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lidiya [134]
2 years ago
12

1. Predict whether the energy required to remove an electron from magnesium and potassium would be more or less than that requir

ed to remove an electron from sodium. Explain your prediction in terms of the forces and interactions involved in this process. \
FILL OUT THE TABLE PLEASE WITH A GOOD ANSWER FOR BRAINLIEST!

Physics
1 answer:
Inga [223]2 years ago
8 0

In both cases less energy is required

But comparetively Mg require more energy than K

Let's see the electron configuration of Both

  • [Mg]=1s²2s²2p⁶3s²=[Ne]3s²
  • [K]=1s²2s²2p⁶3s²3p⁶4s¹=[Ar]4s¹

K has only one valence electron so very less ionization enthalpy so less energy required

Mg has 2 so more IE hence more energy required

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Answer:

Mass is how much "stuff" is in something. Weight is a force that is measured in Newtons. A pan balance is a type of scale. A measuring cup is used to measure volume, such as cups and grams (depending if you use the metric system or not).

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3 years ago
In a simple circuit, if you replace one resistor with a resistor of a higher value will the reading the ammeter increase, decrea
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The ammeter reading (current) will decrease.
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4 years ago
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A jet - powered car called the spirit of America required 9600meters to stop from its highest speed . If the car decelerated at
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v^{2} =  u^{2}  +  2ar
0 = u^2 + 2*(-2)*9600
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3 years ago
The distance between Miami and Oriando is about 220 miles .A pilot flying from
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6 0
4 years ago
A torque of 36.5 N · m is applied to an initially motionless wheel which rotates around a fixed axis. This torque is the result
vivado [14]

Answer:

21.6\ \text{kg m}^2

3.672\ \text{Nm}

54.66\ \text{revolutions}

Explanation:

\tau = Torque = 36.5 Nm

\omega_i = Initial angular velocity = 0

\omega_f = Final angular velocity = 10.3 rad/s

t = Time = 6.1 s

I = Moment of inertia

From the kinematic equations of linear motion we have

\omega_f=\omega_i+\alpha_1 t\\\Rightarrow \alpha_1=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha_1=\dfrac{10.3-0}{6.1}\\\Rightarrow \alpha_1=1.69\ \text{rad/s}^2

Torque is given by

\tau=I\alpha_1\\\Rightarrow I=\dfrac{\tau}{\alpha_1}\\\Rightarrow I=\dfrac{36.5}{1.69}\\\Rightarrow I=21.6\ \text{kg m}^2

The wheel's moment of inertia is 21.6\ \text{kg m}^2

t = 60.6 s

\omega_i = 10.3 rad/s

\omega_f = 0

\alpha_2=\dfrac{0-10.3}{60.6}\\\Rightarrow \alpha_1=-0.17\ \text{rad/s}^2

Frictional torque is given by

\tau_f=I\alpha_2\\\Rightarrow \tau_f=21.6\times -0.17\\\Rightarrow \tau=-3.672\ \text{Nm}

The magnitude of the torque caused by friction is 3.672\ \text{Nm}

Speeding up

\theta_1=0\times t+\dfrac{1}{2}\times 1.69\times 6.1^2\\\Rightarrow \theta_1=31.44\ \text{rad}

Slowing down

\theta_2=10.3\times 60.6+\dfrac{1}{2}\times (-0.17)\times 60.6^2\\\Rightarrow \theta_2=312.03\ \text{rad}

Total number of revolutions

\theta=\theta_1+\theta_2\\\Rightarrow \theta=31.44+312.03=343.47\ \text{rad}

\dfrac{343.47}{2\pi}=54.66\ \text{revolutions}

The total number of revolutions the wheel goes through is 54.66\ \text{revolutions}.

3 0
3 years ago
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