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bearhunter [10]
3 years ago
11

What is the speed of a horse in meters per second that runs a distance of 1.2 miles in 2.4 minutes​

Physics
1 answer:
Ilia_Sergeevich [38]3 years ago
4 0
Time t=2.4 minutes=2.4×60=144 seconds
distance s=1.2 miles=1.2×1609=1930.8 meters
speed v=s/t=1930.8÷144=[tex] \frac{1930.8}{144} = \frac{160.9}{12} =[/13.408m/s ~nearly]
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A strong lightning bolt transfers about 25 C to earth. how many electrons are transferred?
Aneli [31]

Answer:

n = 1.563x {10}^{20}

8 0
3 years ago
Um ônibus percorre a distância de 480 km, entre Santos e Curitiba, com velocidade escalar média de 80 km/h. De Curitiba a Floria
Mariulka [41]

Answer:

10 h

Explanation:

velocidade é a taxa de variação da distância no tempo. é a razão entre a distância e o tempo

de Santos e Curitiba:

distância (d) de 480 km, velocidade (s) de 80 km/h

s=\frac{d}{t}\\ t=\frac{d}{s} =\frac{480}{80}=6h

de Curitiba e Florianópolis:

distância (d) de 300 km, velocidade (s) de 75 km/h

s=\frac{d}{t}\\ t=\frac{d}{s} =\frac{300}{75}=4h

tempo médio de ônibus entre Santos e Florianópolis = 6h + 4h = 10h

7 0
3 years ago
Read 2 more answers
According to the first law of thermodynamics, the total amount of energy in the universe __________.
lubasha [3.4K]

Answer:

is constant

Explanation:

Energy cannot be destroyed or created, but can transfer from places to places and in different forms.

6 0
3 years ago
Difference between health education and physical health education​
Zina [86]

Answer:

Health education that promotes an understanding of how to maintain personal health. Physical and health education focuses on both learning about and learning through physical activity.

Explanation:

Basically you have to know what each of them mean .

6 0
3 years ago
Water flows through a cylindrical pipe of varying cross-section. The velocity is 3.0 m/s at a point where the pipe diameter is 1
Setler79 [48]

Answer:

The flow rate is  R =2.357 *10^{-4} \ m^3/s

Explanation:

From the question we are told that

    The velocity is v  =  3.0 \ m/s

   The  diameter of the pipe is  d =  1.0 \ cm  = 0.01 \ m

 

The  radius of the pipe is mathematically represented as

            r =  \frac{d}{2}

substituting values

            r =  \frac{0.01}{2}

           r =  0.005 \ m

The flow rate is mathematically represented as

       R  =  v  * A

Where is the cross-sectional area of the pipe which is mathematically evaluated as

      A   = \pi r^2

substituting values

      A   =  3.142 *  (0.005)^2

     A   = 7.855 *  10^{-5} \ m^2

So

    R  =  3.0  *  7.855 *10^{-5}

    R  =  2.357*10^{-4} \ m^3 /s

8 0
3 years ago
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