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2 years ago

1. It is common for people to face divorce, job layoffs, and debt at which of the following stages of life?

Physics

2 answers:

kicyunya [14]2 years ago

8 0

1 C 2 C Hope this helps

Elodia [21]2 years ago

8 0

<h2>Answer 1:</h2>

<u>The correct choice is</u><u> (C) later middle years</u>

<h2>Explanation:</h2>

Later middle years is almost 30 to 40 years of a person's life. It is the most critical period of life because every person wants to get a successful life i this stage. We marry in this age and due to certain failures people get divorced too. Some people lose their jobs due to various reasons and many people also suffer from high debts because they want to stabilize themselves by starting a business if they don't get a suitable job.

<h2>Answer 2:</h2>

<u>The correct answer is (</u><u>C) the type of groups we associate with when we become adults </u>

<h2>Explanation:</h2>

It is better said that a man is known by the company he keeps. So therefore it is a proven fact that all people are mostly influenced by their friends and social circle where they spend most of their time. The life style that we get and the behavior that we show is dependent upon the social circle and family members where we live. Therefore as we move in our lives the events we experience are largely influenced by the type of groups we associate with when we become adults

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A charge of 4 nc is placed uniformly on a square sheet of nonconducting material of side 17 cm in the yz plane. (a) what is the

Ratling [72]

The charge density of the sheet is 1.384×10⁻⁷C/m².

Charge density is defined as the charge per unit area.

The sheet is a square of length l=17 cm.

Calculate the area A of the sheet .

$A=l^2=(17 cm)^2= (17*10^-^2m)^2=0.0289 m^2$

The charge Q on the sheet is

$Q=4nC=4*10^-^9C$

The charge density σ is given by,

$\sigma=\frac{Q}{A}$

Substitute 4×10⁻⁹C for Q and 0.0289 m² for A.

$\sigma=\frac{Q}{A}\\ =\frac{4*10^-^9C}{0.0289 m^2} \\ =1.389*10^-^7C/m^2$

Thus, the charge density of the sheet is <u>1.384×10⁻⁷C/m².</u>

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3 years ago

A force of 6.0 N gives a 2.0 kg block an acceleration of 3.0

Semenov [28]

Explanation:

The Net Force of the object can be written by:

Fnet = ma

where m is the mass of the object in <em>kg</em>

a is the acceleration of the object in <em>m/s^2</em>

Hence by applying the formula we get:

Fnet = (2.0)(3.0)

= 6N

We also know that Net force is also the sum of all forces acting on an object. In this case Friction and the Pushing Force is acting on the object. Hence we can write that:

Fnet = Pushing Force + (-Friction)

6N = 6N - Friction

Friction = 0N

Hence the<u> </u><u>f</u><u>orce of friction is 0N.</u>

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2 years ago

A flat surface is in a uniform magnetic field. Given only the area of the surface and the magnetic flux through the surface, it

Tasya [4]

Answer:

Given the area A of a flat surface and the magnetic flux through the surface $\Phi$ it is possible to calculate the magnitude $\frac{\Phi}{A}=B\ cos \theta$ .

Explanation:

The magnetic flux gives an idea of how many magnetic field lines are passing through a surface. The SI unit of the magnetic flux $\Phi$ is the weber (Wb), of the magnetic field B is the tesla (T) and of the area A is ( $m^{2}$ ). So 1 Wb=1 T.m².

For a flat surface S of area A in a uniform magnetic field B, with $\theta$ being the angle between the vector normal to the surface S and the direction of the magnetic field B, we define the magnetic flux through the surface as:

$\Phi=B\ A\ cos\theta$

We are told the values of $\Phi$ and B, then we can calculate the magnitude

$\frac{\Phi}{A}=B\ cos\theta$

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2 years ago

In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co

labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

$m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m$

The final velocity of cart 1 after collision is given as:

$v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\ Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s$

The final velocity of cart 2 after collision is given as:

$v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\ Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s$

b) Using the law of conservation of energy:

$\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm$

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3 years ago

Help with this question please :>

zheka24 [161]

Answer: A if thats not right its C

Explanation:

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3 years ago

Read 2 more answers