$T_{K} = 2000\,^{\circ}C+273.15\,K$

$T_{K} = 2273.15\,K$

The equivalent temperature of 2000 ºC is 2273.15 K.

d) The equivalent temperature on the Réaumur scale is calculated by applying this expression:

$T_{Re} = \frac{4}{5}\cdot T_{C}$ (Eq. 4)

Where:

$T_{C}$ - Temperature, measured in degrees Celsius.

$T_{Re}$ - Temperature, measured in degrees Réaumur.

If we know that $T_{C} = 2000\,^{\circ}C$ , then the temperature is:

$T_{Re} = \frac{4}{5}\cdot (2000\,^{\circ}C)$

$T_{Re} = 1600\,^{\circ}Re$

The equivalent temperature of 2000 ºC is 1600 ºRe.

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3 years ago

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rewona [7]

Hi! I want to know this

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3 years ago

A motorist drives north for 36.0 minutes at 96.0 km/h and then stops for 15.0 minutes. He then continues north, traveling 130 km

Over [174]

Answer:

a) $d=187.6km$

b) $v=61.5km/h$

Explanation:

First we convert our minutes to hours so we work always in the same units.

$36min=36min(\frac{1h}{60min})=0.6h$

$15min=15min(\frac{1h}{60min})=0.25h$

Where we used the fact that 1 hour are 60 min, thus the multiplying factor is equal to 1 (not altering the time, just changing the units).

a) On the first part the motorist travels a distance $d_1=v_1t_1=(96km/h)(0.6h)=57.6km$ , and on the second part he travels $d_2=130km$ .

The total displacement is $d=d_1+d_2=57.6km+130km=187.6km$

b) The average velocity is the relation between the total displacement and the time taken to cover it. Our total time is t=0.6h+0.25h+2.2h=3.05h, thus we have:

$v=\frac{187.6km}{3.05h}=61.5km/h$

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3 years ago

In a double-slit experiment, it is observed that the distance between adjacent maxima on a remote screen is 1.0 cm. What happens

kobusy [5.1K]

Answer:

C) It increases to 2.0 cm

Explanation:

In a double-slit diffraction experiment, the distance on the screen between two adjacent maxima is given by

$\Delta y = \frac{\lambda D}{d}$

where

$\lambda$ is the wavelength of the wave

D is the distance of the screen from the slits

d is the separation between the slits

In this problem, the initial distance between adjacent maxima is 1.0 cm. Later, the slit separation is cut in a half, which means that the new slit separation is

$d'=\frac{d}{2}$

Substituting into the equation, we find that the new separation between the maxima is

$\Delta y' = \frac{\lambda D}{d/2}=2(\frac{\lambda D}{d})=2\Delta y$

So, the distance increases by a factor 2: therefore, the new separation between the maxima will be 2.0 cm.

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