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s2008m [1.1K]
3 years ago
7

A red horse and a black horse raced on a 1-mile long circular racetrack. The red horse

Physics
2 answers:
Lyrx [107]3 years ago
5 0

Answer:

Explanation:

B- The red horse's average speed was greater than the black horse's average speed.

Red average speed = 1/120 = 0.00833 mi/s

Black average speed = 1/150 = 0.00667 mi/s

we only know about average speed based on the information given. Either horse could have had higher or lower, even negative,  instantaneous speed during some phase of the race.

GalinKa [24]3 years ago
4 0

Answer

your format is incorrect!!

Explanation:

there is no such thing as a red horse. there are chestnuts, bays and red roans, just no reds. usually though, blacks are faster scientifically than red roans.

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Along the remote Racetrack Playa in Death valley, California, stones sometimes gouge out prominent trails in the desert floor, a
REY [17]

Answer:

   F = 196 N

Explanation:

For this exercise we will use Newton's second law,  we define a reference system with the x axis in the direction of movement of the stones and the y axis vertically

Y axis  

       N- W = 0

       N = mg

X axis

       F -fr = ma

In this case, they ask us for the force to keep moving, so the stones go at constant speed, which implies that the acceleration is zero.

       F- fr = 0

       F = fr

the friction force has the equation

       fr = μ N

       fr = μ mg

we substitute

        F = μ mg

let's calculate

         F = 0.80 9.8 25

         F = 196 N

8 0
3 years ago
Fill in the blank: In the Northern Hemisphere, June 21 has ______________ than December 21.
Pavel [41]
Answer: I believe is A

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3 0
3 years ago
What measures the distance between two consecutive crests of a wave?
aleksley [76]
The answer is d. Wavelength
4 0
3 years ago
What are the forces that act on the ball?​
scoundrel [369]

Answer:

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4 0
3 years ago
I’m not sure how to solve this
spayn [35]

Answer:

Option 10. 169.118 J/KgºC

Explanation:

From the question given above, the following data were obtained:

Change in temperature (ΔT) = 20 °C

Heat (Q) absorbed = 1.61 KJ

Mass of metal bar = 476 g

Specific heat capacity (C) of metal bar =?

Next, we shall convert 1.61 KJ to joule (J). This can be obtained as follow:

1 kJ = 1000 J

Therefore,

1.61 KJ = 1.61 KJ × 1000 J / 1 kJ

1.61 KJ = 1610 J

Next, we shall convert 476 g to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

476 g = 476 g × 1 Kg / 1000 g

476 g = 0.476 Kg

Finally, we shall determine the specific heat capacity of the metal bar. This can be obtained as follow:

Change in temperature (ΔT) = 20 °C

Heat (Q) absorbed = 1610 J

Mass of metal bar = 0.476 Kg

Specific heat capacity (C) of metal bar =?

Q = MCΔT

1610 = 0.476 × C × 20

1610 = 9.52 × C

Divide both side by 9.52

C = 1610 / 9.52

C = 169.118 J/KgºC

Thus, the specific heat capacity of the metal bar is 169.118 J/KgºC

6 0
3 years ago
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