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ASHA 777 [7]
3 years ago
8

Running, swimming, and biking are what type of exercise aerobic or anaerobic

Physics
2 answers:
fomenos3 years ago
8 0

Answer:

aerobic is the correct answer

sesenic [268]3 years ago
7 0
They are all cardiovascular exercises.
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Sav [38]
PM me for full answer, please. If it's not too late.
3 0
3 years ago
Assume that the blocks are accelerating, and the x components of their accelerations at a certain moment are a1x and a2x. find t
Serga [27]

we know that center of mass is given as

r = (m₁ r_{1x} + m₂ r_{2x})/(m₁ + m₂)

taking derivative both side relative to "t"

dr/dt = (m₁ dr_{1x}/dt + m₂ dr_{2x}/dt)/(m₁ + m₂)

v = (m₁ v_{1x} + m₂ v_{2x})/(m₁ + m₂)

taking derivative again relative to "t" both side

dv/dt = (m₁ dv_{1x}/dt + m₂ dv_{2x}/dt)/(m₁ + m₂)

a= (m₁ a_{1x} + m₂ a_{2x})/(m₁ + m₂)

3 0
3 years ago
Read 2 more answers
Two 30 uC charges lie on the x-axis, one at the origin and the other at 9 m. A third point is located at 27 m. What is the poten
alukav5142 [94]

Answer:

25000 V

Explanation:

The formula for potential is

V = Kq/r

Potential at B due to the charge placed at origin O

V1 = K q / OB

V_{1}= \frac{9 \times 10^{9} \times 30 \times 10^{-6}}{27}

V1 = 10000 V

Potential at B due to the charge placed at A

V2 = K q / AB

V_{2}= \frac{9 \times 10^{9} \times 30 \times 10^{-6}}{18}

V2 = 15000 V

Total potential at B

V = V1 + V2 = 10000 + 15000 = 25000 V

4 0
3 years ago
A speeding car collides with an unlucky bug flying across the road. Which explains why the impact doesn’t equally damage the car
velikii [3]
The bug was a lot smaller than the car, that's for sure. The car is bigger and sturdier, while the bug is smaller and frail. The bug is so frail, that rather that putting a dent in the car, it splatters all over the car. The bug is very damaged (obviously), while the car just needs a good wash.
8 0
4 years ago
an object is placed 20cm from a converging lens. If the real image is formed 80cm from the object, what is the focal length of t
masya89 [10]

Explanation:

Here,

object distance(u)=20cm

image distance(v)=80cm-20cm=60cm

focal length(f)=?

we know ,

(1/f)=(1/v)+(1/u)

1/f = 1/60 + 1/20

1/f = (1+3)/60

60 =4f

f=15cm

8 0
2 years ago
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