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Ipatiy [6.2K]
3 years ago
13

FREEE POINTS BOI !!!! COME AND GET EM!!! XD (owo)

Physics
2 answers:
timurjin [86]3 years ago
5 0

Answer:

yea boiiiiiiiiiiiiiii

aalyn [17]3 years ago
4 0

30 POINTS THX A LOTTTTTT

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What is the closest distance the electrodes used in an NCV test can be placed on a nerve in order to measure the voltage change
sammy [17]

Answer:

0.1 m

Explanation:

The closest distance the electrodes used in an NCV test in oerder to measure

the voltage change as a response to the stimulus is 0.1 m.

This is because the shortest observable time period is not less than the action-potential time response of 1 mili second the length traveled by the sensation during this time is 1 m sec x 100 m / s =0.1 m, which is the shortest distance the electrodes could be positioned on the nerve.

6 0
3 years ago
Is the color spectrum simply a small segment of the electromagnetic spectrum?
ch4aika [34]

Answer:

yup, u r correct

Explanation:

3 0
3 years ago
The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
Sav [38]

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=\dfrac{2c}{3b}.

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = \dfrac{2c}{3b},

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

7 0
3 years ago
The old rubber boot has two leaks. The top of the boot is 0.3 m higher than the leaks. What is the velocity of the water coming
Otrada [13]

Answer:

Leak 1 = 3.43 m/s

Leak 2 = 2.42 m/s

Explanation:

Given that the top of the boot is 0.3 m higher than the leaks. 

Let height H = 0.3m and the acceleration due to gravity g = 9.8 m/s^2

From the figure, the angle of the leak 1 will be approximately equal to 45 degrees. While the leak two can be at 90 degrees.

Using the third equation of motion under gravity, we can calculate the velocity of leak 1 and 2

Find the attached files for the solution and figure

7 0
3 years ago
A car moving in a straight line starts at X=0 at t=0. It passesthe point x=25.0 m with a speed of 11.0 m/s at t=3.0 s. It passes
Agata [3.3K]

Answer:

Average velocity v = 21.18 m/s

Average acceleration a = 2 m/s^2

Explanation:

Average speed equals the total distance travelled divided by the total time taken.

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

Average acceleration equals the change in velocity divided by change in time.

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

Where;

v1 and v2 are velocities at time t1 and t2 respectively.

And x1 and x2 are positions at time t1 and t2 respectively.

Given;

t1 = 3.0s

t2 = 20.0s

v1 = 11 m/s

v2 = 45 m/s

x1 = 25 m

x2 = 385 m

Substituting the values;

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

v = (385-25)/(20-3)

v = 21.18 m/s

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

a = (45-11)/(20-3)

a = 2 m/s^2

8 0
3 years ago
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