What unit do we use for pressure

The carbon monoxide molecule (CO) consists of a carbon atom and an oxygen atom separated by a distance of

valkas [14]

Answer:

6.46*10^-11m

Explanation:

We can use the equation of the center of the mass of two atoms

$Xcm=\frac{mcxc+moxo}{mc+mo}$

If we take the origin at the center of the molecule carbon monoxice, xc will be 0, so

$Xcm=\frac{moxo}{mc+mo}$ = $\frac{xo}{mc/mo+1}$

$Xcm=\frac{1.13.10^-3}{0.750mo/mo+1}$ =6.46*10^-11m

7 0

3 years ago

What must happen for an ion To form

Gwar [14]

Answer:

Lose or gain

Explanation:

Ions are formed when atoms lose or gain electrons in order to fulfill the octet rule and have full outer valence electron shells. When they lose electrons, they become positively charged and are named cations. When they gain electrons, they are negatively charged and are named anions.

3 0

4 years ago

You own a yacht which is 14.5 meters long. It is motoring down a canal at 10.6 m/s. Its bow (the front of the boat) is just abou

Alex73 [517]

Answer:

4.198 seconds

Explanation:

The total length of the yacht is 30m and the width of the bridge is 14.5m.

The speed of the yacht is 10.6m/s.

We can assume the current of the canal has a speed of 0m/s.

To calculate the answer we will break it into two parts

1) The time it takes for the bow to travel from the start of the bridge to the end

Formula Speed = Distance / Time

Speed = 10.6m/s

Distance = 30m

Time = T1

T1 = 2.8301s

2) The time it takes for the stern to cross the end of the bridge after the bow

Formula Speed = Distance / Time

Speed = 10.6m/s

Distance = 14.5m

Time = T2

T2 = 1.3679s

3) The Total Time is

Total = T1 + T2

Total = 4.198s

5 0

3 years ago

A 30 gram bullet is shot upward at a wooden block. The bullet is launched at the speed vi. It travels up 0.40 m to strike the wo

Lerok [7]

Answer:

Explanation:

Mass of bullet m = .03 kg

Mass of wooden block M = 0.5 kg

Since the center of mass of the wooden block with the bullet in it travels up a distance of 0.60 m before reaching its maximum height

Velocity of wooden block + bullet just after impact = √2gH

=√(2 x 9.8 x 0.6)

= 3.43 m / s

Let the launch velocity of bullet be v₁

If v₂ be the velocity with which bullet hits the block

Applying law of conservation of momentum

.03 x v₂ = .530 x 3.43

v₂ = 60.6 m /s

if v₁ be initial velocity

v₂² = v₁² - 2 gh

v₁² = v₂² + 2 gh

= 60.6 ² + 2 x 9.8 x 0.4

v₁ = 60.65 m /s this is launch speed.

b )

Initial kinetic energy of bullet

= 1/2 m v²

= .5 x .03 x 3680

= 55 J

Potential energy of bullet + block = 0

Total energy = 5 J

c)

Kinetic energy of bullet block system

1/2 m v²

= .5 x .53 x 3.43

= 3.11 J

d )

Loss of energy in the impact = Total mechanical energy lost from beginning to end?

3.11 J - 5

= 1.89 J

6 0

3 years ago

A mass MM uniform solid cylinder of radius RR and a mass MM thin uniform spherical shell of radius RR roll without slipping. If

vampirchik [111]

Answer:

vcyl / vsph = 1.05

Explanation:

The kinetic energy of a rolling object can be expressed as the sum of a translational kinetic energy plus a rotational kinetic energy.
The traslational part can be written as follows:

$K_{trans} = \frac{1}{2}* M* v_{cm} ^{2} (1)$

The rotational part can be expressed as follows:

$K_{rot} = \frac{1}{2}* I* \omega ^{2} (2)$

where I = moment of Inertia regarding the axis of rotation.
ω = angular speed of the rotating object.
If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:

$v = \omega * R (3)$

For a solid cylinder, I = M*R²/2 (4)
Replacing (3) and (4) in (2), we get:

$K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2} (5)$

Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:

$K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2} +\frac{1}{4}* M* v_{cmc}^{2} = \frac{3}{4}* M* v_{cmc} ^{2} (6)$

Repeating the same steps for the spherical shell:

$I_{sph} = \frac{2}{3} * M* R^{2} (7)$

$K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2} (8)$

$K_{sph} = \frac{1}{2}* M* v_{cms} ^{2} +\frac{1}{3}* M* v_{cms}^{2} = \frac{5}{6}* M* v_{cms} ^{2} (9)$

Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.
And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.
Rearranging, and taking square roots on both sides, we get:

$\frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)$

This means that the solid cylinder is 5% faster than the spherical shell, which is due to the larger moment of inertia for the shell.

3 0

3 years ago