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3 years ago

As a car moves along the road, the distance traveled each second is measured and recorded. What is the

Physics

1 answer:

scoundrel [369]3 years ago

8 0

If the car moves along the distance it will be 16 of the line graph where is independent of the graph

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What is most likely happen if there was an error during translation of protein?

Neko [114]

A defective protein would be produced leading to either that protein cell being killed or developing into a cancer

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3 years ago

I need help with the question 8B

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The answer of this is C

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3 years ago

Your friend just challenged you to a race through an obstacle course. You know in order to beat him, you must run 30 meters with

seropon [69]

Answer:

Velocity = 0.5 m/s South (A)

Explanation:

You need to determine the average rate of velocity.

The equation you will use is velocity = displacement/time

The displacement is 30m South.

The time is 60 seconds.

Plug into the equation Velocity = 30m South/60 s

Velocity = 0.5 m/s South

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3 years ago

A long solenoid with 8.22 turns/cm and a radius of 7.00 cm carries a current of 19.4 mA. A current of 3.59 A exists in a straigh

daser333 [38]

Answer:

a. 3.039cm

b.magnetic field is $B=2.958\times10^{-5}T$

Explanation:

Direction of the solenoid magnetic field is along the axis of the solenoid. and magnetic field due to the wire perpendicular to that due to the solenoid.. Magnetic field at r is given by:

$\overrightarrow B = \overrightarrow B_s+ \overrightarrow B_w,\ \ \ \ \ \overrightarrow B_s\perp \overrightarrow B_w$

Angle of net magnetic field from axial direction is given by:

$tan\ \theta=\frac{B_w}{B_s}$ ,

Field due to solenoid:

$B_s=\mu_onI_s, \ \ \ \ n=(8.22 t/cm)(100cm/m)=822turn/m$

Field due to wire:

$B_w=\frac{\mu_oI_w}{2\pi r}$

Therefore, r:

$tan\ \theta=\frac{B_w}{B_s}\\\\=\frac{\mu_oI_w}{2\pi r(\mu_o nI_s)}\\\\r=\frac{I_w}{2\pi nI_stan \ \theta}\\\\r=\frac{3.59A}{2\pi\times822\times19.4\times10^{-3}A \ tan 49.7\textdegree}\\\\r=3.039cm$

Hence, the radial distance is 3.039cm

b.The magnetic field strength is given by:

$B=\sqrt{B_w^2+B_s^2}\\\\tan 49.7\textdegree=\frac{B_w}{B_s}\\\\1.179=\frac{B_w}{B_s}\\\\B_w=1.179B_s\\\\B=\sqrt{(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{3}A)+1.179(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{-3}A)}\\\\B=2.958\times10^{-5}T$

Hence, the magnetic field is $B=2.958\times10^{-5}T$

7 0

3 years ago

The speed of a box traveling on a horizontal friction surface changes from vi = 13 m/s to vf = 11.5 m/s in a distance of d = 8.5

ValentinkaMS [17]

The average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s is 3.24 W.

<h3>Acceleration of the box</h3>

The acceleration of the box is calculated as follows;

vf² = vi² + 2as

a = (vf² - vi²)/2s

a = (11.5² - 13²) / (2 x 8.5)

a = -2.16 m/s²

<h3>Time of motion of the box</h3>

The time taken for the box to travel is calculated as follows;

a = (vf - vi)/t

t = (vf - vi) / a

t = (11.5 - 13) / (-2.16)

t = 0.69 s

<h3>Average power supplied by the friction</h3>

P = Fv

P = (ma)(vf - vi)

P = (1 x -2.16) x (11.5 - 13)

P = 3.24 W

Thus, the average power supplied to the box by friction while it slows from 13 m/s to 11.5 m/s is 3.24 W.

Learn more about average power here: brainly.com/question/19415290

#SPJ1

7 0

1 year ago