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4 years ago

If element x has 44 protons how many electrons does it have?

1 answer:

5 0

The correct answer is '44'.

In fact, the atom of an element has an equal number of protons and electrons: therefore, if element X has 44 protons, it should have 44 electrons as well.

For curiosity: the chemical element with 44 protons and 33 electrons is Ruthenium.

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An electron is projected with horizontal speed 105m / s in a downwardly directed 304N / C electric field. Find the vertical posi

Vinil7 [7]

Answer:

-107 m

Explanation:

Sum of forces in the y direction:

∑F = ma

-qE = ma

a = -qE/m

a = -(1.60×10⁻¹⁹ C) (304 N/C) / (9.11×10⁻³¹ kg)

a = -53.4×10¹² m/s²

Given in the y direction:

v₀ = 0 m/s

a = -53.4×10¹² m/s²

t = 2×10⁻⁶ s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (2×10⁻⁶ s) + ½ (-53.4×10¹² m/s²) (2×10⁻⁶ s)²

Δy = -107 m

4 0

3 years ago

(eText prob. 4.25 with some values changed) Air enters a diffuser of a jet engine operating at steady state at 2.65 psia, 389◦R,

krek1111 [17]

Answer:

$V_2 = 45.44m/s$

Explanation:

We have to many data in different system, so we need transform everything to SI, that is

$P_1 = 2.65 Psi = 18.271 kPa\\T_1= 389\°R = 216 K\\V_1 = 869ft/s = 264m/s\\T_2 = 450\°R = 250K$

When we have all this values in SI apply a Energy Balance Equation,

$\dot{Q}_{cv}-\dot{W}_{cv}+\dot{m}[(h_1-h_2)+(\frac{V_1^2-V_2^2}{2})+g(z_1-z_2)]=0$

Solving for V_2

$V_2 = \sqrt{V_1^2+2(h_1-h_2)}$

From the table of gas properties we calculate for $T_1 = 216K$ and $T_2 = 250K$

$h_1 = 209.97+(219.97-209.97)(\frac{216-210}{220-210})$

$h_1 = 215.97kJ/kg$

For T_2;

$h_2 = 250.05kJ/kg$

Substituting in equation for V_2

$V_2 = \sqrt{V_1^2+2(h_1-h_2)}$

$V_2 = \sqrt{265^2+2(215.97-250.05)*10^3}\\V_2 = 45.44m/s$

4 0

3 years ago

The acceleration of an object is inversely proportional to its mass, so if you decrease its mass while keeping the net force the

MaRussiya [10]

Answer:

True

Explanation:

Since acceleration is inversely proportional to mass, decreasing mass will make the object lighter, and thus easier to speed up. So acceleration increases as mass decreases and vice versa

5 0

2 years ago

Read 2 more answers

Consider a machine of mass 70 kg mounted to ground through an isolation system of total stiffness 30,000 N>m, with a measured

kvv77 [185]

Answer:

a)0.0229 m

b)0.393 rad

c)1.57

d)707.6 N

e)0.298 m/s

Explanation:

Given:

Mass of the machine, m=70 kg
Stiffness of the system, k=30000 N/m
Damping ratio=0.2
Damping force, F=450 N
Angular velocity $\omega=13\ \rm rad/s$

a)We know that the amplitude X at steady state is given by

$X=\dfrac{\dfrac{F_0}{m}}{\sqrt{\omega_n^2-\omega^2)^2 +(2\rho \omega_n\omega)^2}}\\$

Where

$\omega_n=\sqrt{\dfrac{k}{m}}\\\\=\sqrt{\dfrac{30000}{70}}\\\\=20.7\ \rm rad/s$

$\omega=13\ \rm rad/s$
$\rho=0.2$

$F_0=450\ \rm N$
$m=70\ \rm kg$

$X=\dfrac{\dfrac{450}{70}}{\sqrt{20.7^2-13^2)^2 +(2\times 0.2\times20.7\times13)^2}}\\[tex]X=0.0229\ \rm m$

b) The phase shift of the motion is given by

$\tan\phi=\dfrac{2\rho \omega_n \omega }{\omega_n^2-\omega^2}\\\\\dfrac{2\times0.2\times20.7\times13 }{20.7^2-\13^2}\\\\\phai=0.393\\$

c)Transmissibility ratio is given by

$T.R.=\sqrt{\dfrac{1+(2\rho r)^2}{(1-r^2)^2+{(2\rho r)^2}}}\\\\T.R.=\sqrt{\dfrac{1+(2\times0.2\times0.628)^2}{(1-0.628^2)^2+{(2\times0.2\times0/628)^2}}}\\\\=1.57$

d)The magnitude of the force transmitted to the ground is

$F_T=(T.R)\times F_0\\\\=450\times1.57\\\\=707.6\ \rm N$

e)The maximum velocity is given by $V_{max}$

$V_{max}=\omega A_0\\\\=13\times 0.0229\\\\=0.298\ \rm m/s$

6 0

3 years ago

A solid metal sphere with radius 0.430 m carries a net charge of 0.270 nC . Part A Find the magnitude of the electric field at a

rodikova [14]

Answer:

8.46 N/C

Explanation:

Using Gauss law

$E=\frac {kQ}{r^{2}}$

Gauss's Law states that the electric flux through a surface is proportional to the net charge in the surface, and that the electric field E of a point charge Q at a distance r from the charge

Here, K is Coulomb's constant whose value is $9\times 10^{9} Nm^{2}/C^{2}$

r = 0.43 + 0.106 = 0.536 m

$E=\frac {9\times 10^{9}\times 0.270\times 10^{-9}}{0.536^{2}}=8.4581755402094007\approx 8.46 N/C$

8 0

4 years ago