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DENIUS [597]
3 years ago
15

If you run 12m/s for 15 minutes, how far will you go explain?

Physics
1 answer:
Aneli [31]3 years ago
7 0
To determine the distance that you would travel by.

Simply convert 15 minutes to seconds which would be 900 seconds.

Then solve for distance, which would be

D = r • t
D = 12 m/s • 900 s

D =10 800 m.

This is the distance that you had ran for.
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Betelgeuse is how many times larger than the sun
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Betelgeuse is one of the largest known stars and is probably at least the size of the orbits of Mars or Jupiter around the sun. That's a diameter about 700 times the size of the Sun or 600 million miles. For a star it has a rather low surface temperature (6000 F compared to the Sun's 10,000 F).
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PLEASE HELP!! DUE SOON!! THANKS BRAINLIEST OFFERED!!
pav-90 [236]

Answer:

real, inverted, and smaller than the object

Explanation:

When the object is placed beyond the center of curvature, the image will formed between the focus and the center of curvature. The size of the image is diminished and its nature is real and inverted.

The whole description is shown in the attached figure. It is clear that the size of the image is smaller than the object.

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3 years ago
What is measurement ?​
Akimi4 [234]

Answer:

Measurement is the assignment of a number to a characteristic of an object or event

Explanation:

6 0
2 years ago
Read 2 more answers
What minimum distance should you separate two sources emitting the same waves with wavelength 5mm in phase such that you obtain
Makovka662 [10]

To solve this problem we will apply the concept related to destructive interference (from the principle of superposition). This concept is understood as a superposition of two or more waves of identical or similar frequency that, when interfering, create a new wave pattern of less intensity (amplitude) at a point called a node. Mathematically it can be described as

d = n \frac{\lambda}{2}

Where,

d = Path difference

\lambda= wavelength

n = Any integer which represent the number of repetition of the spectrum

In this question the distance between the two source will be minimum for the case of minimum path difference, then n= 1

d = \frac{\lambda}{2}

d = \frac{5*10^{-3}}{2}

d = 2.5mm

Therefore the minimum distance that should you separate two sources emitting the same waves is 2.5mm

8 0
3 years ago
Water flows through a cast steel pipe (k = 50 W m.K, ε = 0.8) with an outer diameter of 104mm and 2 mm wall thickness. Calculate
masha68 [24]

Answer:

The heat loss per unit length is   \frac{Q}{L}   = 2981 W/m

Explanation:

From the question we are told that

     The outer diameter of the pipe is d = 104mm = \frac{104}{1000} = 0.104 m

     The thickness is  D = 2mm = \frac{2}{1000} = 0.002m  

      The temperature  of water is  T = 90^oC = 90 + 273 = 363K  

      The outside air temperature is T_a = -10^oC = -10 +273 = 263K

        The water side heat transfer coefficient is z_1 = 300 W/ m^2 \cdot K

       The  heat transfer coefficient is  z_2 = 20 W/m^2 \cdot K

The heat lost per unit length is mathematically represented as

           \frac{Q}{L}   = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1}  +  \frac{ln [\frac{d}{D} ]}{z_2}}

Substituting values

         \frac{Q}{L}   = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300}  +  \frac{ln [\frac{0.104}{0.002} ]}{20}}

           \frac{Q}{L}   = \frac{628}{0.2107}

           \frac{Q}{L}   = 2981 W/m

6 0
3 years ago
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