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Goryan [66]
2 years ago
10

Wendy has been going to the gym more lately, and she tends to be very sore the next day. She is pleased with the results she is

seeing and is taking small steps so that she does not injure herself. What is the BEST way to eliminate the soreness?
A.
Stop working out until it goes away.
B.
Change the kind of exercise she is doing.
C.
Spend more time stretching after her workout.
D.
Ramp up her exercise so that she does not have time to be sore.

PLEASE NO LINKS
Physics
1 answer:
Aneli [31]2 years ago
5 0

Answer:

c

this was on the other question i just answered. but its all good

Explanation:

Have a great day! Byeeee

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The answer is a 4.2m

Explanation:

Given data

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Leslie incorrectly balances an equation as 2C4H10 + 12O2 → 8CO2 + 10H2O.
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Three beads are placed along a thin rod. The first bead, of mass m1 = 23 g, is placed a distance d1 = 1.1 cm from the left end o
Mila [183]

Answer:

a) x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3}  ) }{m_{1}+m_{2}+m_{3} }

b) x = 4.47 cm

c) x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }

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Explanation:

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x=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3} }{m_{1}+m_{2} +m_{3}}

Where m is the mass of beads and x is the distances, if x₁ = d₁, x₂ = d₂ and x₃ = d₃

x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3}  ) }{m_{1}+m_{2}+m_{3} }

b) If

m₁ = 23g

m₂ = 15 g

m₃ = 58 g

d₁ = 1.1 cm

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d₃ = 3.2 cm

x=\frac{23*1.1+15*(1.1+1.9)+58(1.1+1.9+3.2) }{23+15+58 } =4.47cm

c) The center of the mass of the beads realtive to the center of bead is:

x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }

d) x=\frac{23*(-1.9)+(15*0)+(58*3.2) }{23+15+58 } =1.48cm

6 0
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The potential difference in a simple circuit is 2 v and the resistance is 2 ω . what current flows in the circuit? answer in uni
balandron [24]
Hey

Potential Difference given is : 2V

Resistance is : 2 ohms

By Ohm's Law, one can easily utilize the relation :

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Where, { v , i , r } are the potential difference, current and Resistance Respectively.


Hence,
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Hence, the Current is 1 Ampere
7 0
2 years ago
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