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3 years ago

Dispositivo que muestra tanto la dirección como la magnitud de la corriente eléctrica.

Physics

1 answer:

slamgirl [31]3 years ago

5 0

Answer:

<em>Galvanómetro</em>

Explanation:

Un galvanómetro es un dispositivo eléctrico utilizado para <em>detectar la presencia de corriente y voltaje pequeños o para medir su magnitud.</em> Los galvanómetros se utilizan principalmente en puentes y potenciómetros.

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Ross, a college sophomore, has no clear preference for any of the candidates running for student body president. The students he

mylen [45]

Answer:

longing for social inclusion.

Explanation:

Ross here is longing for social inclusion.

He decides to campaign for the his fraternity brother not by choice or will but by peer pressure and social inclusion because most of the students campaigned for Henry so he supports Henry as well. Moreover, he did not be feel left out and he did not have a clear preference as well.

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3 years ago

A 60 kg hockey player skating at 20 m/s gets slammed with a force of 950 N. How long does it take him to stop completely?

irakobra [83]

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2.4

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2 years ago

Distributions of electric charges in a cell play a role in moving ions into and out of a cell. In this situation, the motion of

Y_Kistochka [10]

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3 years ago

A rabbit is hopping along at an approximately constant speed of 3.9 m/s. The rabbit passes a crouched cat ready to chase the rab

Firlakuza [10]

Answer:

t = 7,8 s

Explanation:

From the instant, the rabbit passes the cat. The cat star running acceleration of 0,5 m/s² .

When the cat arrives at the speed of 3,9 m/s the cat catches the rabbit

Then for the cat arrives at 3,9 m/s nedds

v = vo + a*t vo = 0 then v = a*t

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4 0

2 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago