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2 years ago

7

As a student is performing a double slit experiment to determine the wavelength of a light source, she realizes that the nodal l

ines are too close together to be accurately measured. To increase the distance between the nodal lines, she could:

1 answer:

aleksandr82 [10.1K]2 years ago

6 0

To increase the distance between the nodal lines as it may be difficult to determine the wavelength of the light source because the nodal lines are too close, the student should spread the fringe pattern by having to be able to decrease the separation of the slit.

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A hot iron horseshoe (mass = 0.35 kg ), just forged, is dropped into 1.40 L of water in a 0.45 kg iron pot initially at 22.0°C.

olga_2 [115]

Answer:

$420$ °C

Explanation:

$m_{h}$ = mass of the horseshoe = 0.35 kg

$m_{w}$ = mass of the water = 1.40 L = 1.40 kg

$m_{i}$ = mass of the iron pot = 0.45 kg

$c_{i}$ = specific heat of iron = 450 J kg⁻¹ °C⁻¹

$c_{w}$ = specific heat of water = 4186 J kg⁻¹ °C⁻¹

$T_{hi}$ = initial temperature of the horseshoe = ?

$T_{wi}$ = initial temperature of the water = 22 °C

$T_{pi}$ = initial temperature of the iron pot = 22 °C

$T_{f}$ = final temperature = 32 °C

Using conservation of Heat

$m_{h}c_{i}(T_{hi} - T_{f}) = m_{w}c_{w}(T_{f} - T_{wi}) + m_{i}c_{i}(T_{f} - T_{pi})$

$(0.35)(450)(T_{hi} - 32) = (1.40)(4186)(32 - 22) + (0.45)(450)(32 - 22)$

$(157.5)(T_{hi} - 32) = 60629$

$T_{hi} - 32 = 384.95$

$T_{hi} = 420$ °C

7 0

3 years ago

A long cylindrical capacitor is made of a central wire of radius a = 2.50 mm surrounded by a conducting shell of radius b = 7.50

asambeis [7]

Answer:

The capacitance per unit length is $5.06\times10^{-11}\ F/m$

(b) is correct option.

Explanation:

Given that,

Radius a= 2.50 mm

Radius b=7.50 mm

Dielectric constant = 3.68

Potential difference = 120 V

We need to calculate charge per length for the capacitance

Using formula of charge per length

$\lambda=\dfrac{4\pi\epsilon_{0}\Delta V}{2 ln(\dfrac{r_{2}}{r_{1}})}$

Put the value into the formula

$\lambda=\dfrac{120}{9\times10^{9}\times2 ln(\dfrac{7.50\times10^{-3}}{2.50\times10^{-3}})}$

$\lambda=6.068\times10^{-9}\ C/m$

We know that,

$\lambda=\dfrac{Q}{L}$

We need to calculate the capacitance per unit length

Using formula of capacitance per unit length

$C=\dfrac{\dfrac{Q}{L}}{\Delta V}$

$C=\dfrac{6.068\times10^{-9}}{120}$

$C=5.06\times10^{-11}\ F/m$

Hence, The capacitance per unit length is $5.06\times10^{-11}\ F/m$

7 0

3 years ago

Buoyancy is a force that always acts in an

Margarita [4]

Buoyancy is a force that always acts in an upward direction exerted by a fluid on a body placed in the fluid

Hope this helps :)

7 0

3 years ago

Read 2 more answers

You can obtain a rough estimate of the size of a molecule by the following simple experiment. Let a droplet of oil spread out on

Marat540 [252]

Answer:

The diameter of the molecule of oil is $1.405 10 ^{-9} \ m$

Explanation:

We define density as

$\rho = \frac{mass}{volume}$

So, the volume for our oil will be

$volume = \frac{mass}{\rho}$

$volume = \frac{7.62 \ 10^{-7} \ kg}{ 912 \ \frac{kg}{m^3} }$

$volume = \frac{7.62 \ 10^{-7} \ kg}{ 912 \ \frac{kg}{m^3} }$

$volume = 8.355 \ 10 ^{-10} \ m^3$

the volume for a cylinder with radius r and height h is

$volume = \pi r^2 h$

So, we can obtain the height of the droplet of oil as:

$h = \frac{volume}{\pi r^2}$

the radius is

$r=43.5 \ cm = 0.435 \ m$

$h = 1.405 10 ^{-9} \ m^3$

And this is the diameter of the oil molecule.

6 0

3 years ago

When drawing a Bohr model for an element that has 16 electrons, how many electrons would be placed in the third energy level?

lubasha [3.4K]

There would be 6 electrons placed on the third energy level.

5 0

2 years ago