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2 years ago

Search about non-ohmic devices in 3 pages

Physics

1 answer:

Tomtit [17]2 years ago

7 0

Answer:

electronic diode,

Explanation:

Non-ohmic conductors are materials that do not obey ohm's law and they are electronic diode, transistors, tungsten, thermistors and vacuum tube etc.

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A 2190 kg car moving east at 10.5 m/s collides with a 3220 kg car moving east. The cars stick together and move east as a unit a

Bezzdna [24]

To solve this problem it is necessary to apply the concepts related to the conservation of the Momentum describing the inelastic collision of two bodies. By definition the collision between the two bodies is given as:

$m_1v_1+m_2v_2 = (m_1+m_2)V_f$

Where,

$m_{1,2}$ = Mass of each object

$v_{1,2}$ = Initial Velocity of Each object

$V_f$ = Final Velocity

Our values are given as

$m_1 = 2190Kg$

$v_1 =10.5m/s$

$m_2 = 3220kg$

$V_f = 4.74m/s$

Replacing we have that

$m_1v_1+m_2v_2 = (m_1+m_2)V_f$

$(2190)(10.5)+(3220)v_2 = (2190+3220)(4.74)$

$v_2 = 0.8224m/s$

Therefore the the velocity of the 3220 kg car before the collision was 0.8224m/s

8 0

3 years ago

A amusement park moves riders in a circle at a rate of 6.0m/s if the radius is 9.0 meters what is the acceleration of the ride

oksano4ka [1.4K]

Centripetal acceleration is (speed-squared) / (radius)

CA = (6 m/s)² / (9 m)

CA = (36 m²/s²) / (9 m)

CA = (36/9) (m²/m·s²)

<em>Centripetal acceleration = 4 m/s²</em>

4 0

2 years ago

An electron is moving east in a uniform electric field of 1.55 N/C directed to the west. At point A, the velocity of the electro

valkas [14]

Answer:

Final velocity of electron, $v=6.45\times 10^5\ m/s$

Explanation:

It is given that,

Electric field, E = 1.55 N/C

Initial velocity at point A, $u=4.52\times 10^5\ m/s$

We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :

$v^2=u^2+2as$ ........(1)

a is the acceleration, $a=\dfrac{F}{m}$

We know that electric force, F = qE

$a=\dfrac{qE}{m}$

Use above equation in equation (1) as:

$v^2=u^2+\dfrac{2qEs}{m}$

$v^2=(4.52\times 10^5\ m/s)^2+2\times \dfrac{1.6\times 10^{-19}\ C\times 1.55\ N/C}{9.1\times 10^{-31}\ kg}\times 0.395\ m$

v = 647302.09 m/s

$v=6.45\times 10^5\ m/s$

So, the final velocity of the electron when it reaches point B is $6.45\times 10^5\ m/s$ . Hence, this is the required solution.

3 0

2 years ago

g A meteoroid is in a circular orbit 600 km above the surface of a distant planet. The planet has the same mass as Earth but has

Nikitich [7]

Answer:

Explanation:

gravitational acceleration of meteoroid

= GM / R²

M is mass of planet , R is radius of orbit of meteoroid from the Centre of the planet .

R = (.9 x 6370 + 600 )x 10³ m

= 6333 x 10³ m

M , mass of the planet = 5.97 x 10²⁴ kg .

gravitational acceleration of meteoroid

= GM / R²

= (6.67 x 10⁻¹¹ x 5.97 x 10²⁴ kg / (6333 x 10³ m)²

9.92m/s²

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2 years ago

PLEASE HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! IM DESPERATE!!!!!!!!!!!!!! EVEN IF YOU DONT KNOW , GIVE A GUESS. What would

dezoksy [38]

Well, A = T or U C = G G = C T or U = A So it would be like this; DNA Sequence: GCTAATTGCATCCGA The Complementary Sequence: CGATTAACGTAGGCT Hope this helped :)

3 0

2 years ago

Search about non-ohmic devices in 3 pages​

Search about non-ohmic devices in 3 pages