PHYSICS COLLEGE

Which of the following is correct? *PLEASE HELP MEEEE
1 cm = 100 m
1 mm = 100 cm
100 mm = 1 cm
1 m = 100 cm

Answers

Answer 1
Answer:

Answer:

The last one

1m = 100 cm

Explanation:

If you do not trust me look it up

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A celestial body moving in an ellipical orbit around a star

Answers

A celestial body moving in an elliptical orbit around a star is a planet.

Depending on its size, composition, and the eccentricity of its orbit, that scanty description could apply to a planet, an asteroid, a comet, a meteoroid, or another star.

An electromagnetic wave with frequency 65.0hz travels in an insulating magnetic material that has dielectric constant 3.64 and relative permeability 5.18 at this frequency. the electric field has amplitude 7.20×10−3v/m.what is the intensity of the wave in a medium\?

Answers

The intensity of the electromagnetic wave which travels in an insulating magnetic material in a medium is  5.766×10⁻⁸ W/m².

What is the intensity of the wave?

The intensity of a wave is the total power delivered  per unit area. It can be given as,

I=(P)/(A)

It can also be given as,

I=(E^2)/(2)\sqrt{(k\varepsilon_o)/(\mu_r\mu_o)}

Here, (\mu_r) is relative permeability, (\mu_0) is physical constant, (k) is dielectric constant, (E) is the amplitude of electric field, and \varepsilon_o is the permittivity of free space.

Here, the electromagnetic wave with frequency 65.0hz travels in an insulating magnetic material that has dielectric constant 3.64 and relative permeability 5.18 at this frequency.

As the electric field has amplitude 7.20×10−3v/m. Thus, put the values in the above formula to find the intensity as,

I=((7.20*10^(-3))^2)/(2)\sqrt{(3.64*8.85*10^(-12))/(5.18*(4\pi*10^(-7)))}\nI=5.766*10^(-8)\rm W/m^2

Hence, the intensity of the electromagnetic wave which travels in an insulating magnetic material in a medium is  5.766×10⁻⁸ W/m².

Learn more about the intensity of the wave here;

brainly.com/question/18109453
Ans: Intensity = I = 5.8 * 10^(-8) W/m^2

Explanation:
First you need to find out the speed of Electromagnetic wave:

Since
v = \sqrt{ (1)/(\mu\epsilon) }


v = \sqrt{ (1)/(\mu_r\mu_ok\epsilon_o) }
\mu_r = 5.18 \n\mu_o = 4 \pi * 10^(-7) \nk = 3.64 \n\epsilon_o = 8.85 * 10^(-12)

Plug in the values:
v = \sqrt{ (1)/((5.18)(4\pi*10^(-7))(3.64)(8.85*10^(-12))) }
v = 6.9 * 10^7 m/s

Now that we have "v", we can use the following formula to find the intensity of wave:

I = (k\epsilon_o*v*E_(max)^2)/(2)

I = ((3.64)(8.85*10^(-12))*(6.9*10^7)*(7.20*10^(-3))^2)/(2)

Intensity = I = 5.8 * 10^(-8) W/m^2

Of the four most important pathways by which stress affects health, the first one to occuris usually related to physiology

Answers

Answer: Physiologic response to fear is very similar to that of PTSD and stress. Fear is accompanied by increased heart rate due to the release of adrenaline, sympathetic nervous system is aroused. The release of adrenaline also causes increased sweating, pulse and blood pressure. In line with this, the parasympathetic nervous system experiences reduced activity such as decrease in digestion.

A substance that does NOT conduct an electric current when it forms a solution is a(n) ____. a electrolyte

b nonelectrolyte

c liquid

d solid

Answers

Answer:

B. Nonelectrolyte.

Explanation:

Nonelectrolytes do not dissociate into ions in solution, hence, nonelectrolyte solutions don't conduct electricity.

A non-electrolyte doesn’t conduct electric current even when it forms a solution.

Answer: Option B

Explanation:

Where electrolytes are defined as the compounds that can conduct electric current with mobile ions existing in its solution, non-electrolytes are the compounds that don’t behave the same either in the aqueous solution or in the molten state.  

This is all because these compounds don’t produce mobile ions to flow from one electrode to the other and hence conduct electric flow in the solution. Sugar and ethanol are the best examples of non-electrolytes that don’t induce electric current even after getting dissolved in water.

Q.1- Find the distance travelled by a particle moving in a straight line with uniform acceleration, in the 10th unit of time.

Answers

Answer:

If the acceleration is constant, the movements equations are:

a(t) = A.

for the velocity we can integrate over time:

v(t) = A*t + v0

where v0 is a constant of integration (the initial velocity), for the distance traveled between t = 0 units and t = 10 units, we can solve the integral:

\int\limits^(10)_0 {A*t + v0} \, dt = ((A/2)10^2 + v0*10) = (A*50 + v0*10)

Where to obtain the actual distance you can replace the constant acceleration A and the initial velocity v0.

A 75 kg man starts climbing a ladder that leans against a wall. If the weight of the ladder is negligible, determine how far up the ladder the man can climb before the ladders starts to slip. The coefficient of friction on both surfaces is μS=0.25

Answers

The man can climb \bold { X (max) = 0.25* L* tan \alpha }, before  the ladders starts to slip.  

   

A - point at the top of the ladder  

B - point at the bottom of the ladder  

C - point where the man is positioned in the ladder  

L- the length of the ladder  

α - angle between ladder and ground  

x - distance between C and B  

 

The forces act on the ladder,  

Horizontal reaction force (T) of the wall against the ladder  

Vertical (upward) reaction force (R) of ground against the ladder.  

Frictionalhorizontal ( to the left ) force (F)  

Vertical( downwards) of the man,

mg = 75 Kg x 9.8 m/s² = 735 N  

in static conditions,  

∑Fx = T - F = 0                   Since,  T = F  

∑Fy = mg - R = 0                Since,  735 - R = 0, R = 735  

∑ Torques(b) = 0  

In point B the torque produced by forces R and F is Zero  

Then:  

∑Torques(b) = 0        

And the arm lever for each force,  

mg = 735  

   

Since, ∑Torques(b) = 0    

 \bold {735* x* cos\alpha = F* L* sin\alpha }     Since,T = F  

 

\bold {F = \frac {735* x* cos\apha }{L* sin\alpha }}      \bold { \frac {cos \alpha } { sin\alpha }= cot\alpha =\frac 1{tan\alpha}}  

\bold {F = \frac {735* x* cos\apha }{L }}    

\bold {F = 735* x* tan\alpha }}  

F < μR the ladder will starts slipping over the ground  

μ(s) = 0.25    

 

\bold { X (max) = 0.25* L* tan \alpha }

Therefore, the man can climb \bold { X (max) = 0.25* L* tan \alpha }, before  the ladders starts to slip. \

To know more about  Torque,

brainly.com/question/6855614  

Answer:

x (max) = 0,25*L*tanα

Explanation:

Letá call  

A: point at the top of the ladder

B: the point at the bottom of the ladder

C: point where the man is up the ladder

L the length of the ladder

α angle between ladder and ground

"x" distance between C and B

Description

The following forces act on the ladder

Point A: horizontal (to the right)  reaction (T) of the wall against the     ladder

Point B : Vertical (upwards) reaction (R)  of ground against the ladder

               frictional horizontal ( to the left ) force (F)

Point C : Weight (vertical downwards)) of the man mg

mg = 75 Kg * 9,8 m/s²       mg = 735 [N]

Then in static conditions:

∑Fx = T - F = 0    ⇒   T = F

∑Fy = mg - R = 0       ⇒   735 - R = 0     ⇒  R = 735

∑Torques(b) = 0

Note: In point B the torque produced by forces R and F are equal to 0

Then:

∑Torques(b) = 0      

And the arm lever for each force is:

mg = 735

d₁ for mg     and d₂  for T

cos α = d₁/x     then    d₁ = x*cosα

sin α  = d₂ / L   then    d₂ = L*sinα

Then:

∑Torques(b) = 0     ⇒   735*x*cosα  - T*L*sinα = 0

735*x*cosα =  T*L*sinα

T = F then       735*x*cosα = F*L*sinα

F = (735)*x*cosα/L*sinα         cos α / sinα = cotgα = 1/tanα

F = (735)*x*cotanα/L     or   F = (735)*x/L*tanα

When F < μ* R  the ladder will stars slippering over the ground

μ(s) = 0,25           0,25*R = 735*x/L*tanα

x   = 0,25*R*tanα*L/735

But R = mg = 735 then

0,25*L*tanα = x

Then  x (max) = 0,25*L*tanα
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