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Xelga [282]
2 years ago
9

A small charged ball lies within the hollow of a metallic spherical shell of radius R. For three situations, the net charges on

the ball and shell, respectively, are (1) +4q, 0; (2) –6q, +10q; (3) +16q, –12q. Rank the situations according to their charge on (a) the inner surface of the shell and (b) the outer surface, most positive first
Physics
1 answer:
Katena32 [7]2 years ago
3 0

Answer:

Part a)

2)  >  1)  >  3)

Part b)

1)  =  2)  =  3)

Explanation:

Due to charge induction the magnitude of charge on the inner surface of the outer shell is having same charge as that of the small sphere inside but the sign of charge must be opposite.

So here we can say

1)+ 4q, 0

so inner surface has charge - 4q and outer surface charge is +4q

2) -6q , +10q

so inner surface charge is +6q, outer surface charge is +4q

3) +16q , -12q

so inner surface charge is -16q, outer surface charge is +4q

Part a)

situations in which inner surface charge is arranged in decreasing order is given as

2)  >  1)  >  3)

Part b)

Situations in which outer surface charge is arranged in decreasing order is given as

1)  =  2)  =  3)

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lesantik [10]

Answer: The light bulb produces the continuous light. At minimum wavelength the spectrum have maximum intensity.

Explanation:

According to Wein's displacement law, the wavelength is inversely proportional to the temperature.

The intensity depends on the frequency. The frequency is inversely proportional to the wavelength.

Therefore, when the temperature of the light bulb will be maximum then the wavelength will be minimum. At minimum wavelength the spectrum have maximum intensity.

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2 years ago
What resistance must be connected in parallel with a 633-Ω resistor to produce an equivalent resistance of 205 Ω?
alukav5142 [94]

Answer:

303 Ω

Explanation:

Given

Represent the resistors with R1, R2 and RT

R1 = 633

RT = 205

Required

Determine R2

Since it's a parallel connection, it can be solved using.

1/Rt = 1/R1 + 1/R2

Substitute values for R1 and RT

1/205 = 1/633 + 1/R2

Collect Like Terms

1/R2 = 1/205 - 1/633

Take LCM

1/R2 = (633 - 205)/(205 * 633)

1/R2 = 428/129765

Take reciprocal of both sides

R2 = 129765/428

R2 = 303 --- approximated

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2 years ago
For a group class project, students are building model roller coasters. Each roller coaster needs to begin at the top of the fir
abruzzese [7]

Case A :

A .75 kg 65 N/m 1.2 m

m = mass of car = 0.75 kg

k = spring constant of the spring = 65 N/m

h = height of the hill = 1.2 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (65) (0.25)² + (0.75 x 9.8 x 1.2) = (0.5) (0.75) v²

v = 5.4 m/s



Case B :

B .60 kg 35 N/m .9 m

m = mass of car = 0.60 kg

k = spring constant of the spring = 35 N/m

h = height of the hill = 0.9 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (35) (0.25)² + (0.60 x 9.8 x 0.9) = (0.5) (0.60) v²

v = 4.6 m/s




Case C :

C .55 kg 40 N/m 1.1 m

m = mass of car = 0.55 kg

k = spring constant of the spring = 40 N/m

h = height of the hill = 1.1 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (40) (0.25)² + (0.55 x 9.8 x 1.1) = (0.5) (0.55) v²

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Case D :

D .84 kg 32 N/m .95 m

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k = spring constant of the spring = 32 N/m

h = height of the hill = 0.95 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (32) (0.25)² + (0.84 x 9.8 x 0.95) = (0.5) (0.84) v²

v = 4.6 m/s


hence closest is in case C at 5.1 m/s




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2 years ago
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Alborosie

Answer:

<em>0.45 mm</em>

Explanation:

The complete question is

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B) 0.63 mm

C.) 0.68 mm

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Area of the wire in the fuse = I/ρ

Where I is the current through the fuse

ρ is the current density of the fuse

Area = 1/620 = 1.613 x 10^-3 cm^2

We know that 10000 cm^2 = 1 m^2, therefore,

1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2

Recall that this area of this wire is gotten as

A = \frac{\pi d^{2} }{4}

where d is the diameter of the wire

1.613 x 10^-7 = \frac{3.142* d^{2} }{4}

6.448 x 10^-7 = 3.142 x d^{2}

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