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allochka39001 [22]
3 years ago
13

Racewalking involves more impact than running. true false

Physics
2 answers:
ahrayia [7]3 years ago
6 0

Answer:

Given statement is false.

Explanation:

We have been given a statement  that racewalking involves more impact than running.

The given statement is false.

Running involves more effort  that is number of calories burn will be more. So, running involves more impact.

Running  is good for cardiovascular health that is heart health.

natta225 [31]3 years ago
5 0
I think the statement is false. Racewalking involves less impact than running. It <span> is a long-distance discipline within the sport of athletics. Although it is a foot race, it is different from running in that one foot must appear to be in contact with the ground.</span>
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Please write 4-5 sentences, please!! How would you describe the climate of Colorado? Find 1 piece of evidence from online to def
andreyandreev [35.5K]

Answer:

The climate in colorado is combination of high elevation, midlatitude, and continental interior geography results in a cool, dry, and invigorating climate. The average annual temperature for the state is 43.5 degrees Fahrenheit (F), which is 13.7 degrees below the global mean

Explanation:

4 0
3 years ago
A car battery with a 12 V emf and an internal resistance of 0.11 Ω is being charged with a current of 56 A. What are (a) the pot
denpristay [2]

Answer:

Part a)

V = 18.16 V

Part b)

P_r = 345 Watt

Part c)

P = 672 Watt

Part d)

V = 5.84 V

Part e)

P_r = 345 Watt

Explanation:

Part a)

When battery is in charging mode

then the potential difference at the terminal of the cell is more than its EMF and it is given as

\Delta V = E + i r

here we have

E = 12 V

i = 56 A

r = 0.11

now we have

\Delta V = 12 + (0.11)(56) = 18.16 V

Part b)

Rate of energy dissipation inside the battery is the energy across internal resistance

so it is given as

P_r = i^2 r

P_r = 56^2 (0.11)

P_r = 345 W

Part c)

Rate of energy conversion into EMF is given as

P_{emf} = i E

P_{emf} = (56)(12)

P_{emf} = 672 Watt

Now battery is giving current to other circuit so now it is discharging

now we have

Part d)

V = E - i r

V = 12 - (56)(0.11)

V = 12 - 6.16 = 5.84 V

Part e)

now the rate of energy dissipation is given as

P_r = i^2 r

P_r = 56^2 (0.11)

P_r = 345 W

7 0
3 years ago
14. Convert 22 degrees celsius to fahrenheit.<br> Please show your work
forsale [732]
(22°C × 9/5) + 32 = 71.6°F
6 0
3 years ago
Read 2 more answers
Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 80.5 N, Jill pulls with 81.7 N in the
Gnesinka [82]

Answer:

F = 233.52 N,  θ' = 351.41º

Explanation:

In this exercise we must find the net force applied on the donkey.

For this we use Newton's second law, where we create a reference frame with the horizontal x axis

let's decompose the forces

Jack

        = 80.5 N

Jill

       cos 45 = F_{2x} / F₂2

       sin 45 = F_{2y} / F₂2

       F_{2x} = F₂ cos 45

       F_{2y} = F₂ sin 45

       F_{2x} = 81.7 cos 45 = 57.77 N

       F_{2y} = 81.7 sin 45 = 57.77 N

Jane

      cos (270 + 45) = F_{3x} / F₃3

      sin 315 = F_{3y} / F₃

      F_{3x} = 131 cos 315 = 92.63 N

      F_{3y} = 131 sin 315 = -92.63 N

the force can be found in each axis

X axis

         F_{x} = F_{1x} + F_{2x} + F_{3x}

         F_{x} = 80.5 +57.77 + 92.63

         F_{x} = 230.9 N

Axis y

         F_{y} = F_{1y} + F_{2y} + F_{3y}

         F_{y} = 0 + 57.77 -92.63

         F_{y} = -34.86 N

we can give the result in two ways

a) F = (230.9 i ^ - 34.86 j ^) N

b) in the form of module and angle

we use the Pythagorean theorem

         F = √(Fₓ² + F_{y}²

        F = √(230.9² + 34.86²)

        F = 233.52 N

let's use trigonometry for the angle

        tan θ = \frac{F_y}{F_x} }

        θ = tan⁻¹ (\frac{F_y}{F_x} })

        θ = tan⁻¹ (-34.86 / 230.9)

        θ = -8.59º

if we measure this angle from the positive side of the x-axis counterclockwise

          θ' = 360 -θ

          θ‘= 360- 8.59

          θ' = 351.41º

5 0
2 years ago
What is the stretch when you pull with a force of 25 N on a spring with a spring constant of 8 N/m? *
Pani-rosa [81]

Hooke's Law

\tt F=k.\Delta x

k = spring constant

x = stretch

F = force

Input the value

\tt \Delta x=\dfrac{F}{k}=\dfrac{25}{8}=3.125\rightarrow 3.13\:m

7 0
2 years ago
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