PHYSICS COLLEGE

A pair of thin spherical shells with radius r and R, r < R are arranged to share a center. What is the capacitance of the system. If a potential difference V is created between the shells, how much energy is stored between them?

Answers

Answer 1

Answer:

Answer:

Capacitance = ( 4π×∈×r×R ) / (R-r)

energy store = ( 4π×∈×r×R )×V² / (R-r)

Explanation:

given data

radius = r

radius = R

r < R

to find out

capacitance and how much energy store

solution

we consider here r is inner radius and R is outer radius

so now apply capacitance C formula that is

C = Q/V .................1

here Q is charge and V is voltage

we know capacitance have equal and opposite charge so

V = $\int\limits^R_r {E} \, dx$

here E = Q / 4π∈k²

V = Q / 4π∈ $\int\limits^R_r {1/k^2} \, dx$

V = Q / 4π∈ × ( 1/r - 1/R )

V = Q(R-r) / ( 4π×∈×r×R )

so from equation 1

C = Q/V

Capacitance = ( 4π×∈×r×R ) / (R-r)

and

energy store is 1/2×C×V²

energy store = ( 4π×∈×r×R )×V² / (R-r)

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A 75-m-long train begins uniform acceleration from rest. The front of the train has a speed of 18 m/swhen it passes a railway worker who is standing 125 m from where the front of the train started. What will be the speed of the last car as it passes the worker?

Answers

Answer:22.76 m/s

Explanation:

Given

Train length(L)=75 m

Front of train after travelling 125 m is 18 m/s

Time taken by the front of train to cover 125 m

$v^2-u^2=2as$

$18^2-0=2* a* 125$

$a=1.296 m/s^2$

Speed of the last part of train when it passes the worker i.e. front of train has to travel has to travel a distance of 125+75=200 m

$v^2-u^2=2as$

$v^2=2* 1.296* 200$

$v=√(518.4)=22.76 m/s$

A uniform, 4.5 kg, square, solid wooden gate 2.0 m on each side hangs vertically from a frictionless pivot at the center of its upper edge. A 1.2 kg raven flying horizontally at 4.5 m/s flies into this door at its center and bounces back at 1.5 m/s in the opposite direction. What is the angular speed of the gate just after it is struck by the unfortunate raven?

Answers

Answer:

Explanation:

Mass of the gate, $m_1 = 4.5 kg$

Mass of the raven, $m_2 = 1.2 kg$

Initial speed of raven, $v_1 = 4.5 m/s$

Final speed of raven, $v_2 = - 1.5 m/s$

Moment of Inertia of the gate about the axis passing through one end:

$I = (1)/(3) m_1 a^2\nI = (1)/(3) *4.5 * 2^2\nI = 6 kg m^2$

Angular momentum of the gate, $L = I \omega$

$L = 5.33 \omega$

Using the law of conservation of angular momentum:

$m_2 v_f (a/2) + I\omega = m_2v_i (a/2)\nI\omega = m_2 (a/2)(v_i - v_f)\n$

A 12.0 kg mass, fastened to the end of an aluminum wire with an unstretched length of 0.50 m, is whirled in a vertical circle with a constant angular speed of 120 rev/min. The cross-sectional area of thewire is 0.014 cm2. Calculate the elongation of the wire when the mass is a) at its lowest point of the path and b) at the highest point of its path

Answers

Answer:

a) the elongation of the wire when the mass is at its lowest point on the path = 0.5 cm

b) the elongation of the wire when the mass is at its highest point on the path = 0.42 cm

Explanation:

Given that;

the angular speed $\omega = 120 \ rev/min$

Then converting it to rad/s ; we have:

= $(120 \ rev/min )((2 \ \pi \ rad )/(1 \ rev) ) ((1 \ min )/(60 \ s) )$

= 12.57 rad/s

The cross-sectional area of the wire A = 0.014 cm²

A = (0.014 cm²) ( $(10^(-4) \ m^2)/(1 \ cm^3)$ )

A = $0.014*10^(-4) \ m^2$

mass (m) = 12.0 kg

R = 0.5 m

g = 9.8 m/s²

To calculate for the mass when its at the lowest point of the path; we use the Newton's second law of motion; which is expressed as:

$T - mg = ma_(rad)$

where;

$a_rad = ( radical \ acceleration ) = R \omega^2$

Now; we can rewrite our equation as;

$T -mg = m R \omega ^2$

$T = mR \omega^2 + mg$

$T = m( R \omega^2 + g)$

Replacing our given values ; we have:

$T = 12.0( 0.5(12.57)^2 + 9.8)$

$T = 12.0( 0.5(158.0049) + 9.8)$

$T = 12.0( 79.00245 + 9.8)$

$T = 12.0( 88.80245)$

T = 1065.6294 N

T ≅ 1066 N

Determining the elongation $\delta l$ in the wire by using the equation

Y = $(Tl)/(AY)$

Making $\delta l$ the subject of the formula; we have

$\delta l = (Tl)/(AY)$

where ;

l = length of the wire

T =Tension in the wire

A = cross - sectional area

Y = young's modulus

Then;

$\delta l = ((1066 N)(0.5m))/((0.014*10^(-4)m^2)(7.0*10^(10)Pa))$

$\delta l$ = $( 0.00544 m ) *((10 ^2 cm)/(1m) )$

$\delta l$ = 0.5 cm

Thus, the elongation of the wire when the mass is at its lowest point on the path = 0.5 cm

Using Newton's second law of motion also for the mass at its highest point of the path; we have:

$T +mg = m R \omega ^2$

$T = mR \omega^2- mg$

$T = m( R \omega^2 - g)$

Replacing our given values ; we have:

$T = 12.0( 0.5(12.57)^2 - 9.8)$

$T = 12.0( 0.5(158.0049)-9.8)$

$T = 12.0( 79.00245 - 9.8)$

$T = 12.0( 69.20245)$

T = 830.4294 N

T = 830 N

Determining the elongation $\delta l$ in the wire by using the equation

Y = $(Tl)/(AY)$

Making $\delta l$ the subject of the formula; we have

$\delta l = (Tl)/(AY)$

where ;

l = length of the wire

T =Tension in the wire

A = cross - sectional area

Y = young's modulus

Then;

$\delta l = ((830 N)(0.5m))/((0.014*10^(-4)m^2)(7.0*10^(10)Pa))$

$\delta l$ = $( 0.00424 m ) *((10 ^2 cm)/(1m) )$

$\delta l$ = 0.42 cm

Thus, the elongation of the wire when the mass is at its highest point on the path = 0.42 cm

How does activity on the Sun affect natural phenomena on Earth?

Answers

Answer and Explanation:

The Sun is the main source of energy on the earth if there will be no availability of Sun energy then life is impossible om the earth besides this the Sun warms our planet. The heating of ocean and atmosphere is mainly sue to Sun energy .Sun has also a great impact on the weather we can say that Sun is weather deciding on the earth our climate is totally dependent on the how much energy we got in form of radiation from earth.

Students have four identical, hollow, uncharged conducting spheres, W, X, Y, and Z.Sphere Z is given a positive charge of +40 C. Sphere Z is touched first to sphere W, then sphere X, and finally to sphere Y. What is the resulting charge on sphere Y?

a. +5 με

b. +10 μC

c. +20 μC

d. +40 με

Answers

Explanation:

because they made contact that means their new force will be the same

Final answer:

Sphere Z is initially charged with +40 C. When it is touched to three other spheres, the charge is evenly distributed among them. The resulting charge on sphere Y is +10 μC.

Explanation:

The initial charge on sphere Z is +40 C. When sphere Z is touched to sphere W, the charge is evenly distributed between the two spheres, resulting in each sphere having a charge of +20 C. Then, when sphere Z is touched to sphere X, the total charge is evenly distributed between all three spheres, resulting in each sphere having a charge of +13.33 C. Finally, when sphere Z is touched to sphere Y, the total charge is evenly distributed between all four spheres, resulting in each sphere having a charge of +10 C. Therefore, the resulting charge on sphere Y is +10 μC (option b).

Learn more about Conducting spheres here:

brainly.com/question/12444946

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A 2.0 cm thick brass plate (k_r = 105 W/K-m) is sealed to a glass sheet (kg = 0.80 W/K m), and both have the same area. The exposed face of the brass plate is at 80°C, while the exposed face of the glass is at 20 °C. How thick is the glass if the glass brass interface is at 65 C? Ans. 0.46 mm

Answers

I need a good phisics teacher for one day online please123

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