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4 years ago

What is the difference between a neutral atom and an ion?

Physics

1 answer:

nasty-shy [4]4 years ago

6 0

Answer:here u go

Explanation:

Atoms vs. Ions. Atoms are neutral; they contain the same number of protons as electrons. By definition, an ion is an electrically charged particle produced by either removing electrons from a neutral atom to give a positive ion or adding electrons to a neutral atom to give a negative ion.

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A spring stretches 0.150 m when a 0.30 kg mass is hung from it. The spring is then stretched an additional 0.100 m from this equ

DochEvi [55]

Answer:

a) $k=19.6N/m$

b) $V_m=0.81m/s$

c) $a_m=6.561m/s^2$

d) $K.E=0.096J$

e) $T=0.78sec$ & $F=1.29sec$

f) $mx'' + kx' =0$

Explanation:

From the question we are told that:

Stretch Length $L=0.150m$

Mass $m=0.30kg$

Total stretch length $L_t=0.150+0.100=>0.25$

Generally the equation for Force F on the spring is mathematically given by

$F=-km\\\\k=F/m\\\\k=\frac{m*g}{x}\\\\k=\frac{0.30*9.8}{0.15}$

$k=19.6N/m$

b)Generally the equation for Max Velocity of Mass on the spring is mathematically given by

$V_m=A\omega$

Where

A=Amplitude

$A=0.100m$

And

$\omega=angulat Velocity\\\\\omega=\sqrt{\frac{k}{m}}\\\\\omega=\sqrt{\frac{19.6}{0.3}}\\\\\omega=8.1rad/s$

Therefore

$V_m=A\omega\\\\V_m=8.1*0.1$

$V_m=0.81m/s$

Generally the equation for Max Acceleration of Mass on the spring is mathematically given by

$a_m=\omega^2A$

$a_m=8.1^2*0.1$

$a_m=6.561m/s^2$

Generally the equation for Total mechanical energy of Mass on the spring is mathematically given by

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}*0.3*0.8^2$

$K.E=0.096J$

Generally the equation for the period T is mathematically given by

$\omega=\frac{2\pi}{T}$

$T=\frac{2*3.142}{8.1}$

$T=0.78sec$

Generally the equation for the Frequency is mathematically given by

$F=\frac{1}{T}$

$F=1.29sec$

Generally the Equation of time-dependent vertical position of the mass is mathematically given by

$mx'' + kx' =0$

Where

'= signify order of differentiation

7 0

3 years ago

Arigid body must rotate about an axis in order for it to have angular momentum about that axis. True False

kompoz [17]

Answer:

False

Explanation:

Let's consider the definition of the angular momentum,

$\vec{L} = I \vec{\omega}$

where $I = \int\limits_m r^2 dm = \lim_{n \to \infty} \sum\limits_{i=1}^n m_i r_i^2$ is the moment of inertia for a rigid body. Now, this moment of inertia could change if we change the axis of rotation, because "r" is defined as the distance between the puntual mass and the nearest point on the axis of rotation, but still it's going to have some value. On the other hand,

$\vec{\omega} = \frac{\vec{r} \times \vec{v}}{r^2}$ so $\vec{\omega} \neq 0$ unless $\vec{r}$ ║ $\vec{v}$ .

In conclusion, a rigid body could rotate about certain axis, generating an angular momentum, but if you choose another axis, there could be some parts of the rigid body rotating around the new axis, especially if there is a projection of the old axis in the new one.

7 0

3 years ago

Which is the resulting direction of a surface wave

Sliva [168]

In a surface wave (like water) the water goes up and down, but the wave travels across (parallel to) the surface.

3 0

3 years ago

Water from a fire hose is directed horizontally against at a rate of 50.0 kg/s and a speed of 42.0 m/s.

ss7ja [257]

Answer:

Force, |F| = 2100 N

Explanation:

It is given that,

Water from a fire hose is directed horizontally against at a rate of 50.0 kg/s, $\dfrac{m}{t}=50\ kg/s$

Initial speed, v = 42 m/s

The momentum is reduced to zero, final speed, v = 0

The relation between the force and the momentum is given by :

$F=\dfrac{p}{t}$

$F=\dfrac{mv}{t}$

$F=50\ kg/s\times 42\ m/s$

|F| = 2100 N

So, the magnitude of the force exerted on the wall is 2100 N. Hence, this is the required solution.

8 0

3 years ago

A green truck is moving to the right. A red truck is moving to the left with a speed of 6 m/s. The mass of the red truck is 1,00

Contact [7]

Answer:

2 m/s

Explanation:

$m_1$ = Mass of red truck = 1000 kg

$m_2$ = Mass of green truck= 3000 kg

$u_1$ = Initial Velocity of red truck = 6 m/s

$u_2$ = Initial Velocity of green truck

$v$ = Velocity with which they move together = 0

For elastic collision

$m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow m_1u_1 + m_2u_2 =0\\\Rightarrow m_1u_1 + m_2u_2 =(m_1 + m_2)v\\\Rightarrow m_1u_1 + m_2u_2 =0\\\Rightarrow u_2=-\frac{m_1u_1}{m_2}\\\Rightarrow u_2=-\frac{1000\times 6}{3000}\\\Rightarrow u_2=-2\ m/s$

Velocity of the green truck is 2 m/s

8 0

3 years ago