A charge Q = 1.96 10-8 C is surrounded by an equipotential surface with a surface area of 1.18 m2. what is the electric potential at this surface?

Answers

Answer 1
Answer:

Answer:

V = 575.6 Volts

Explanation:

As we know that surface area of the equi-potential surface is given as

A = 1.18 m^2

so we will say

A = 4\pi r^2

1.18 = 4\pi r^2

r = 0.31 m

Now the potential due to a point charge is given as

V = (kQ)/(r)

V = ((9* 10^9)(1.96 * 10^(-8)))/(0.31)

V = 575.6 Volts


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What do you think most likely cause the differences seen in the two graphs for cold spring and normal wearher

Answers

Temperature variations in a graph occur as a result of changing environmental conditions and changing temperature.  

what is temperature ?

Temperature is a physical quantity which measures hotness and coldness of a body. Temperature measures the degree of vibration of molecule in a body. Temperature is measured in centigrade (°C), Fahrenheit (°F) and Kelvin (K) in which Kelvin (K) is a SI unit of temperature. Absolute scale of temperature means Kelvin scale of temperature. relation between Kelvin(K) and centigrade (°C), °C= K - 273.15 from equation, 273.15 K means 0 °C, which is freezing point of water (ice). when we give temperature to the body, its molecule or atom absorbs thermal energy and vibrate about their mean position. Amplitude of vibration get increases as we go on increasing temperature and for higher temperature force of attraction between molecules gets weaker. Hence for higher temperature, due to weaken the force of attraction between molecule, solid goes into liquid state. and further increase in temperature liquid goes into gaseous state.

To know more about Temperature :

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#SPJ2.

Answer:

The tempature changes, and the envronment chnages because of this, therefore making tempature changes in a graph.

Explanation:

sorry if this isnt good

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

The graph to the right shows the change in Canada‘s harvest of Atlantic cod from 1950-2004 what year shows the clearest evidence of a collapse of fishing stocks?A.1965
B.1985
C.1995
D.2005

Answers

The correct answer is C. 1995

Explanation:

The graph shows the changes in the harvest of Atlantic cod. In general, this graph illustrates how the peak occurred in the 1980s but then there was a sudden and sharp decline in 1995. Indeed, 1995 is the year with the lowest number of harvested cod as in this year there were approximately least than 10 thousand metric tonnes of cods. Also, this year shows the collapse of fishing stocks or that the population of this fish collapsed, which made it impossible to harvest as many fish as in previous years. According to this, the year that shows the collapse of fishing stocks is 1995.

Which are electromagnetic waves? check all that apply.earthquake waves
infrared waves
ocean waves
radio waves
untraviolet waves

Answers

Since electromagnetic waves do not require a medium for their transmission, the electromagnetic waves are radio waves, ultraviolet waves and infrared waves.

What are electromagnetic waves?

Electromagnetic waves or radiations are waves which occur as a result of the interaction between the electric and magnetic fields.

Electromagnetic waves do not require a material medium for their transmission and as such can travel through a vacuum.

Some examples of electromagnetic waves are radio waves, ultraviolet waves, microwaves, infrared waves etc.

Therefore, the electromagnetic waves are radio waves, ultraviolet waves and infrared waves.

Learn more about electromagnetic waves at: brainly.com/question/25847009

The electromagnetic waves are:
Radio waves
Ultraviolet waves
And Infrared waves
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The equations for single-slit and multiple-slit interference both contain the variable θ. For the multiple-slit case, the angle is: a. the angular location of the first order minimum in the diffraction pattern. Which means at this point the light experiences constructive interference.
b. the angular location of the first order minimum in the diffraction pattern. Which means at this point the light experiences destructive interference.
c. the angular location of bright interference maxima in the pattern. Which means at this point the light experiences constructive interference.
d. the angular location of bright interference maxima in the pattern. Which means at this point the light experiences destructive interference.

Answers

Answer:

the answers the correct one is c

Explanation:

The diffraction pattern for a slit is

         a sin θ = m λ

Where a is the width of the slit, λ the wavelength, m the order of destructive interference and θ the angle where the interference occurs.

The expression for multi-slit diffraction (diffraction grating) is

          d sin θ = m λ

Where d is the distance between slits, λ the wavelength m the order of the diffraction maximums and θ the angle for these maximums.

When we compare the expressions of the answers the correct one is c

1. The workpart in a turning operation is 88 min in diameter and 400 mm long. A feed of 0.25 mm/rev is used in this operation. If cutting speed is 3.5 m/s, the too should be changed in every 3 workparts, but if the cutting speed is 2.5 m/sec, the tool can be used to produce 20 pieces between the tool changes. Determine the cutting speed that will allow the tool to be used for 50 parts between tool changes.

Answers

Find the given attachments