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4 years ago

The moon is halfway through the waxing crescent phase. In approximately how many days will it be in the last-quarter phase?

1 answer:

4 0

Let's say the Moon's complete cycle of phases is about 28 days (4 weeks) ... just so the arithmetic is easier. (Close enough. It's actually 29.53 days.)

If it's halfway through the waxing crescent, then it'll reach first quarter in 3 days or so.

From there, it'll take the next 7 days to become Full, and another 7 days to reach last quarter.

Total from half-waxing-crescent to last quarter . . . about 17 days.

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Write a hypothesis about the effect of temperature and surface area on the rate of chemical reactions using this format: “If . .

grandymaker [24]

Effect of temperature.

"If the temperature of the substance is increased then the rate of chemical reaction is also increased because the kinetic energy is greater."

Effect of surface area.

"If the surface area is increased then the rate of reaction is increased because there will be more active sites for the reaction to occur.

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4 years ago

Read 2 more answers

If the range of the projectile is 4.3 m, the time-of-flight is T = 0.829 s, and air resistance is negligible, determine the foll

ankoles [38]

Answer

given,

range of the projectile = 4.3 m

time of flight = T = 0.829 s

$v =\dfrac{d}{T}$

$v =\dfrac{4.3}{0.829}$

v = 5.19 m/s

vertical component of velocity of projectile

v_y = gt'

$v_y = 9.8 \times {\dfrac{T}{2}}$

$v_y = 9.8 \times {\dfrac{0.829}{2}}$

$v_y =4.06\ m/s$

a) Launch angle

$\theta = tan^{-1}(\dfrac{v_y}{v})$

$\theta = tan^{-1}(\dfrac{4.06}{5.19})$

θ = 38°

b) initial speed of projectile

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{5.19^2 + 4.06^2}$

v = 6.59 m/s

c) maximum height reached by the projectile

$y_{max}=v_{avg}t'$

$y_{max}=\dfrac{1}{2}v_y\dfrac{T}{2}$

$y_{max}=\dfrac{1}{2}\times(g\dfrac{T}{2})\times\dfrac{T}{2}$

$y_{max}=\dfrac{gT^2}{8}$

7 0

4 years ago

Q.1 Lorna built the circuit diagram below. All the bulbs are identical. (a) Complete the table below by writing on or off for ea

irga5000 [103]

Answer:

on and off

Explanation:

if there are switches, it can change if the electricity can get to the bulb or not. if it appears that there is no pathway for the electricity to get to the light bulb, it is of, if there is a pathway, its on

8 0

3 years ago

Help in physics please :(((

Arlecino [84]

Answer:

I am sorry I can't draw graphical ok how to draw the graph where what is your position the displacement of time and work 7 kilometres east in 2 hours and what will happen to the time and 72 in 1 hour what is the displacement you after take the displacement formula that is total time taken divided by the distance travelled ok displacement and distance travelled is different about its terms ok

8 0

3 years ago

monochromatic light from a distant source is incident on a slit 0.75 mm wide. on screen 2 m away, the distance from the central

hjlf

Displacement from the center line for minimum intensity is 1.35 mm , width of the slit is 0.75 so Wavelength of the light is 506.25.

<h3>How to find Wavelength of the light?</h3>

When a wave is bent by an obstruction whose dimensions are similar to the wavelength, diffraction is observed. We can disregard the effects of extremes because the Fraunhofer diffraction is the most straightforward scenario and the obstacle is a long, narrow slit.

This is a straightforward situation in which we can apply the

Fraunhofer single slit diffraction equation:

y = mλD/a

Where:

y = Displacement from the center line for minimum intensity = 1.35 mm

λ = wavelength of the light.

D = distance

a = width of the slit = 0.75

m = order number = 1

Solving for λ

λ = y + a/ mD

Changing the information that the issue has provided:

λ = 1.35 * 10^-3 + 0.75 * 10^-3 / 1*2

=5.0625 *10^-7 = 506.25

Wavelength of the light 506.25.

To learn more about Wavelength of the light refer to:

brainly.com/question/15413360

#SPJ4

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1 year ago