All categories

3 years ago

In which stage of loss and grief does a person realize that, despite the pain, life must go on?

Physics

2 answers:

suter [353]3 years ago

6 0

Answer: you just should go on

Explanation:

Gnesinka [82]3 years ago

4 0

Answer:

acceptance

Explanation:

The five stages, denial, anger, bargaining, depression and acceptance are a part of the framework that makes up our learning to live with the one we lost. They are tools to help us frame and identify what we may be feeling. But they are not stops on some linear timeline in grief. Once a person has realized that life must go on, they have reached the acceptance stage.

You might be interested in

When a mass of 0.350 kg is attached to a vertical spring and lowered slowly, the spring stretches 12.0 cm. The mass is now displ

pantera1 [17]

Answer:

The period is $T = 0.700 \ s$

Explanation:

From the question we are told that

The mass is $m = 0.350 \ kg$

The extension of the spring is $x = 12.0 \ cm = 0.12 \ m$

The spring constant for this is mathematically represented as

$k = \frac{F}{x}$

Where F is the force on the spring which is mathematically evaluated as

$F = mg = 0.350 * 9.8$

$F =3.43 \ N$

$k = \frac{3.43 }{ 0.12}$

$k = 28.583 \ N/m$

The period of oscillation is mathematically evaluated as

$T = 2 \pi \sqrt{\frac{m}{k} }$

substituting values

$T = 2 * 3.142* \sqrt{\frac{0.35 }{28.583} }$

$T = 0.700 \ s$

7 0

3 years ago

Find the moment of inertia about each of the following axes for a rod that is 0.360 {cm} in diameter and 1.70 {m} long, with a m

mamaluj [8]

The complete question is;

Find the moment of inertia about each of the following axes for a rod that is 0.36 cm in diameter and 1.70m long, with a mass of 5.00 × 10 ^(−2) kg.

A) About an axis perpendicular to the rod and passing through its center in kg.m²

B) About an axis perpendicular to the rod and passing through one end in kg.m²

C) About an axis along the length of the rod in kg.m²

Answer:

A) I = 0.012 kg.m²

B) I = 0.048 kg.m²

C) I = 8.1 × 10^(-8) kg.m²

Explanation:

We are given;

Diameter = 0.36 cm = 0.36 × 10^(−2) m

Length; L = 1.7m

Mass;m = 5 × 10^(−2) kg

A) For an axis perpendicular to the rod and passing through its center, the formula for the moment of inertia is;

I = mL²/12

I = (5 × 10^(−2) × 1.7²)/12

I = 0.012 kg.m²

B) For an axis perpendicular to the rod and passing through one end, the formula for the moment of inertia is;

I = mL²/3

So,

I = (5 × 10^(−2) × 1.7²)/3

I = 0.048 kg.m²

C) For an axis along the length of the rod, the formula for the moment of inertia is; I = mr²/2

We have diameter = 0.36 × 10^(−2) m, thus radius;r = (0.36 × 10^(−2))/2 = 0.18 × 10^(−2) m

I = (5 × 10^(−2) × (0.18 × 10^(−2))^2)/2

I = 8.1 × 10^(-8) kg.m²

3 0

3 years ago

Estimate frequency of vibration of your arm. Let the length of the arm be 0.57 m. Consider the arm as a simple pendulum and assu

skad [1K]

Answer:

0.80865 Hz

1.23662 seconds

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

l = Length of arm = 0.57 m

Length of simple pendulum is given by

$L=\dfrac{2}{3}l\\\Rightarrow L=\dfrac{2}{3}\times 0.57\\\Rightarrow L=0.38\ m$

The frequency is given by

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{g}{L}}\\\Rightarrow f=\dfrac{1}{2\pi}\sqrt{\dfrac{9.81}{0.38}}\\\Rightarrow f=0.80865\ Hz$

The frequency is 0.80865 Hz

The time period is given by

$T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{0.80865}\\\Rightarrow T=1.23662\ s$

The time period is 1.23662 seconds

3 0

2 years ago

What kind of weather is signaled by cumulus clouds

Novay_Z [31]

Cumulonimbus clouds are associated by severe weather. They are thunderstorm clouds that bring heavy rain, snow, hail, lightning and even tornadoes.

5 0

3 years ago

Examples of energy balance: physical education

mart [117]

Answer:children burn calories to being a student

Explanation:That mean when a children getting ready to go to high school

7 0

3 years ago