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3 years ago

the nucleus and the subatomic particles in the nucleus, what is the charge of the nucleus of an atom?

1 answer:

4 0

The Nucleus contains Protons and Neutrons.

The Neutrons does not have a charge.

The Protons are positively charge.

Hence the charge on the Nucleus, would be the charge of the proton, which is positive.

Hence Nucleus is Positively Charged.

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I what is the resistance of

Lynna [10]

The resistance of a conductor is given by:
$R= \frac{\rho L}{A}$
where
$\rho$ is the resistivity of the material
L is the length of the conductor
A is its cross-sectional area

We can use this formula to solve both parts of the problem.

a) The length of the copper wire is L=1.0 m. Its diameter is d=0.50 mm, so its radius is
$r= \frac{d}{2}=0.25 mm=0.25 \cdot 10^{-3} m$
And its cross-sectional area is
$A=\pi r^2 = \pi (0.25 \cdot 10^{-3}m)^2 = 1.96 \cdot 10^{-7} m^2$
The copper resistivity is $\rho=1.68 \cdot 10^{-8} \Omega m$ , therefore the resistance of this piece of wire is
$R= \frac{\rho L}{A}= \frac{(1.68 \cdot 10^{-8} \Omega m)(1.0 m)}{1.96 \cdot 10^{-7} m^2}= 8.57 \cdot 10^{-2} \Omega$

b) The length of this piece of iron is L=10 cm=0.10 m. Its cross-sectional size is L=1.0 mm=0.001 m, so its cross-sectional area is
$A=L^2 = (0.001 m)^2 =1 \cdot 10^{-6}m^2$
The iron resistivity is $\rho = 9.71 \cdot 10^{-8} \Omega m$ , therefore the resistance of this piece of wire is
$R= \frac{\rho L}{A}= \frac{(9.71 \cdot 10^{-8} \Omega m)(0.10 m)}{1.0 \cdot 10^{-6} m^2}=9.71 \cdot 10^{-3} \Omega$

3 0

3 years ago

Explain what happens to particles in diffusion. What causes diffusion?

vlabodo [156]

Diffusion:
 in diffusion particles move from area of higher concentration to the area of lower concentration..
causes:
the movement of particles allow them ti disperse,,, when molecules are close to each other they collide with each other and push each other apart so they spread in the whole area... :)

8 0

3 years ago

Read 2 more answers

A player kicks a football from ground level with a velocity of 26.2m/s at an angle of 34.2° above the horizontal. How far back f

Amanda [17]

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.

For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

<h3>Explanation</h3>

How long does it take for the ball to reach the goal?

Let the distance between the kicker and the goal be $x$ meters.

Horizontal velocity of the ball will always be $26.2\times\cos{34.2\textdegree}$ until it lands if there's no air resistance.

The ball will arrive at the goal in $\displaystyle \frac{x}{26.2\times\cos{34.2\textdegree}}$ seconds after it leaves the kicker.

What will be the height of the ball when it reaches the goal?

Consider the equation

$\displaystyle h(t) = -\frac{1}{2}\cdot g\cdot t^{2} + v_{0,\;\text{vertical}} \cdot t + h_0$ .

For this soccer ball:

$g = 9.81\;\text{m}\cdot\text{s}^{-2}$ ,
$v_{0,\;\text{vertical}} = 26.2\times \sin{34.2\textdegree{}}\;\text{m}\cdot\text{s}^{-2}$ ,
$h_0 = 0$ since the player kicks the ball "from ground level."

$\displaystyle t=\frac{x}{26.2\times\cos{34.2\textdegree}}$

when the ball reaches the goal.

$\displaystyle h= - 9.81 \times \frac{x^2}{(26.2\times\cos{34.2\textdegree})^2} + (26.2 \times \sin{34.2\textdegree})\times\frac{x}{26.2\times\cos{34.2\textdegree}} \\\phantom{h} = -\frac{9.81}{(26.2\times\cos{34.2\textdegree})^2}\cdot x^{2} + \frac{\sin{34.2\textdegree}}{\cos{34.2\textdegree}}\cdot x$ .

Solve this quadratic equation for $x$ , $x > 0$ .

$x = 65.1$ meters when $h = 0$ meters.
$x = 6.54$ or $58.5$ meters when $h = 4$ meters.

In other words,

For the ball to go straight into the goal, the kicker needs to be no more than 6.54 meters away from the goal.
For the ball to arc into the goal, the kicker needs to be between 58.5 and 65.1 meters away from the goal.

3 0

3 years ago

A tug-of-war game is played by five c/hildren: three on one team and two on the other. How much force will the two child team ha

dem82 [27]

Answer:

no, 3 porces is more tha 2 so the power between the 3 should be more than 2

Explanation:

4 0

2 years ago

Calculate the final velocity right after a 117 kg rugby player who is initially running at 7.45 m/s collides head‑on with a padd

Free_Kalibri [48]

Answer:

v_f = 0.87 m/s

Explanation:

We are given;

F_avg = -17700 N (negative because it's backward)

m = 117 kg

Δt = 5.50 × 10^(−2) s

v_i = 7.45 m/s

Now, formula for impulse is given by;

I = F•Δt = - 17700 x 5.50 × 10^(−2) = - 973.5 kg.m/s

From impulse momentum theory, we know that;

Change in momentum of particle is equal to impulse.

Thus,

Δp = I = m•v_f - m•v_i

Thus,

-973.5= 117(v_f - 7.45)

Thus,

-973.5/117 = (v_f - 7.45)

-8.3205 + 7.45 = v_f

v_f = - 0.87 m/s

We'll take absolute value as;

v_f = 0.87 m/s

5 0

3 years ago