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3 years ago

What are all the moon phases

2 answers:

8 0

New Moon
Waxing Crescent
First Quarter
Waxing Gibbous
Full Moon
Waning Gibbous
Last Quarter
Waning Crescent

;)

photoshop1234 [79]3 years ago

5 0

New moon, waxing crescent, first quarter, waxing gibbous, full moon, waning gibbous, third quarter (or last quarter) , waning crescent .

You might be interested in

If a material contains three elements but not joined in a fixed proportion it is a

iren [92.7K]

My answer -

 a molecule The clues are that it is elements, so that means atoms, and that it is in a fixed proportion like a molecular formula.

Happy to help you have a great day

4 0

3 years ago

How many football fields (including the 10 yards in each end zone) would it take to make a mile?

Inessa05 [86]

Answer:

Explanation:

1760 yd/mi / 120 yd/field = 14⅔ fields/mi

4 0

2 years ago

How old are the hydrogen atoms in water?

topjm [15]

Answer: 4.5 billion years

Explanation:

Some of the water molecules in your drinking glass were created more than 4.5 billion years ago, according to new research.

That makes them older than the Earth, older than the solar system — even older than the sun itself.

8 0

3 years ago

Read 2 more answers

Determine the amount of work done by the engine of a car with a mass of 2500 kg when it accelerates from 45 mph to 65 mph. (Use

Y_Kistochka [10]

Answer:

amount of work done, W = 549.36 kJ

Given:

mass of a car engine, m = 2500 kg

initial velocity, u = 45 mph

final velocity, v = 65 mph

1 mile = 1609

Solution:

We know that 1 hour = 3600 s

Now, velocities in m/s are given as:

u = 45 mph = $\frac{45\times 1609}{3600}$ = 20.11 m/s

v = 65 mph = $\frac{65\times 1609}{3600}$ = 29.05 m/s

Now, the amount of work done, W is given by the change in kinetic energy of the car and is given by:

W = $\frac{1}{2}m\Delta v^{2}$

W = $\frac{1}{2}m\times (v^{2} - u^{2})$

W = $\frac{1}{2}2500\times (29.05^{2} - 20.11^{2})$

W = 549.36 kJ

3 0

3 years ago

he capacitor can withstand a peak voltage of 550 VV . If the voltage source operates at the resonance frequency, what maximum vo

anygoal [31]

Answer:

The maximum voltage is 39.08 V.

Explanation:

Given that,

Voltage = 550 V

Suppose, In an L-R-C series circuit, the resistance is 400 ohms, the inductance is 0.380 Henry, and the capacitance is $1.20\times10^{-2}\mu F$

We need to calculate the resonant frequency

Using formula of resonant frequency

$f=\dfrac{1}{2\pi\sqrt{LC}}$

Put the value into the formula

$f=\dfrac{1}{2\pi\sqrt{0.380\times1.20\times10^{-8}}}$

$f=2356.8\ Hz$

We need to calculate the maximum current

Using formula of current

$I=\dfrac{V_{c}}{X_{c}}$

$I=2\pi f C\times V_{c}$

Put the value into the formula

$I=2\pi\times2356.8\times1.20\times10^{-8}\times550$

$I=0.0977\ A$

We need to calculate the impedance of the circuit

Using formula of impedance

$Z=\sqrt{R^2+(X_{L}-X_{C})^2}$

At resonant frequency , $X_{L}=X_{C}$

So, Z = R

We need to calculate the maximum voltage

Using formula of voltage

$V=IR$

Put the value into the formula

$V=0.0977\times400$

$V=39.08\ V$

Hence, The maximum voltage is 39.08 V.

4 0

3 years ago