Ask question

Ask question

All categories

2 years ago

7

A light beam shines through a slit and illuminates a distant screen. The central bright fringe on the screen is 1.00 cm wide, as

measured between the dark fringes that border it on either side. Which of the following actions would decrease the width of the central fringe? ( There may be more than one correct answer)
a- increase the wavelength of the light
b-decrease the wavelength of the light
c-increase the width of the slit
d-put the apparatus all under water

1 answer:

Pachacha [2.7K]2 years ago

4 0

Answer:

B, C, D.

Explanation:

You might be interested in

If a trapezoidal channel has side slopes of 1:1, hydraulic depth is 5 feet, the bottom width is 8 feet, flow is 2,312 cubic feet

Evgesh-ka [11]

Answer:

$S = \dfrac{1}{2.5}$

Explanation:

given,

side slope = 1 : 1

hydraulic depth(y) = 5 ft

bottom width (b)= 8 ft

x = 1

Q = 2,312 ft³/s

n = 0.013

slope of channel = ?

$R = \dfrac{A}{P}$

$R = \dfrac{y(b + xy)}{b+2y\sqrt{1+x^2}}$

$R = \dfrac{5(8+ 5)}{8+2\times 5\sqrt{1+1^2}}$

R = 4.69 m

using manning's equation

$Q = \dfrac{1}{n}AR^{2/3} S^{1/2}$

$2312= \dfrac{1}{0.013}\times (5(8+5))\times 4.69^{2/3} S^{1/2}$

$2312=14009.37\times S^{1/2}$

S = 0.406

$S = \dfrac{1}{2.5}$

5 0

2 years ago

A student placed 700g of water at 28° c in the freezer.after 6 minutes and 15 seconds the water was transformed to ice.

Maru [420]

Answer:

is this the full questions

7 0

2 years ago

HELP PLS!<br> (LOOK AT THE PICTURE)

Arturiano [62]

The answer is b !!!! Hope it helps

4 0

2 years ago

10.0 ppm 10.4 ppm 10.2 ppm 10.8 ppm 10.1 ppm 10.5 ppm 10.1 ppm Using the new instrument on the first day the technicians got a v

iren [92.7K]

Answer:

The datapoint 9.0 ppm is outlier at the 90% confidence level.

Explanation:

The old data has following values

mean=10.5 mm

standard deviation 0.2 mm

Now the mean of new values is calculated as following

$mean=\frac{10+10.4+10.2+10.8+10.1+10.5+10.1}{7}\\mean=10.3 ppm$

So the value as 9.0 ppm can be considered easily as outlier in this regard.

3 0

2 years ago

2200 kg semi truck driving down the highway has lost control. The truck rolls across the median and into oncoming traffic. The t

serious [3.7K]

Answer:

The semi truck travels at an initial speed of 69.545 meters per second downwards.

Explanation:

In this exercise we see a case of an entirely inellastic collision between the semi truck and the car, which can be described by the following equation derived from Principle of Linear Momentum Conservation: (We assume that velocity oriented northwards is positive)

$m_{S}\cdot v_{S}+m_{C}\cdot v_{C} = (m_{S}+m_{C})\cdot v$ (1)

Where:

$m_{S}$ , $m_{C}$ - Masses of the semi truck and the car, measured in kilograms.

$v_{S}$ , $v_{C}$ - Initial velocities of the semi truck and the car, measured in meters per second.

$v$ - Final speed of the system after collision, measured in meters per second.

If we know that $m_{S} = 2200\,kg$ , $m_{C} = 2000\,kg$ , $v_{C} = 45\,\frac{m}{s}$ and $v = -15\,\frac{m}{s}$ , then the initial velocity of the semi truck is:

$m_{S}\cdot v_{S} = (m_{S}+m_{C})\cdot v -m_{C}\cdot v_{C}$

$v_{S} = \frac{(m_{S}+m_{C})\cdot v - m_{C}\cdot v_{C}}{m_{S}}$

$v_{S} = \left(1+\frac{m_{C}}{m_{S}} \right)\cdot v - \frac{m_{C}}{m_{S}} \cdot v_{C}$

$v_{S} = v +\frac{m_{C}}{m_{S}}\cdot (v-v_{C})$

$v_{S} = -15\,\frac{m}{s}+\left(\frac{2000\,kg}{2200\,kg} \right) \cdot \left(-15\,\frac{m}{s}-45\,\frac{m}{s} \right)$

$v_{S} = -69.545\,\frac{m}{s}$

The semi truck travels at an initial speed of 69.545 meters per second downwards.

3 0

1 year ago