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3 years ago

Compare and contrast formation of with the formation of mgcl2

1 answer:

3 0

Answer: Both of the compounds contain primary bonding; however, the bonding is of different types. Magnesium chloride (MgCl₂) contains ionic bonds, which are stronger and involve the complete transfer of electrons. Meanwhile, ethane (C₂H₆) also bonds by electron transfer but the electrons are not completely transferred from one atom to another. Instead, they are shared, forming what is known as a covalent bond.

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A woman pushes an oak chest across an oak floor at a constant speed of 0.450 m/s. The chest has a mass of 40.0 kg, and the coeff

natima [27]

Answer:

A. 243 N

Explanation:

Friction is the force that opposes the relative motion between systems that are in contact.

This friction force that opposes the motion of the oak chest across the oak surface will be equal and opposite to that exerted by the woman.

First find the normal force which is the force that would point directly upwards to support weight of the block.

Normal force, N= mg where m is the mass of the chest and g is the acceleration due to gravity.

Given m=40 kg and g=9.80 m/s²

N force=40×9.80 =392N

Then find the force of friction which is given by the formula;

F=μN where μ is friction coefficient for the oak chest and N is the normal force on the chest

Given μ=0.620 and N force = 392 N then it will be;

F=0.620× 392 =243.04 N

Answer : 243 N

6 0

3 years ago

Nahzvsbsnsnsmsskskssks

prisoha [69]

Answer:

some one might report you

Explanation:

3 0

3 years ago

Read 2 more answers

Why current remains same in series combination of resistors in all resistors and p.d. remains different?

IgorC [24]

Current at all points of a series circuit must be the same, because there's no place in the circuit where electrons are being manufactured, and no place where they're leaking out and falling on the floor. The nimber of electrons that leaves the loop is the same number that entered it. I'm not sure what is nmeant by "p.d. remains different" .

4 0

4 years ago

In hydrogen, the transition from level 2 to level 1 has a rest wavelength of 121.6 nm.1).Find the speed for a star in which this

soldier1979 [14.2K]

Answer:

1). $v = - 2960526m/s$

2). Toward us

3). $v = - 493421m/s$

4). Toward us

5). $v = 1480263m/s$

6). Away from us

7). $v = 3207236m/s$

8). Away from us

Explanation:

Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving toward the observer, or to the red part of the spectrum when it is moving away from the observer (that is known as the Doppler effect).

The wavelength at rest is 121.6 nm ( $\lambda_{0} = 121.6nm$ )

$Redshift: \lambda_{measured} > \lambda_{0}$

$Blueshift: \lambda_{measured} < \lambda_{0}$

Then, for this particular case it is gotten:

Star 1: $\lambda_{measured} = 120.4nm$

Star 2: $\lambda_{measured} = 121.4nm$

Star 3: $\lambda_{measured} = 122.2nm$

Star 4: $\lambda_{measured} = 122.9nm$

Star 1:

$Blueshift: 120.4nm < 121.6nm$

Toward us

Star 2:

$Blueshift: 121.4nm < 121.6nm$

Toward us

Star 3:

$Redshift: 122.2nm > 121.6nm$

Away from us

Star 4:

$Redshift: 122.9nm > 121.6nm$

Away from us

Due to that shift the velocity of the star can be determine by means of Doppler velocity.

$v = c\frac{\Delta \lambda}{\lambda_{0}}$ (1)

Where $\Delta \lambda$ is the wavelength shift, $\lambda_{0}$ is the wavelength at rest, v is the velocity of the source and c is the speed of light.

$v = c(\frac{\lambda_{measured}- \lambda_{0}}{\lambda_{0}})$ (2)

Case for star 1 $\lambda_{measured} = 120.4 nm$ :

$v = (3x10^{8}m/s)(\frac{120.4nm-121.6nm}{121.6nm})$

$v = - 2960526m/s$

Notice that the negative velocity means that is approaching to the observer.

Case for star 2 $\lambda_{measured} = 121.4 nm$ :

$v = (3x10^{8}m/s)(\frac{121.4nm-121.6nm}{121.6nm})$

$v = - 493421m/s$

Case for star 3 $\lambda_{measured} = 122.2 nm$ :

$v = (3x10^{8}m/s)(\frac{122.2nm-121.6nm}{121.6nm})$

$v = 1480263m/s$

Case for star 4 $\lambda_{measured} = 122.9 nm$ :

$v = (3x10^{8}m/s)(\frac{122.9nm-121.6nm}{121.6nm})$

$v = 3207236m/s$

4 0

4 years ago

A spring hangs from the ceiling with an unstretched length of x 0 = 0.35 m . A m 1 = 6.3 kg block is hung from the spring, causi

Jlenok [28]

Answer:

x₂=0.44m

Explanation:

First, we calculate the length the spring is stretch when the first block is hung from it:

$\Delta x_1=0.50m-0.35m=0.15m$

Now, since the stretched spring is in equilibrium, we have that the spring restoring force must be equal to the weight of the block:

$k\Delta x_1=m_1g$

Solving for the spring constant k, we get:

$k=\frac{m_1g}{\Delta x_1}\\\\k=\frac{(6.3kg)(9.8m/s^{2})}{0.15m}=410\frac{N}{m}$

Next, we use the same relationship, but for the second block, to find the value of the stretched length:

$k\Delta x_2=m_2g\\\\\Delta x_2=\frac{m_2g}{k}\\\\\implies \Delta x_2=\frac{(3.7kg)(9.8m/s^{2})}{410N/m}=0.088m$

Finally, we sum this to the unstretched length to obtain the length of the spring:

$x_2=0.35m+0.088m=0.44m$

In words, the length of the spring when the second block is hung from it, is 0.44m.

3 0

4 years ago