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4 years ago

9

As the clump of colored fluid at the bottom of the lava lamp reaches the top, which has increased?

2 answers:

bagirrra123 [75]4 years ago

6 0

the heat and mass...

Elan Coil [88]4 years ago

4 0

Answer:

As the clump of colored fluid at the bottom of the lava lamp reaches the top, which has increased?

Its Density is the answer

Explanation:

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When connected to a battery, a lightbulb glows brightly. If the battery is reversed and reconnected to the bulb, the bulb will g

attashe74 [19]

Answer: c) with the same brightness

Explanation: The load in this case the bulb, is not polarized ( it has no positive and negative points) thus any connection relative to the battery (source) will have no effect on it brightness.

Also, brightness is a function of current and in this case the voltage ( from battery) and resistance of load (bulb) is constant, and according to ohms law (V=IR) if the current is constant at the first connection, it will be the same at the reversed connection.

8 0

3 years ago

6. What is the molar mass of H2O?

kotykmax [81]

Answer:

a) 18.015

Explanation:

I Googled it, hope this helped

7 0

3 years ago

Monochromatic light is incident on two slits separated by 0,2 mm. An interference pattern is observed on a screen 3,7 m away. Th

zalisa [80]

Answer:

λ = 864 nm

Explanation:

To find the wavelength of the light you use the following formula, which determines the position of the m-th fringe in an interference pattern:

$y_m=\frac{m\lambda D}{d}$ (1)

ym: position of a bright fringe

D: distance from the slits to the screen = 3,7 m

d: distance between slits = 0,2mm = 0,2 *10^-3 m

m: order of the fringe

λ: wavelength of the light

You have the distance from the central peak to the third fringe (0,048m). Then, you can use the equation (1) with m=3 and solve for the wavelength:

$y_3=\frac{3\lambda D}{d}\\\\\lambda=\frac{dy_3}{3D}=\frac{(0,2*10^{-3}m)(0,048m)}{3(3,7m)}=8,64*10^{-7}m\\\\\lambda=864*10^{-9}m=864nm$

henc, the wavelength of the light is 864nm

5 0

3 years ago

The distance between the ruled lines on a diffraction grating is 1900 nm. The grating is illuminated at normal incidence with a

SashulF [63]

Answer:

3.28 degree

Explanation:

We are given that

Distance between the ruled lines on a diffraction grating, d=1900nm= $1900\times 10^{-9}m$

Where $1nm=10^{-9} m$

$\lambda_2=400nm=400\times10^{-9}m$

$\lambda_1=700nm=700\times 10^{-9}m$

We have to find the angular width of the gap between the first order spectrum and the second order spectrum.

We know that

$\theta=sin^{-1}(\frac{m\lambda}{d})$

Using the formula

m=1

$\theta_1=sin^{-1}(\frac{1\times700\times 10^{-9}}{1900\times 10^{-9}})$

$\theta=21.62^{\circ}$

Now, m=2

$\theta_2=sin^{-1}(\frac{2\times400\times 10^{-9}}{1900\times 10^{-9}})$

$\theta_2=24.90^{\circ}$

$\Delta \theta=\theta_2-\theta_1$

$\Delta \theta=24.90-21.62$

$\Delta \theta=3.28^{\circ}$

Hence, the angular width of the gap between the first order spectrum and the second order spectrum=3.28 degree

6 0

3 years ago

A projectile is fired over level ground with an initial velocity that has a vertical component of 20 m/s and a horizontal compon

Anettt [7]

First of all, let's write the equation of motions on both horizontal (x) and vertical (y) axis. It's a uniform motion on the x-axis, with constant speed $v_x=30 m/s$ , and an accelerated motion on the y-axis, with initial speed $v_y=20 m/s$ and acceleration $g=9.81 m/s^2$ :
$S_x(t)=v_xt$
$S_y(t)=v_y t- \frac{1}{2} gt^2$
where the negative sign in front of g means the acceleration points towards negative direction of y-axis (downward).

To find the distance from the landing point, we should find first the time at which the projectile hits the ground. This can be found by requiring
$S_y(t)=0$
Therefore:
$v_y t - \frac{1}{2}gt^2=0$
which has two solutions:
$t=0$ is the time of the beginning of the motion,
$t= \frac{2 v_y}{g} = \frac{2\cdot 20 m/s}{9.81 m/s^2}=4.08 s$ is the time at which the projectile hits the ground.

Now, we can find the distance covered on the horizontal axis during this time, and this is the distance from launching to landing point:
$S_x(4.08 s)=v_x t=(30 m/s)(4.08 s)=122.4 m$

4 0

3 years ago