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Mila [183]
4 years ago
11

An observer sees a full moon in the night sky. What will the Moon's phase be 10 days later?

Physics
1 answer:
algol [13]4 years ago
5 0
I'm pretty sure the moon would be a crescent within the 10 day period because it takes about 28 days for the moon to go through the different stages.
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A substance maintaining a uniform appearence throught is called
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A substance maintaining a uniform appearance throught is called homogeneous
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What are the dark areas on the surface of the Sun?
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Sunspots,despite being called sunspots they are still thousands of degrees hot with an average cycle of 11years per minimum and the climax of these phenomenon s
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How many photons will be required to raise the temperature of 1.8 g of water by 2.5 k ?'?
tatyana61 [14]
Missing part in the text of the problem: 
"<span>Water is exposed to infrared radiation of wavelength 3.0×10^−6 m"</span>

First we can calculate the amount of energy needed to raise the temperature of the water, which is given by
Q=m C_s \Delta T
where
m=1.8 g is the mass of the water
C_s = 4.18 J/(g K) is the specific heat capacity of the water
\Delta T=2.5 K is the increase in temperature.

Substituting the data, we find
Q=(1.8 g)(4.18 J/(gK))(2.5 K)=18.8 J=E

We know that each photon carries an energy of
E_1 = hf
where h is the Planck constant and f the frequency of the photon. Using the wavelength, we can find the photon frequency:
\lambda =  \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{3 \cdot 10^{-6} m}=1 \cdot 10^{14}Hz

So, the energy of a single photon of this frequency is
E_1 = hf =(6.6 \cdot 10^{-34} J)(1 \cdot 10^{14} Hz)=6.6 \cdot 10^{-20} J

and the number of photons needed is the total energy needed divided by the energy of a single photon:
N= \frac{E}{E_1}= \frac{18.8 J}{6.6 \cdot 10^{-20} J} =2.84 \cdot 10^{20} photons
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3 years ago
Which property of a substance can be determined using a pH indicator?
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uring the investigation of a traffic accident, police find skid marks 89.9 m long. They determine the coefficient of friction be
aivan3 [116]

Answer:

V_{0}=29.68m/s

Explanation:

In order to solve this problem, we must first do a drawing of the situation (see attached picture).

When the brakes of the car were applied, we can see that there was only one horizontal force affecting the vehicle's movement, which was the force of friction. When analyzing the free body diagram we can apply Newton's laws to determine the equations we will use to solve this.

\sum F_{x}=ma

so:

-f=ma

we also know that:

f=N\mu_{k}

so

-N\mu_{k}=ma

we can now solve for the acceleration, so we get:

a=\frac{-N\mu_{k}}{m}

we don't know what the normal force is, so we can find it out by analyzing the vertical forces applied to the car:

\sum F_{y}=0

so we get:

N-W=0

which means that:

N=W

we also know that:

W=mg

so

N=mg

we can substitute this into the first equation so we get:

a=\frac{-mg\mu_{k}}{m}

which simplifies to:

a=-g\mu_{k}

we can now substitute the provided data:

a=(-9.8m/s^{2})(0.5)

which yields:

a=-4.9m/s^{2}

once we got the acceleration, we can use kinematics formulas to solve this, we got the following formula:

a=\frac{V_{f}^{2}-V_{0}^{2}}{2x}

we know the final velocity must be zero, since that's where the car got to a stop, so the formula then becomes:

a=-\frac{V_{0}^{2}}{2x}

we can now solve for the initial velocity, which yields:

V_{0}=\sqrt{-2xa}

so we can now substitute the daa we know, so we get:

V_{0}=\sqrt{-2(89.9m)(-4.9m/s^{2})}

so we get:

V_{0}=29.68m/s

So from this we know that the velocity of the car must have been of at least 29.68m/s when the brakes were applied.

4 0
4 years ago
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