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STatiana [176]
3 years ago
12

What element has 3 valence electrons and 4 energy levels​

Physics
1 answer:
Tems11 [23]3 years ago
6 0
Gallium is the answer
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A boxer hits punching bag and gives it a change in momentum of 12 kg multiplied by m divided by s over 7.0ms what is the magnitu
mr_godi [17]

Answer: 1700

Explanation:

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2 years ago
List 3 to 5 ways to reduce friction
pychu [463]
<span>Make the surfaces smoother. Rough surfaces produce more friction and smooth surfaces reduce friction
Lubrication is another way to make a surface smoother
Make the object more streamlined
Reduce the forces acting on the surfaces
<span>Reduce the contact between the surfaces.</span></span>
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3 years ago
Read 2 more answers
A piece of silicon sample has a resistivity of 0.1 ω.Cm. Its thickness is 100µm. The electron mobility is 1350cm 2 v -1sec-1. Wh
lawyer [7]

The answer is Rh = 135 cm^3 and B = 0.05185 wh/m^2


Explanation:

Resitivity of silicon = 0.1

thickness = 100um

so, I = ma

Required to find out concentration of electron , we know that

Rh = up

By putting in the values,

Rh = 1350 x 0.1

Rh = 135 cm^3


Now consider,

Rh = 1 / Rh.q

= 1 / Rh . q

= 1 / 135 x1.609 x10^-19

= 4.6037 x 10^16 / cm^3

Vh = BIRh / w

B = Vh w/ IRh

B = -70 x10^-6 x 100 x10^-6 / 1x 10^-3 x 135 x 10^-6

B = 0.05185 wh / m^2


5 0
3 years ago
2. A test reveals that 150 J of work is required to lift an object 3 m at a
Nuetrik [128]

Answer:

50N

Explanation:

W=Fd

150=F(3)

50N=F

7 0
3 years ago
A student places an object with a mass of m on a disk at a position r from the center of the disk. The student starts rotating t
koban [17]

Answer:

The coefficient of static friction between the object and the disk is 0.087.

Explanation:

According to the statement, the object on the disk experiments a centrifugal force due to static friction. From 2nd Newton's Law, we can represent the object by the following formula:

\Sigma F_{r} = \mu_{s}\cdot N = m\cdot \frac{v^{2}}{R} (1)

\Sigma F_{y} = N - m\cdot g = 0 (2)

Where:

N - Normal force from the ground on the object, measured in newtons.

m - Mass of the object, measured in newtons.

g - Gravitational acceleration, measured in meters per square second.

v - Linear speed of rotation of the disk, measured in meters per second.

R - Distance of the object from the center of the disk, measured in meters.

By applying (2) on (1), we obtain the following formula:

\mu_{s}\cdot m\cdot g = m\cdot \frac{v^{2}}{R}

\mu_{s} = \frac{v^{2}}{g\cdot R}

If we know that v = 0.8\,\frac{m}{s}, g = 9.807\,\frac{m}{s^{2}} and R = 0.75\,m, then the coefficient of static friction between the object and the disk is:

\mu_{s} = \frac{\left(0.8\,\frac{m}{s} \right)^{2}}{\left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.75\,m)}

\mu_{s} = 0.087

The coefficient of static friction between the object and the disk is 0.087.

5 0
3 years ago
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