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3 years ago

How is electronegativity related to the position of elements on the periodic table and covalent bonding?

Physics

1 answer:

andrezito [222]3 years ago

6 0

Electrons in a polar covalent bond are shifted toward the more electronegative atom

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Two forces act on a 1250 kg sailboat as it moves through the water with an initial velocity of 11 m/s. The forward force of the

Naddik [55]

Answer:

The velocity of the boat 15 seconds later is 5.6 meters per second.

Explanation:

We assume that sailboat can be modelled as particle, so that we use solely translations equations. It is noticed that the sailboat is move by action of the wind and drag force of the water is opposed to such force, but the last force has a greater magnitude than the first one, meaning that net force is less than zero.

From Newton's Laws we have the following equation of equilibrium for the sailboat:

$\Sigma F = F - f = m\cdot a$ (Eq. 1)

Where:

$F$ - Force from the wind exerted on the sailboat, measured in newtons.

$f$ - Drag force of the water, measured in newtons.

$m$ - Mass of the sailboat, measured in kilograms.

$a$ - Net acceleration of the sailboat, measured in meters per square second.

If we know that $F = 3.90\times 10^{3}\,N$ , $f = 4.35\times 10^{3}\,N$ and $m = 1250\,kg$ , then the net acceleration of the sailboat is:

$a = \frac{F-f}{m}$

$a = \frac{3.90\times 10^{3}\,N-4.35\times 10^{3}\,N}{1250\,kg}$

$a = -\frac{9}{25}\,\frac{m}{s^{2}}$

If sailboat decelerates uniformly, then we can get the final velocity of the boat by using this equation of motion:

$v = v_{o}+a\cdot t$ (Eq. 2)

Where:

$v_{o}$ , $v$ - Initial and final velocities of the sailboat, measured in meters per second.

$t$ - Time, measured in seconds.

If we get that $v_{o} = 11\,\frac{m}{s}$ , $a = -\frac{9}{25}\,\frac{m}{s^{2}}$ and $t = 15\,s$ , then the final velocity of the sailboat is:

$v = 11\,\frac{m}{s} +\left(-\frac{9}{25}\,\frac{m}{s^{2}} \right)\cdot (15\,s)$

$v = 5.6\,\frac{m}{s}$

The velocity of the boat 15 seconds later is 5.6 meters per second.

3 0

3 years ago

A 1.50-m cylindrical rod of diameter 0.500 cm is connected to a power supply that maintains a constant potential difference of 1

nydimaria [60]

Answer:

a) 1.06*10^-5

b) 0.00105 °C^-1

Explanation:

Given that

Length of the cylinder, L = 1.5 m

Radius of the cylinder, r = 0.25 cm

Voltage across the rod, V = 15 V

I• at Temperature T• = 20° C is 18.5 A

I at Temperature T = 90° C is 17.2 A

See attachment for calculations

5 0

3 years ago

Marie is puzzled by her findings she has done several meticulous calculations and has gotten the numbers .37 rad, .89 rad and 1.

Irina-Kira [14]

Answer:

I think the answer is a

Explanation:

for it to be accurate has be to exactly 0.9 rad

it is not precise because the answer she is getting is different everytime and not even close. For instance,

It would have been precise if she had gotten 0.37 rad in every attempt. or 0.89 every attempt...

6 0

3 years ago

The acceleration due to gravity on jupiter is 2.5 times whats on earth. An object of mass 10kg is taken to Jupiter. What is the

Damm [24]

The mass is still 10 kg. But instead of weighing 98N as it does on Earth, it weighs 245N on Jupiter.

4 0

3 years ago

A 200-kg object and a 500-kg object are separated by 4.00 m. (a) find the net gravitational force exerted by these objects on a

pantera1 [17]

Check the attached file for the solution for this problem.

5 0

3 years ago