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4 years ago

7

WILL NAME BEST ANSWER THE BRAINLIEST (if you answer well)

1 answer:

mylen [45]4 years ago

7 0

The best and most correct answer among the choices provided by the question is third choice. What happened during the set-up was the e<span>xcess protons move from the positive rod to the negative rod.</span> I hope my answer has come to your help. God bless and have a nice day ahead!

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Two copper rods are separated by a small gap at B. Rod AB has a diameter of 200mm and rod BC has a diameter of 150mm. Find the f

gulaghasi [49]

Answer:

A)3.8196 * $10^{9}$ Newtons B) 2.153 * $10^{9}$ Newtons

Explanation:

Force = Pressure * Area.

Pressure is given as 120GPa

Area = Cross Sectional Area of Rod = Area Of A circle = π $R^{2}$ .

Area Expansivity β = $\frac{Change in Area}{Original Area * Temperature Rise}$

Area Expansivity β = 2α

α = 16.9x10-6/ ° C.

Radius of Rod AB = Half 0f Diameter, 200mm = $\frac{0.2m}{2}$ = 0.1 m

Radius of Rod BC = Half Of Diameter, 150mm = $\frac{0.15}{2}$ = 0.075 m.

Note: To convert from millimeter to meter you divide by 1000.

Area of Rod AB = π * $0.1^{2}$ = 0.0314 $m^{2}$

Area of Rod BC = π * $0.075^{2}$ = 0.0177 $m^{2}$ .

Change in Area = 2α * Original Area * Temperature Rise. Therefore for

Rod AB = 2 * 16.9* $10^{-6}$ * 0.0314 * 30 = 3.183 * $10^{-5}$ $m^{2}$ .

For Rod BC we have = 2 * 16.9* $10^{-6}$ * 0.0177 * 30 = 1.794∈-5 $m^{2}$ .

The forces on each rods will be given by.

Force AB = Pressure * Area of Rod AB

= 120GPa * 3.183 * $10^{-5}$ gives 3.8196 * $10^{9}$ Newtons.

Force BC = Pressure * Area of Rod BC

= 120GPa * 1.794 * $10^{-5}$ gives 2.153 * $10^{9}$ Newtons

7 0

3 years ago

When a ball is launched from the ground at a 45° angle to the horizontal, it falls back to the ground 50 m from the launch point

Inessa [10]

Answer:

Explanation:

Given

angle through which ball is launched $=45^{\circ}$

Range of ball=50 m

Range of projectile is $=\frac{u^2sin2\theta }{g}$

$50=\frac{u^2sin90}{9.8}$

u=22.136 m/s

If ball is thrown straight upward

$v^2-u^2=2as$

$0-(22.136)^2=2(-9.8)s$

$s=\frac{22.136^2}{2\times 9.8}$

s=25 m

(b)For Projectile time of flight is

$t=\frac{2usin\theta }{g}$

$t=\frac{2\times 22.136\times sin45}{9.8}$

t=3.19 s

7 0

3 years ago

5 . W h ic h o f th e fo l lo w in g is N O T a fu n d a m e n ta l S .I u n it? A . M e te r B . A m p e re C . K e lv in D . R

Flauer [41]

Answer:

D. Radian

Explanation:

hope it helps

5 0

3 years ago

Read 2 more answers

When a front wheel drops off the roadway you should?

nikdorinn [45]

<span>braking and returning suddenly to the roadway</span>

6 0

3 years ago

cycling at 13.0 m/s on your new aero carbon fiber bike, how much times would it take a pro cyclist to ride in 120 km? Give your

sp2606 [1]

We have that the time in seconds, minutes, and hours is

$t=1.083*10^{-4}s$

$T_{min}=1.805*10^{-6}min$

$T_{hours}=3.0083*10^{-8}hours$

From the Question we are told that

Velocity $v=13.0m/s$

Distance $d=120 km$

Generally the equation for the Time is mathematically given as

$t=\frac{13}{120*10^3}\\\\t=1.083*10^{-4}s$

Therefore

$T_{min}=1.083*10^{-4}s/60$

$T_{min}=1.805*10^{-6}min$

And

$T_{hours}=T_{min}/60$

$T_{hours}=3.0083*10^{-8}hours$

For more information on this visit

brainly.com/question/12319416?referrer=searchResults

4 0

3 years ago