a light bulb is 4.1 m from a surface. how much luminous flux must the bulb produce if the illuminance required is 22 lx?

A car is moving on a straight road in a fixed direction at a constant speed of v = 62 km/h with respect to the road. You wish to

Angelina_Jolie [31]

Answer:

A) $V=62 i + 0 j + 0 k$

B) $X=0.93 km$

Explanation:

A) In this situation we are talking about a car moving only in the X- axis, hence the velocity of the car is:

$V=62 i + 0 j + 0 k$

Where the unit vectors $i$ , $j$ and $k$ represent the components $x$ , $y$ and $z$ in the cartesian plane.

In this sense, each unit vector is defined to have a magnitude of exactly one (1).

B) Velocity is defined as the variation of position in time, if this car is moving only along the x direction we will have:

$V=\frac{X}{t}$

Clearing the position:

$X=V.t$

$X=(62 km/h)(0.015 h)$

$X=0.93 km$

5 0

2 years ago

Motion maps for two objects, Y and Z, are shown.

Vesna [10]

Answer:

The answer is B) 3 seconds

Explanation:

I just took the test on 2020 edge and got it right

5 0

2 years ago

Consider a double-slit with a distance between the slits of 0.04 mm and slit width of 0.01 mm. Suppose the screen is a distance

scZoUnD [109]

Answer:

The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

Explanation:

Given that,

Distance between the slits = 0.04 mm

Width = 0.01 mm

Distance between the slits and screen = 1 m

Wavelength = 600 nm

We need to calculate the distance between the places where the intensity is zero due to the double slit effect

For constructive fringe

First minima from center

$x_{1}=\dfrac{\lambda D}{2d}$

Second minima from center

$x_{2}=\dfrac{3\lambda D}{2d}$

The distance between the places where the intensity is zero due to the double slit effect

$\Delta x_{d}=x_{2}-x_{1}$

$\Delta x_{d}=\dfrac{3\lambda D}{2d}-\dfrac{\lambda D}{2d}$

$\Delta x_{d}=\dfrac{\lambda D}{d}$

Put the value into the formula

$\Delta x_{d}=\dfrac{600\times10^{-9}\times1}{0.04\times10^{-3}}$

$\Delta x_{d}=0.015 =15\times10^{-3}\ m$

$\Delta x_{d}=15\ mm$

Hence, The distance between the places where the intensity is zero due to the double slit effect is 15 mm.

8 0

2 years ago

To practice Problem-Solving Strategy 17.1 for wave interference problems. Two loudspeakers are placed side by side a distance d

Nimfa-mama [501]

Complete Question

The compete question is shown on the first uploaded question

Answer:

The speed is $v = 350 \ m/s$

Explanation:

From the question we are told that

The distance of separation is d = 4.00 m

The distance of the listener to the center between the speakers is I = 5.00 m

The change in the distance of the speaker is by $k = 60 cm = 0.6 \ m$

The frequency of both speakers is $f = 700 \ Hz$

Generally the distance of the listener to the first speaker is mathematically represented as

$L_1 = \sqrt{l^2 + [\frac{d}{2} ]^2}$

$L_1 = \sqrt{5^2 + [\frac{4}{2} ]^2}$

$L_1 = 5.39 \ m$

Generally the distance of the listener to second speaker at its new position is

$L_2 = \sqrt{l^2 + [\frac{d}{2} ]^2 + k}$

$L_2 = \sqrt{5^2 + [\frac{4}{2} ]^2 + 0.6}$

$L_2 = 5.64 \ m$

Generally the path difference between the speakers is mathematically represented as

$pD = L_2 - L_1 = \frac{n * \lambda}{2}$

Here $\lambda$ is the wavelength which is mathematically represented as

$\lambda = \frac{v}{f}$

=> $L_2 - L_1 = \frac{n * \frac{v}{f}}{2}$

=> $L_2 - L_1 = \frac{n * v}{2f}$

Here n is the order of the maxima with value of n = 1 this because we are considering two adjacent waves

=> $5.64 - 5.39 = \frac{1 * v}{2*700}$

=> $v = 350 \ m/s$

7 0

2 years ago

Steam at 0.6 MPa, 200 oC, enters an insulated nozzle with a velocity of 50 m/s. It leaves at a pressure of 0.15 MPa and a veloci

Rudiy27

Answer:

x2 = 0.99

Explanation:

from superheated water table

at pressure p1 = 0.6MPa and temperature 200 degree celcius

h1 = 2850.6 kJ/kg

From energy equation we have following relation

$\dot m( h1+\frac{v1^2}{2}+ gz1 )+ Q = \dot m( h2+\frac{v2^2}{2}+ gz1) + W$

$\dot m( h1+\frac{v1^2}{2}) = \dot m( h2+\frac{v2^2}{2})$

$h1+\frac{v1^2}{2} = h2+\frac{v2^2}{2}$

$2850.6 + [\frac{50^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}] = h2 +[ \frac{600^2}{2} * \frac{1 kJ/kg}{1000 m^2/S^2}]$

h2 = 2671.85 kJ/kg

from superheated water table

at pressure p2 = 0.15MPa

specific enthalpy of fluid hf = 467.13 kJ/kg

enthalpy change hfg = 2226.0 kJ/kg

specific enthalpy of the saturated gas hg = 2693.1 kJ/kg

as it can be seen from above value hf>h2>hg, so phase 2 is two phase region. so we have

quality of steam x2

h2 = hf + x2(hfg)

2671.85 = 467.13 +x2*2226.0

x2 = 0.99

6 0

2 years ago