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3 years ago

How to make snf62- lewis structure in aleks?

1 answer:

4 0

Sn is in group 4 so has 4 electrons in the outer shell (draw these as dots) Add 1 more electron from each bond to F (+6)

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A mechanic needs to replace the motor for a merry-go-round. The merry-go-round should accelerate from rest to 1.5 rad/s in 7.0 s

marta [7]

Answer

given,

initial speed of merry-go-round = 0 rad/s

final speed of merry-go-round = 1.5 rad/s

time = 7 s

Radius of the disk = 6 m

Mass of the merry-go-round = 25000 Kg

Moment of inertia of the disk

$I = \dfrac{1}{2}MR^2$

$I = \dfrac{1}{2}\times 25000\times 6^2$

I = 450000 kg.m²

angular acceleration

$\alpha = \dfrac{\omega_f-\omega_0}{t}$

$\alpha = \dfrac{1.5-0}{7}$

$\alpha =0.214\ rad/s^2$

we know,

$\tau= I \alpha$

$\tau= 450000\times 0.214$

$\tau=96300\ N.m$

8 0

4 years ago

26. The equation E=hf describes the energy of each photon in a beam of light. If Planck’s constant, h, were larger, would photon

Elenna [48]

The equation of the energy of a photon is E=h*f.

If we increase the Planck's constant h, the energy would increase.

For example, lets double the value of Planck's constant and name it H:

H=2*h. Now lets put that into the equation for energy that we will call E₂:

E₂=H*f=2*h*f=2*E.

So we can clearly see that E₂=2*E or that if we double Planck's constant, the energy also doubles.

6 0

3 years ago

An electron of mass 9.11 1031 kg has an initial speed of 3.00 105 m/s. It travels in a straight line, and its speed increases to

elena55 [62]

Explanation:

It is given that,

Mass of an electron, $m=9.11\times 10^{-31}\ kg$

Initial speed of the electron, $u=3\times 10^5\ m/s$

Final speed of the electron, $v=7\times 10^5\ m/s$

Distance, d = 5 cm = 0.05 m

(a) The acceleration of the electron is calculated using the third equation of motion as :

$a=\dfrac{v^2-u^2}{2d}$

$a=\dfrac{(7\times 10^5)^2-(3\times 10^5)^2}{2\times 0.05}$

$a=4\times 10^{12}\ m/s^2$

Force exerted on the electron is given by :

$F=m\times a$

$F=9.11\times 10^{-31}\times 4\times 10^{12}$

$F=3.64\times 10^{-18}\ N$

(b) Let W is the weight of the electron. It can be calculated as :

$W=mg$

$W=9.11\times 10^{-31}\times 9.8$

$W=8.92\times 10^{-30}\ N$

Comparison,

$\dfrac{F}{W}=\dfrac{3.64\times 10^{-18}}{8.92\times 10^{-30}}$

$\dfrac{F}{W}=4.08\times 10^{11}$

Hence, this is the required solution.

8 0

4 years ago

Identifying Advantages of Parallel Circuits

melisa1 [442]

Answer: If one bulb goes out the other bulbs stay lit.

If there is a break in one branch of the circuit, current can still flow through the other branches.

Explanation:

3 0

3 years ago

Read 2 more answers

A 6.4-N force pulls horizontally on a 1.5-kg block that slides on a smooth horizontal surface. This block is connected by a hori

Elena L [17]

-- Although it's not explicitly stated in the question,we have to assume that
the surface is frictionless. I guess that's what "smooth" means.

-- The total mass of both blocks is (1.5 + 0.93) = 2.43 kg. Since they're
connected to each other (by the string), 2.43 kg is the mass you're pulling.

-- Your force is 6.4 N.
Acceleration = (force)/(mass) = 6.4/2.43 m/s²
 That's about 2.634 m/s² 

(I'm going to keep the fraction form handy, because the acceleration has to be
used for the next part of the question, so we'll need it as accurate as possible.)

-- Both blocks accelerate at the same rate. So the force on the rear block (m₂) is

Force = (mass) x (acceleration) = (0.93) x (6.4/2.43) = 2.45 N.

That's the force that's accelerating the little block, so that must be the tension
in the string.

7 0

3 years ago