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3 years ago

13

A wire delivers 12.0 C of charge in 4.0 s. What is the current in the wire? 3.0 A 8.0 A 16 A 48 A

2 answers:

poizon [28]3 years ago

6 0

Q = Charge delivered by the wire = 12.0 C

t = time taken for the charge to be delivered = 4.0 s

i = current flowing in the wire due to the flow of charge = ?

Current is defined as the time rate at which charge pass through a particular point.

Current in the wire is given as

i = Q/ t

inserting the values

i = (12.0) /(4.0)

i = 3 A

hence the correct choice is 3.0 A

const2013 [10]3 years ago

5 0

3.0 A i got it off Quizlet and there usually always right lol can't submit tho my answers to short.... Dot dot dot

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A natural atom can be negatively charged by

svet-max [94.6K]

Answer:

By Gaining Electrons

Explanation:

A nuetral atom is negative when it gains electrons, and it can be positive when it loses electrons.

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2 years ago

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In a real world trampoline, how do gymnasts keep a constant bounce height

kkurt [141]

By using common factors of physics: weight, gravity, and stability.
Weight would keep them at a constant height.
Gravity helps the weight and how much force it propels the person, or objects, into the air.
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3 years ago

A particle starts from the origin at t = 0 with an initial velocity of 4.8 m/s along the positive x axis.If the acceleration is

tia_tia [17]

Answer:

Part a)

Velocity = 6.9 m/s

Part b)

Position = (3.6 m, 5.175 m)

Explanation:

Initial position of the particle is ORIGIN

also it initial speed is along +X direction given as

$v_x = 4.8 m/s$

now the acceleration is given as

$\vec a = -3.2 \hat i + 4.6 \hat j$

when particle reaches to its maximum x coordinate then its velocity in x direction will become zero

so we will have

$v_f = v_i + at$

$0 = 4.8 - 3.2 t$

$t = 1.5 s$

Part a)

the velocity of the particle at this moment in Y direction is given as

$v_f_y = v_i + at$

$v_f_y = 0 + 4.6(1.5)$

$v_f_y = 6.9 m/s$

Part b)

X coordinate of the particle at this time

$x = v_x t + \frac{1}{2}a_x t^2$

$x = 4.8(1.5) - \frac{1}{2}(3.2)(1.5^2)$

$x = 3.6 m$

Y coordinate of the particle at this time

$y = v_y t + \frac{1}{2}a_y t^2$

$y = 0(1.5) + \frac{1}{2}(4.6)(1.5^2)$

$y = +5.175 m$

so position is given as (3.6 m, 5.175 m)

5 0

3 years ago

Una mujer de masa m está parada en el borde de una mesa giratoria horizontal de momento de inercia I y radio R. La mesa al princ

Dovator [93]

r

$- \frac{mR^2 }{I } \ v$ Answer:

a) w = - $\frac{m r }{I} v$ , b) W = - ½ m_woman R² (1 + m_woman R / I²) v²

Explanation:

a) To solve this exercise, let's use the conservation of angular momentum.

We define a system formed by the table and the woman, therefore the torques are internal and the moment is conserved

initial instant. Before starting to move the woman

L₀ = 0

final instant. After starting to move

L_f = I w + m v r

the moment is preserved

L₀ = L_f

0 = Iw + m v r

w = - $\frac{m r }{I} v$ (1)

the direction of the angular velocity is opposite to the direction of the linear velocity, that is, counterclockwise

b) for this part we use the relationship between work and kinetic energy

W = ΔK

in this case the initial speed is zero and the final speed of the table, using the relationship between linear and angular variables

v = w r

we substitute

W = 0 - ½ I_total w²

I_total = I + m_{woman} R²

W = - ½ (I + m_woman R²) ( $\frac{m_{woman} R}{I} \ v$ ) ²

W = - ½ (m_woman² R² + m_woman³ R³ / I²) v²

W = - ½ m_woman R² (1 + m_woman R / I²) v²

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3 years ago

On which one of the surfaces listed below will a toy car released with the same initial speed travel the farthest? *

irinina [24]

Are you gonna list the surfaces or nah

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2 years ago