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3 years ago

14

F your muscles are not getting enough oxygen during exercise, they will stop moving.

2 answers:

ohaa [14]3 years ago

8 0

If your muscles stop moving dude you've been dead for a while now lol..
The answer is false.

MariettaO [177]3 years ago

5 0

Answer:

FALSE

Explanation:

The lesson says-

"At first, the body can't change quickly enough to give the muscles the amount of oxygen they need. This situation produces an oxygen deficit in the muscles. During this oxygen deficit, the muscles don't have the necessary amount of oxygen they need in order to produce enough energy to keep moving. <u><em>Instead of the muscles just stopping, however, the body has a second system that kicks in during emergencies</em></u>."

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If a car can go from 0 to 60 km/h in 8.0 seconds, what would be its final speed after 5.0 seconds if its starting speed were 50

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Answer:

This question assumes that the car accelerates at the same rate as when it went from 0 to 60km/h

24.29m/s or 87.4km/h

Explanation:

Let's find the acceleration of the car:

let vi=0, vf=60km/h (16.67m/s), Δt = 8.0s

a = (vf-vi)/Δt

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2 years ago

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An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

$F_e=qE=ma$ (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}$

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

$x_{max}=v_o\sqrt{\frac{2d}{a}}$ (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

$x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm$

The horizontal distance traveled by the electron is 9.5cm

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