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3 years ago

Electrons are important to electric current because they are able to__________.

Physics

2 answers:

Lubov Fominskaja [6]3 years ago

7 0

The answer would be:

Move from one atom to another; the atom that looses an electron becomes positively charged and the atom that gains an electron becomes negatively charged.

It is the movement of electrons from on atom to another that enables elements to form chemical bonds. If electrons do not move, they are shared. The atom that loses an electron becomes positive because it would have less electrons than protons. The atom that accepts the electron becomes negative because it would have more electrons than protons.

Just remember that all elements are neutral, or they have the same number of protons and electrons. They become charged when they lose or gain one.

lesya692 [45]3 years ago

5 0

Electrons are important to electric current because they are able to :
move from one atom to another; the atom that loses an electron becomes positively charged and the atom that gains an electron becomes negatively charged

hope this helps

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A block with mass m =6.9 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.29 m.

crimeas [40]

1) The spring constant is 233.2 N/m

2) The oscillation frequency is 0.93 Hz

3) The speed of the block is 3.2 m/s

4) The maximum acceleration is $23.9 m/s^2$

5) The net force on the block is 104.9 N

Explanation:

At equilibrium, the weight of the mass is equal to the restoring force of the spring. Therefore, we can write:

$mg=kx$

where

m = 6.9 kg is the mass hanging on the spring

$g=9.8 m/s^2$ is the acceleration of gravity

k is the spring constant

x = 0.29 m is the stretching of the spring

Solving for k, we find

$k=\frac{mg}{x}=\frac{(6.9)(9.8)}{0.29}=233.2 N/m$

The oscillation frequency of a spring-mass system is given by

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where

k is the spring constant

m is the mass

In this problem,

k = 233.2 N/m is the spring constant

m = 6.9 kg is the mass

Substituting, we find the frequency:

$f=\frac{1}{2\pi}\sqrt{\frac{233.2}{6.9}}=0.93 Hz$

The velocity of the block at time t is given by the equation:

$v(t) = -v_0 cos(\omega t)$

where

$v_0 = 4.1 m/s$ is the initial speed

$\omega$ is the angular frequency

t is the time

The angular frequency can be found from the frequency:

$\omega=2\pi f=2\pi(0.93)=5.84 rad/s$

And substituting t = 0.42 s, we find the velocity of the block at this time:

$v(t)=-4.1 cos ((5.84)(0.42))=3.2 m/s$

The maximum acceleration of the block is given by

$a_{max} = \omega^2 A$ (1)

where

$\omega=5.84 rad/s$ is the angular frequency

A is the amplitude

The amplitude is related to the initial velocity by the equation:

$v_0=\omega A$ (2)

Combining (1) and (2), we find

$a_{max}=v_0 \omega$

And substituting $v_0 = 4.1 m/s$ , we find

$a_{max}=(4.1)(5.84)=23.9 m/s^2$

The acceleration at time t can be found by calculating the derivative of v(t), and it is given by the equation

$a(t) = a_{max} sin(\omega t)$

where

$a_{max}=23.9 m/s^2$ is the maximum acceleration

$\omega=5.84 rad/s$ is the angular frequency

t is the time

Substituting t = 0.42 s,

$a(t)=(23.9)sin((5.84)(0.42))=15.2 m/s^2$

Finally, the net force on the block can be found by using Newton's second law:

$F=ma=(6.9)(15.2)=104.9 N$

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3 years ago

2 kg and 3 kg objects slide together, and then they break apart. If the final velocity of 2 kg is 10 m/s,

lara [203]

Answer:

P = (2 + 3) * V where V is their initial speed (total momentum)

P = 2 * 10 + 3 * Vx where Vx here would be V3

If the initial momentum is not known how can one determine the final velocity of the 3 kg obj.

Also work depends on the sum of the velocities

W (initial) = 1/2 (2 + 3) V^2 the initial kinetic energy

W (final) = 1/2 * 2 * V2^2 + 1/2 * 3 * V3^2

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Answer:

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