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goblinko [34]
3 years ago
5

A tension-force replaces what objects when creating a Free Body Diagram?

Physics
1 answer:
algol133 years ago
3 0

Answer:

Option C D and E

chains cables and ropes.

Explanation:

A free body diagram is a simple sketch showing the forces acting on objects in a physical system.

Usually tensile forces  are those forces that occur on objects that are pulled by a force. To get the correct answer choices we need to identify the objects that are most likely to be pulled by a force in any physical system. Such objects from our options include: chains cables and ropes.

This means that whenever these are acting in a physical system, they are mostly used in tension.

Consequently, whenever a free body diagram of such system is drawn, chains cables and ropes.are usually replaced by tension force arrows

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A 1-mCi source of^60 Co is placed in the center of a cylindrical water-filled tank with an inside diameter of 20 cm and depth of
LenKa [72]

To solve this problem we need the concepts of Energy fluency and Intensity from chemical elements.

The energy fluency is given by the equation

\Psi=4RcE\pi

Where

\Psi =The energy fluency

c = Activity of the source

r = distance

E = electric field

In the other hand we have the equation for current in materials, which is given by

I= I_0 e^{-\mu_{h20}X_{h2o}} e^{-\mu_{Fe}X_{Fe}}

Then replacing our values we have that

I = 1*10^{-3} * 3.3*10^{10} * e ^{-0.06*1.1} e^{-0.058*7.861}

I = 1.3*10^7 Bq

We can conclude in this part that 1.3*10^7Bq is the activity coming out of the cylinder.

Now the energy fluency would be,

\Psi = \frac{cE}{4\pir^2}

\Psi = \frac{1.3*10^7*2*1.25}{4\pi*11^2}

\Psi = 2.14*10^4 MeV/cm^2.s

The  uncollided flux density at the outer surface of the tank nearest the source is \Psi = 2.14*10^4 MeV/cm^2.s

6 0
2 years ago
Siruis, the brightest star in the night sky, has a luminosity of 22. This means that Sirius: A, B, C, D QUESTION
suter [353]

Answer:

i think C

Explanation:

4 0
2 years ago
An airplane flies in a loop (a circular path in a vertical plane) of radius 160 m . The pilot's head always points toward the ce
notka56 [123]

Answer:

a) 39.6 m/s b) 4123 N

Explanation:

a) At the top of the loop, all of the forces point downwards (force of gravity and normal force).

Fnet=ma

ma=m(v^2/R) (centripetal acceleration)

mg=m(v^2/R)

m cancels out (this is why pilot feels weightless) so,

g=(v^2/R)

9.8 m/s^2 = v^2/160 m

v^2=1568 m^2/s^2

v=39.6 m/s

b) At the bottom of the loop, the normal force and the force of gravity point in opposite directions. The normal force is the weight felt.

Convert 300 km/hr to m/s

300 km/hr=83.3 m/s

Convert pilot's weight into mass:

760 N = 77.55 kg

Fnet=ma

n-mg=m(v^2/R)

n=(77.55 kg)(((83.3 m/s)^2)/160 m)+(77.55 kg)(9.8 m/s^2)

n=3363.2 N+760 N=4123 N

5 0
3 years ago
How can you incorporate the information
il63 [147K]

Explanation:

There are many ways to achieve a healthier lifestyle and increase fitness. Healthy eating and constant exercise are essential for maintaining good physical and mental health. Through effective nutrition the human body is able to function better, there is an increase in disposition, improvement in immunity, improvement in blood rates, etc. Physical exercise, on the other hand, assists in strengthening muscles, controlling blood pressure, preventing cardiovascular disease, greater satisfaction, less stress, etc.

Some simple examples that can assist in improving health and fitness are:

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6 0
2 years ago
(a) What is the ionization energy of a hydrogen atom that is in the n = 6 excited state? (b) For a hydrogen atom, determine the
crimeas [40]

Answer:

(a) 0.3778 eV

(b) Ratio = 0.0278

Explanation:

The Bohr's formula for the calculation of the energy of the electron in nth orbit is:

E=\frac {-13.6}{n^2}\ eV

(a) The energy of the electron in n= 6 excited state is:

E=\frac {-13.6}{6^2}\ eV

E=-0.3778\ eV

Ionisation energy is the amount of this energy required to remove the electron. Thus, |E| = 0.3778 eV

(b) For first orbit energy is:

E=\frac {-13.6}{1^2}\ eV

E=-13.6\ eV

Ratio=\frac {E_6}{E_1}

Ratio=\frac {-0.3778}{-13.6}

Ratio = 0.0278

7 0
3 years ago
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