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3 years ago

6

Explain why atomic radius decreases as you move to the right across a period for main-group elements but not for transition elem

ents.Match the words in the left column to the appropriate blanks in the sentences on the right.

1 answer:

Aleksandr-060686 [28]3 years ago

7 0

Answer:

Explained.

Explanation:

Only the first question has been answered

In a period from left to right the nuclear charge increases and hence nucleus size is compressed. Thus, atomic radius decreases.

In transition elements, electrons in ns^2 orbital remain same which is the outer most orbital having 2 electrons and the electrons are added to (n-1) d orbital. So, outer orbital electron experience almost same nuclear attraction and thus size remains constant.

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Explain your answers to 9a and 9b in terms of Newton's laws of motion.

kogti [31]

Answer:i One way to solve the quadratic equation x2 = 9 is to subtract 9 from both sides to get one side equal to 0: x2 – 9 = 0. The expression on the left can be factored:

Explanation:

4 0

3 years ago

A coin slides over a frictionless plane and across an xy coordinate system from the origin to a point with xy coordinates (1.40

leva [86]

Answer:

The work done required on the coin during the displacement is 21.75 w.

Explanation:

Given that,

A coin slides over a friction-less plane i.e friction force = 0.

The co-ordinate of the given point is (1.40 m, 7.20 m).

The position vector of the given point is represented by $1.40 \hat i+7.20 \hat j$ .

The displacement of the coin is

$\vec d=1.40 \hat i+7.20 \hat j$

The force has magnitude 4.50 N and its makes an angle 128° with positive x axis.

Then x component of the force = 4.50 cos128°

The y component of the force = 4.50 sin128°

Then the position vector of the force is

$\vec F=(4.50 cos 128^\circ)\hat i+(4.50 sin 128^\circ)\hat j$

$=-2.77 \hat i+3.56 \hat j$

We know that,

work done is a scalar product of force and displacement.

$W=\vec F.\vec d$

$=(-2.77 \hat i+3.56 \hat j).(1.40 \hat i+7.20 \hat j)$

=(-2.77×1.40+ 3.56×7.20) w

=21.75 w

The work done required on the coin during the displacement is 21.75 w.

6 0

3 years ago

The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .

vlabodo [156]

Answer: $90\°$

Explanation:

The Compton Shift $\Delta \lambda$ in wavelength when the photons are scattered is given by the following equation:

$\Delta \lambda=\lambda_{c}(1-cos\theta)$ (1)

Where:

$\lambda_{c}=2.43(10)^{-12} m$ is a constant whose value is given by $\frac{h}{m_{e}c}$ , being $h$ the Planck constant, $m_{e}$ the mass of the electron and $c$ the speed of light in vacuum.

$\theta)$ the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through $180\°$ :

$\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))$ (2)

$\Delta \lambda_{max}=\lambda_{c}(1-(-1))$

$\Delta \lambda_{max}=2\lambda_{c}$ (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

$\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)$ (4)

$\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)$ (5)

If we want $\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}$ , $1-cos\theta$ must be equal to 1:

$1-cos\theta=1$ (6)

Finding $\theta$ :

$1-1=cos\theta$

$0=cos\theta$

$\theta=cos^{-1} (0)$

Finally:

$\theta=90\°$ This is the scattering angle that will produce $\frac{1}{4}\Delta \lambda_{max}$

7 0

3 years ago

Suppose that an object is moving with constant nonzero acceleration. Which of the following is an accurate statement concerning

Wittaler [7]

Answer: The right answer is b)

Explanation:

By definition, acceleration is the change in velocity (in module or direction) over a given time interval, as follows:

a = (v-v₀) / (t-t₀)

If we take t₀ = 0 (this is completely arbitrary), we can rewrite the equation above, as follows:

v = v₀ + at

We can recognize this function as a linear one, where a represents the slope of the line.

If a is constant, this means that the relationship between the change in velocity and the change in time remains constant, in other words, in equal times, its velocity changes in an equal amount.

Let's suppose that a = 10 m/s/s. (Usually written as 10 m/s²).

This is telling us that each second, the velocity increases 10 m/s.

8 0

3 years ago

Oil of specific gravity 0.75 flows through a smooth contraction in a pipe at a volumetric flow rate of 3.2 cu ft / sec.Find the

pogonyaev

Answer:

The force required to hold the contraction in place is 665.91 N ↑

Explanation:

Given;

specific gravity of oil, γ = 0.75

Volumetric flow rate, V 3.2 Ft³/s = 0.0906 m³/s

$\gamma =\frac{\rho_o}{\rho_w}$

where;

$\rho_o$ is the density of oil

$\rho_w$ is the density of water = 1000 kg/m³

∴density of oil ( $\rho_o$ ) = γ × density of water( $\rho_w$ )

= 0.75 × 1000 kg/m³

= 750kg/m³

Buoyant Force = ρVg

= 750 × 0.0906 × 9.8

= 665.91 N ↑

This force acts upward or opposite gravitational force.

Therefore, the force required to hold the contraction in place is 665.91 N ↑

6 0

3 years ago