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lana66690 [7]
2 years ago
5

A cross-country skier can complete a 7.5 km race in 45 min. What is the skier’s average speed?

Physics
1 answer:
kap26 [50]2 years ago
3 0

Answer:

Speed = 10 km/hr

Explanation:

Speed = Distance / Time

Since we have to convert it into minutes to hours ( because it is km/hr )

45 minutes = 0.75 hours

Therefore,

Speed = 7.5 / 0.75

           = <u>10 km/hr </u>

<u></u>

<u>Hope it helps!</u>

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QUESTION:-A beam of white light shines onto a sheet of white paper. An identical beam of light shines onto a mirror. The light is scattered from the paper and reflected from the mirror.
Describe how scattering by paper and reflection by a mirror are different from each other.


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"The predictions of Einstein’s Theory of General Relativity were tested on a double pulsar system in January of 2004. His equati
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Answer:

99.95%

Explanation:

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2 years ago
Which of the following statements best represents the impact of evolutionary theory on the field of psychology?
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Answer- the correct option for the question would be choice number c which tells us about the natural selection process of human species.

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A composite wall separates combustion gases at 2400°C from a liquid coolant at 100°C, with gas and liquid-side convection coeffi
evablogger [386]

Answer:

\text{heat loss} = 24864.05 \  W/m^2

Explanation:

If

  • T_1, T_2 are temperatures of gasses and liquid in Kelvins,
  • t_1 and t_2 are thicknesses of gas layer and steel slab in meters,
  • h_1, h_2 are convection coefficients gas and liquid in W/m^2 \cdot K,
  • R_c is the contact resistance in m^2 \cdot K/W,
  • and k_1, k_2 are thermal conductivities of gas and steel in W/m \cdot K,

then: part(a):

\text{heat loss } =  \frac{T_1 - T_2} { \frac{1}{h_1} + \frac{t_1}{t_2} + R_c + \frac{t_2}{k_2} + \frac{1}{h_2}}

using known values:

\text {heat loss} = 2486.05 W/m^2

part(b): Using the rate equation :

\text {heat loss} = h_1 (T_1 - T_{s1})

the surface temperature T_{s1} = 1678.438 \ K

and T_{c1} = T_{s1} - \frac {t_1 (\text{heat loss})}{k_1} = 1664.560 \ K

Similarly

T_{c2} = T_{c1} - R_c (\text{heat loss}) = 421.357 \ K

T_{s2} = T_{c2} - \frac {t_2 (\text{heat loss})}{ k_2} = 397.864 \ K

The temperature distribution is shown in the attached image

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