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2 years ago

Describe current in terms of charge and electrons.PLEASE HELP I WILL GIVE BRAINLIEST

Physics

2 answers:

Sauron [17]2 years ago

7 0

Electrons are negatively charged and are found outside in the nucleus in the rings surrounding the nucleus.

oksian1 [2.3K]2 years ago

4 0

Charge: CURRENT is the rate of flow of charge

Electrons: the flow of electrons in a traditional CURRENT flow from the negative(-) to the positive(+)

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A force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work W is done in stretching i

Sunny_sXe [5.5K]

Answer:

114.44 J

Explanation:

From Hook's Law,

F = ke................. Equation 1

Where F = Force required to stretch the spring, k = spring constant, e = extension.

make k the subject of the equation

k = F/e.............. Equation 2

Given: F = 10 lb = (10×4.45) N = 44.5 N, e = 4 in = (4×0.254) = 1.016 m.

Substitute into equation 2

k = 44.5/1.016

k = 43.799 N/m

Work done in stretching the 9 in beyond its natural length

W = 1/2ke²................. Equation 3

Given: e = 9 in = (9×0.254) = 2.286 m, k = 43.799 N/m

Substitute into equation 3

W = 1/2×43.799×2.286²

W = 114.44 J

3 0

3 years ago

Read 2 more answers

Which is not true about the pH scale?

pochemuha

Answer:

The pH scale was invented by Svante Arrhenius is false.

Explanation :

because the pH scale was invented by Soren Sorensen

4 0

2 years ago

Read 2 more answers

A car achieves a velocity 50 meters per second after 5 seconds. What is the Car's acceleration?

Reptile [31]

Answer: 10 m/s

Explanation: Velocity/Time

50/5= 10

7 0

2 years ago

An aluminum wing on a passenger jet is 35 m long when its temperature is 17°C. At what temperature would the wing be 3 cm (0.03

Mnenie [13.5K]

Answer:

53.32°C

Explanation:

Length of the aluminium wing = 35 m

Change in length of aluminium wing = 0.03 m

The linear expansion coefficient of aluminium $\alpha =23.6\times 10^{-6}/^{\circ}C$

We know that change in length is given by $\Delta L=L\alpha \Delta T$

So $0.03=35\times 23.6\times 10^{-6}\Delta T$

$\Delta T=36.32^{\circ}C$

So final temperature $=T_I+\Delta T=17+36.32=53.3196^{\circ}C$

5 0

2 years ago

A 50.0-g object connected to a spring with a force constant of 35.0 n/m oscillates with an amplitude of 4.00 cm on a frictionles

Dimas [21]

A) The total energy of the system is sum of kinetic energy and elastic potential energy:
$E=K+U= \frac{1}{2}mv^2 + \frac{1}{2}kx^2$
where
m is the mass
v is the speed
k is the spring constant
x is the elongation/compression of the spring

The total energy is conserved, so we can calculate its value at any point of the motion. If we take the point of maximum displacement:
$x=A=4.00 cm = 0.04 m$
then the velocity of the system is zero, so the total energy is just potential energy, and it is equal to
$E=U= \frac{1}{2}kA^2 = \frac{1}{2}(35.0 N/m)(0.04 m)^2=0.028 J$

b) When the position of the object is
$x=1.00 cm = 0.01 m$
the potential energy of the system is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.01 m)^2 = 1.75 \cdot 10^{-3} J$
and so the kinetic energy is
$K=E-U=0.028 J - 1.75 \cdot 10^{-3}J =0.026 J$
since the mass is $m=50.0 g=0.05 kg$ , and the kinetic energy is given by
$K= \frac{1}{2}mv^2$
we can re-arrange the formula to find the speed of the object:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.026 J}{0.05 kg} }=1.02 m/s$

c) The potential energy when the object is at
$x=3.00 cm=0.03 m$
is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$
Therefore the kinetic energy is
$K=E-U=0.028 J-0.016 J = 0.012 J$

d) We already found the potential energy at point c, and it is given by
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$

5 0

3 years ago

Describe current in terms of charge and electrons.PLEASE HELP I WILL GIVE BRAINLIEST​

Describe current in terms of charge and electrons.PLEASE HELP I WILL GIVE BRAINLIEST