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____ [38]
3 years ago
14

A red tail flew 200 meters in 25 seconds. What was the speed?

Physics
1 answer:
Zarrin [17]3 years ago
3 0
The red tail flew 8 meters a second
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The moon is 3x10^5 km away from Nepal and the mass of the moon is 7x10^22 kg. Calculate the force with which the Moon pulls ever
hammer [34]

Answer:

Approximately 5.19 \times 10^{-5}\; \rm N.

Explanation:

Let G denote the gravitational constant. (G \approx 6.67 \times 10^{-11} \; \rm N \cdot kg^{-2} \cdot m^{2}.)

Let M and m denote the mass of two objects separated by r.

By Newton's Law of Universal Gravitation, the gravitational attraction between these two objects would measure:

\displaystyle F = \frac{G \cdot M \cdot m}{r^{2}}.

In this question: M = 7 \times 10^{22}\; \rm kg is the mass of the moon, while m = 1\; \rm kg is the mass of the water. The two are r = 3\times 10^{5}\; \rm km apart from one another.

Important: convert the unit of r to standard units (meters, not kilometers) to reflect the unit of the gravitational constant G.

\displaystyle r = 3 \times 10^{5}\; \rm km \times \frac{10^{3}\; \rm m}{1\; \rm km} = 3 \times 10^{8}\; \rm m.

\begin{aligned} F &= \frac{G \cdot M \cdot m}{r^{2}} \\ &= \frac{6.67 \times 10^{-11}\; \rm N \cdot kg^{-2}\cdot m^{2} \times 7 \times 10^{22}\; \rm kg \times 1\; \rm kg}{(3 \times 10^{8}\; \rm m)^{2}} \\ &\approx 5.19 \times 10^{-5} \; \rm N\end{aligned}.

5 0
3 years ago
Sunlight has its maximum intensity at a wavelength of 4.83 x 10'm; wha energy does this correspond to in e?
Lera25 [3.4K]

Answer:

E = 2,575 eV

Explanation:

For this exercise we will use the Planck equation and the relationship of the speed of light with the frequency and wavelength

     E = h f

     c = λ f

Where the Planck constant has a value of 6.63 10⁻³⁴ J s

Let's replace

    E = h c / λ

Let's calculate for wavelengths

    λ = 4.83 10-7 m     (blue)

    E = 6.63 10⁻³⁴ 3 10⁸ / 4.83 10⁻⁷

    E = 4.12 10-19 J

The transformation from J to eV is 1 eV = 1.6 10⁻¹⁹ J

    E = 4.12 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)

    E = 2,575 eV

5 0
4 years ago
Can some help me out on some science ?
Dennis_Churaev [7]
Yea but whats the question
3 0
3 years ago
Read 2 more answers
An open pipe, 0.29 m long, vibrates in the second overtone with a frequency of 1,227 Hz. In this situation, the fundamental freq
Andru [333]

Answer:

f = 409 Hz

Explanation:

We have,

Length of the open organ pipe, l = 0.29 m

Frequency of vibration of second overtone, f_2 = 1227 Hz

It is required to find the fundamental frequency of the pipe. For the open organ pipe, the frequency of second overtone is given by :

f_2=\dfrac{3v}{2l}

v is speed of sound

Let f is the fundamental frequency. It is given by :

f=\dfrac{v}{2l}

The relation between f and f₂ can be written as :

f_2=3f\\\\f=\dfrac{f_2}{3}\\\\f=\dfrac{1227}{3}\\\\f=409\ Hz

So, the fundamental frequency of the pipe is 409 Hz.              

6 0
3 years ago
Why are denser materials found closer to earths center
ICE Princess25 [194]
Denser materials tend to be closer to earths center due to their mass gravity is shown by the equation mg
Which stands for mass x gravity.
5 0
3 years ago
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