One of two nonconducting spherical shells of radius a carries a charge Q uniformly distributed over its surface, the other carries a charge -Q, also uniformly distributed. The spheres are brought together until they touch.A.) What does the electric field look like, both outside and inside the shells?

B.) How much work is needed to move them far apart?

Answers

Answer 1

Answer:

Answer:

Explanation:

A ) The spheres are non conducting , charge will not move on the surface so neutralization of charge by + ve and - ve charge is not possible. Charges will remain intact on them . The electric field inside them will be zero . Electric field outside shell will not be spherically symmetrical . Lines of force will emanate from the surface of positively charged shell outwardly oriented and end at negatively charged shell .

B )

distance between the centres of spherical shell

= 2 a

potential energy of charges

= k q₁ x q₂ / R

= k x - Q x Q / ( 2a )

= - k Q²/ 2a

So work needed to separate them to infinity will be equal to

= k Q²/ 2a

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Answers

Answer:

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Explanation:

See the picture attached

A weather balloon is designed to expand to a maximum radius of 16.2 m when in flight at its working altitude, where the air pressure is 0.0282 atm and the temperature is â65âC. If the balloon is filled at 0.873 atm and 21âC, what is its radius at lift-off?

Answers

Answer:

5.78971 m

Explanation:

$P_1$ = Initial pressure = 0.873 atm

$P_2$ = Final pressure = 0.0282 atm

$V_1$ = Initial volume

$V_2$ = Final volume

$r_1$ = Initial radius = 16.2 m

$r_2$ = Final radius

Volume is given by

$(4)/(3)\pi r^3$

From the ideal gas law we have the relation

$(P_1V_1)/(T_1)=(P_2V_2)/(T_2)\n\Rightarrow (0.873* (4)/(3)\pi r_1^3)/(294.15)=(0.0282(4)/(3)\pi r_2^3)/(208.15)\n\Rightarrow (0.873r_1^3)/(294.15)=(0.0282* 16.2^3)/(208.15)\n\Rightarrow r_1=(0.0282* 16.2^3* 294.15)/(208.15* 0.873)\n\Rightarrow r_1=5.78971\ m$

The radius of balloon at lift off is 5.78971 m

Final answer:

To find the radius of the weather balloon at lift-off, the ideal gas law can be used. Using the equation P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the pressure and volume at lift-off, the radius at lift-off can be calculated to be approximately 4.99 m.

Explanation:

To find the radius of the weather balloon at lift-off, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.

In this case, we know that the number of moles is constant, as the balloon is filled with the same amount of helium at lift-off and in flight. Therefore, we can write the equation as P1V1 = P2V2, where P1 and V1 are the initial pressure and volume, and P2 and V2 are the pressure and volume at lift-off.

Plugging in the given values, we have (0.873 atm)(V1) = (0.0282 atm)(16.2 m)^3. Solving for V1, we find that the volume at lift-off is approximately 110.9 m^3. The radius can then be calculated using the formula for the volume of a sphere: V = (4/3) * π * r^3, where r is the radius.

Therefore, the radius at lift-off is approximately 4.99 m.

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Use a(t) =−32 feet per second squared as the acceleration due to gravity. a ball is thrown vertically upward from the ground with an initial velocity of 56 feet per second. for how many seconds will the ball be going upward?

Answers

Since the ball is moving by uniformly accelerated motion, its vertical velocity at time t is given by
$v(t)= v_0 - a t$
where we took upward as positive direction, and where $v_0$ is the initial velocity, a the acceleration and t the time.

The instant at which $v(t)=0$ is the instant when the ball reverses its velocity (from upward to downward). This means that the difference between the time t at which v(t)=0 and the instant t=0 is the total time during which the ball was going upward:
$0=v_0 - at$
By plugging numbers into the equation, we find
$t= (v_0)/(a)= (56 ft/s)/(32 ft/s^2)=1.75 s$

A large aquarium has portholes of thin transparent plastic with a radius of curvature of 1.95 m and their convex sides facing into the water. A shark hovers in front of a porthole, sizing up the dinner prospects outside the tank.a) If one of the sharks teeth is exactly 46.5 cm from the plastic, how far from the plastic does it appear to be to observers outside the tank? (You can ignore refraction due to the plastic.)b) Does the shark appear to be right side up or upside down?c) If the tooth has an actual length of 5.00 cm, how long does it appear to the observers?

Answers

Answer:

Explanation:

For refraction through a curved surface , the formula is as follows

μ₂ / v - μ₁ / u = (μ₂ -μ₁ )/R , Here μ₂( air) = 1 , μ₁ ( water) = 4/3 , R = 1.95 m

u , object distance = - .465 m

1 / v + 1.333 / .465 = (1 -1.333 )/1.95

1 / v + 2.8667 = - .171

1 / v = - 2.8667 - .171 = - 3.0377

v = - .3292 m

= - 32.92 cm

image will be formed in water.

c ) magnification = μ₁v / μ₂u , μ₁ = 1.33 , μ₂ = 1 , u = 46.5 , v = 32.92 .

= (1.33 x 32.92) / (1 x 46.5)

= .94

size of image of teeth = .94 x 5

= 4.7 cm .

Net force is the sum of all the forces acting on an object. If a spring balance pulls on a body with a force of 10 N, and friction acts on the body in the opposite direction with a force of 1 N, the net force would be 9 N in the direction of the spring balance (10 N – 1 N = 9 N).What is the net force acting on the object when the spring balance pulls the rope with a force of 25 N and friction acts on the body with a force of 20N?

Answers

Answer:

Explanation:

(25 N - 20 N = 5 N)

Let's say you have a plot for Pendulum experiment. Let's assume g for this experiment was measured (from the slope of the plot) to be 9.78 [ms^-2]. The vertical intercept, however, is 0.021 [s^2]. What might this translate to for a measurement of the length offset systematic in all the length measurements? Length Offset =[mm] .011