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3 years ago

Which of the following compounds is not likely to have ionic bonds? Question 5 options: LiF NaCl CH4 MgF2

Physics

1 answer:

Harrizon [31]3 years ago

7 0

Answer:

CH4

Explanation:

CH4 is joined together by a covalent bond, aka a bond between two non-metals. Non-metals are found on the right side of the periodic table and include Carbon (C) and Hydrogen. Although Hydrogen is technically on the left side of the table, it has the characteristics of a non-metal. Futhermore, Ionic bonds generally are between an element on the right joined with an element on the left. This is because ionic bonds want charges that will cancel out to create a neutral molecule.

example: LiF

Li→ Li+

F→F-

(Li+)+(F-)=charges cancel out.

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A wire carries a current of 0.66 A. This wire makes an angle of 58° with respect to a magnetic field of magnitude 5.50 10-5 T. T

Oliga [24]

Answer:

The length of the wire is "1.93 m".

Explanation:

Given:

Current,

I = 0.66 A

Angle,

$\Theta$ = 58°

Magnetic field of magnitude,

F = $5.5\times 10^{-5}$ N

The length of the wire will be:

⇒ $F = B\times I\times L\times Sin \Theta$

or,

⇒ $L = \frac{F}{B\times I\times Sin \theta}$

By putting the values, we get

$=\frac{5.5\times 10^{-5}}{0.66\times 5.1\times 10^{-5}\times Sin \ 58^{\circ}}$

$=\frac{5.5}{2.828}$

$=1.93 \ m$

6 0

3 years ago

[4 marks]<br> 22. Compare and contrast the structure and roles of tendons and ligaments.

sladkih [1.3K]

Answer:

Tendons: Tendons are cord-like, strong, inelastic structures that join skeletal muscles to bones

ligaments: Ligaments are elastic structures which connect bones to bones in joints.

Explanation:

6 0

3 years ago

Which statements best describe displacement? Check all that apply.

xxMikexx [17]

Answer:

the last one, the third one, and the first one.

8 0

3 years ago

Read 2 more answers

Identify the position where the cyclist has maximum kinetic energy.

finlep [7]

You need the picture. However, since I'm doing PLATO as well, here's the answer:

The third position in which the cyclist is going downhill is the position where the cyclist has the maximum kinetic energy. The other two photos show that the biker has more potential energy.

Thanks, let me know if this helped!

3 0

4 years ago

Read 2 more answers

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

5 0

3 years ago