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3 years ago

Why is the current atomic model called the "electron cloud model?

1 answer:

5 0

The electron cloud model is the orbitals. Its called that because the model demonstrates the atom. The electrons are in the orbital cloud.

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Can some one please help me with 7 and 8 I can’t fail this I’ll mark brainless .

Musya8 [376]

I think the first question is talking about the ionic compound Sodium sulfide and it’s formula is Na2S.
And for the second question, i’m pretty sure it’d be a positive charge

3 0

3 years ago

A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim

Kaylis [27]

Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

$F_D = -\frac{1}{2}\rho V^2 C_d A$

Where

$\rho$ is the density of the flow

V = Velocity

$C_d$ = Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

$F=ma=m\frac{dV}{dt}$

Equating both equations we have:

$m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A$

$m(dV)=-\frac{1}{2}\rho C_d A (dt)$

$\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)$

Integrating

$\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)$

$-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t$

Here,

$V_f = 60mph = 26.82m/s$

$V_i = 120.7m/s$

$m= 1600lbf = 725.747Kg$

$\rho = 1.21 kg/m^3$

$C_d = 0.3$

$\Delta t=7s$

Replacing:

$\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)$

$-0.029 = -5.4997r^2$

$r = 2.2963m$

$d= r*2 = 4.592m \approx 15.065ft$

4 0

3 years ago

Your 64-cm-diameter car tire is rotating at 3.3 rev/swhen suddenly you press down hard on the accelerator. After traveling 250 m

umka21 [38]

Answer:

0.76 rad/s^2

Explanation:

First, we convert the original and final velocity from rev/s to rad/s:

$v_o = 3.3\frac{rev}{s} * \frac{2\pi rad}{1rev} =20.73 rad/s$

$v_f = 6.4\frac{rev}{s} * \frac{2\pi rad}{1rev}=40.21 rad/s$

Now, we need to find the number of rads that the tire rotates in the 250m path. We use the arc length formula:

$D = x*r \\x = \frac{D}{r} = \frac{250m}{0.64m/2} = 781.25 rads$

Now, we just use the formula:

$w_f^2-w_o^2=2\alpha*x$

$\alpha =\frac{w_f^2-w_o^2}{2x} = \frac{(40.21rad/s)^2-(20.73rad/s)^2}{2*781.25rad} = 0.76 rad/s^2$

6 0

3 years ago

Read 2 more answers

A brick is thrown from a wall that is 19.2 m above the ground and hits the ground 5.00 s later. What was the initial velocity of

Brums [2.3K]

Answer:

Explanation:

Given the following :

Height of wall = 19.2 m

Time taken to hit the ground = 5 seconds

Acceleration due to gravity (g) = 9.8m/s ( downward motion)

The initial velocity of the object refers to the Velocity of the object at time t = 0

Initial Velocity = g × time

Initial Velocity = 9.8 × 0 = 0

5 0

3 years ago

A box sits on a table. A long arrow labeled F subscript P = 22 N points right. A short arrow labeled F subscript f = ? N points

anygoal [31]

Answer:

b -4 N

may the force be with you

8 0

3 years ago

Read 2 more answers