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2 years ago

A zinc atom loses 3electron to become a zinc oin

1 answer:

6 0

It lost 3 electron, meaning IT will become an ion. Also, since it specifically LOST, it would be a cation, and gain a positive charge.

Answer: True

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A driven RLC circuit is being driven by an AC emf source with a maximum current of 2.75 A and maximum voltage of 150 V. The curr

weqwewe [10]

Answer:

(a). Z = 54.54 ohm

(b). R = 36 ohm

(c). The circuit will be Capacitive.

Explanation:

Given data

I = 2.75 A

Voltage = 150 V

$\phi = 0.85$ rad = 48.72°

(a). Impedance of the circuit is given by

$Z = \frac{V}{I}$

$Z = \frac{150}{2.75}$

Z = 54.54 ohm

(b). We know that resistance of the circuit is given by

$R = \frac{Z}{\sqrt{1 + \tan^{2}\phi } }$

Put the values of Z & $\phi$ in above formula we get

$R = \frac{54.54}{\sqrt{1 + \tan^{2} ( \ 48.72) } }$

R = 36 ohm

(c). Since the phase angle is negative so the circuit will be Capacitive.

3 0

3 years ago

A speeder passes a parked police car at a constant speed of 23.3 m/s. At that instant, the police car starts from rest with a un

KonstantinChe [14]

Answer:

32s

Explanation:

We must establish that by the time the police car catches up to the speeder, both have travelled a certain distance during the same amount of time. However, the police car experiences accelerated motion whereas the speeder travels at a constant velocity. Therefore we will establish two formulas for distance starting with the speeder's distance:

$x=vt=23.3\frac{m}{s}t$

and the police car distance:

$x=vt+\frac{at^{2}}{2}=0+\frac{2.75\frac{m}{s^{2}} t^{2}}{2}=0.73\frac{m}{s^{2}}$

Since they both travel the same distance x, we can equal both formulas and solve for t:

$0 = 0.73\frac{m}{s^{2}}t^{2}-23.3\frac{m}{s} t\\\\0=t(0.73\frac{m}{s^{2}}t-23.3\frac{m}{s} )\\\\$

Two solutions exist to the equation; the first one being $t=0$

The second solution will be:

$0.73\frac{m}{s^{2}}t=23.3\frac{m}{s}\\\\t=\frac{23.3\frac{m}{s}}{0.73\frac{m}{s^{2}}}=32s$

This result allows us to confirm that the police car will take 32s to catch up to the speeder

7 0

3 years ago

A box is being pulled to the right. What is the magnitude of the Kinect frictional force?

Anna35 [415]

The answer to this question is A - 25 N

3 0

2 years ago

Read 2 more answers

Particle A and particle B, each of mass M, move along the x-axis exerting a force on each other. The potential energy of the sys

Archy [21]

Speed of particle B is 2v₀/3 m/s to the left. Particle A and particle B will always have equal speed since they experience equal forces.

<h3>Conservation of energy</h3>

The speed and direction of the particle B is determined by applying the principle of conservation of energy as follows;

K.E₁ + P.E₁ = K.E₂ + P.E₂

$\frac{1}{2} Mv^2_A + \frac{G}{r^2} = \frac{1}{2} Mv^2_B + \frac{G}{r^2} \\\\ \frac{1}{2} Mv^2_A = \frac{1}{2} Mv^2_B\\\\v^2_A = v^2_B\\\\v_A = v_B$

$v_B = \frac{2v_0}{3} \ m/s \ to \ the \ left$

At any given position, the speed of particle A and particle B will be equal, since they experience equal force and they have equal masses.

The complete question is below:

Particle A and particle B, each of mass M, move along the x-axis exerting a force on each other. The potential energy of the system of two particles assosicated with the force is given by the equation U=G/r 2, where r is the distance between the two particles and G is a positive constant. At time t=T1 particle A is observed to be traveling with speed 2vo/3 to the left. The speed and direction of motion of particle B is ?

Learn more about conservation of energy here: brainly.com/question/166559

5 0

1 year ago

Read 2 more answers

An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.7

Kazeer [188]

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

$v = \frac{I}{nqA}$

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

$n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3}$

Now, we can find the drift speed:

$v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s$

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!

4 0

2 years ago