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bezimeni [28]
3 years ago
10

An unknown charge sits on a conducting solid sphere of radius 9.0 cm. If the electric field 14 cm from the center of the sphere

has magnitude 2.7 103 N/C and is directed radially inward, what is the net charge on the sphere?
Physics
2 answers:
romanna [79]3 years ago
7 0

Answer: The net charge on the sphere is -5.9nC

Explanation:

The electric field outside the conducting solid sphere is given as:

E = (kq)/r^2

Here, q is the net charge on the sphere and r is the distance from the center of the sphere.

The net charge is calculated as follows;

q = Er^2/k

E = 2.7*10^3N/C

r = 14cm = 0.14m

k = 8.99*10^9Nm^2/C^2

q = (2.7*10^3)(0.14^2)/(8.99*10^9)

q = 5.9*10^-9C

= 5.9nC

As the electric field directed radially inward, the net charge on the sphere is negative.

Hence the net charge on the sphere is -5.9nC

spin [16.1K]3 years ago
7 0

Answer:

The net charge on the sphere is - 5.9 × 10⁻⁹ C = - 5.9 nC

Explanation:

The electric field (E) produced by a charge of magnitude Q, at a point with distance r away from the charge, is given by

E = kQ/r²

where k = Coulomb's constant = 9.0 x 10⁹ Nm²/C²

The electric field 14 cm (0.14 m) from the sphere is 2.7 × 10³ N/C

2.7 × 10³ = 9.0 x 10⁹ × Q/(0.14²)

Q = (2.7 × 10³ × 0.14²)/(9.0 x 10⁹) = 5.9 × 10⁻⁹ C

Since the charge is directed inwards towards the centre of the sphere, the sign on it will be -ve. Q = - 5.9 × 10⁻⁹ C

b) The electric field 9 cm (0.09 m) from the sphere will be

E = kQ/(0.09²) = (9 × 10⁹ × 5.9 × 10⁻⁹)/(0.09²) = 6555.56 = 6.56 × 10³ N/C

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A coffee cup calorimeter is prepared, containing 100.000 g of water (specific heat capacity = 4.184 J/g K) at initial temperatur
BARSIC [14]

Answer : The molar heat of solution for the salt is 452.9 kJ/mole

Explanation :

First we have to calculate the heat of solution.

q=m\times c\times \Delta T

where,

q = heat of solution = ?

c = specific heat of water = specific heat of solution = 4.184J/g.K=4.184J/g^oC

m = mass of solution = 107.093 g

  • Mass of solution = Mass of water + Mass of salt
  • Mass of solution = 100.000 g + 7.093 g = 107.093 g

\Delta T = change in temperature = T_2-T_1=(80.000-61.128)=18.872^oC=

Now put all the given values in the above formula, we get:

q=107.093g\times 4.184J/g^oC\times 18.872^oC

q=8456.111J=8.456kJ

Now we have to calculate the molar heat of solution for the salt, in kJ/mol.

\Delta H=\frac{q}{n}

where,

\Delta H = molar heat of solution = ?

q = heat required = 8.456 kJ

m = mass of salt = 7.093 g

Molar mass of salt = 379.984 g/mol

\text{Moles of salt}=\frac{\text{Mass of salt}}{\text{Molar mass of salt}}=\frac{7.093g}{379.984g/mole}=0.01867mole

\Delta H=\frac{8.456kJ}{0.01867mole}=452.9kJ/mole

Therefore, the molar heat of solution for the salt is 452.9 kJ/mole

4 0
3 years ago
Draw a free-body diagram of the rod ab. Assume the contact surface at b is smooth.
Romashka [77]

Answer:

See attachment

Explanation:

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3 years ago
In a mass spectrometer used for commercial purposes, uranium ions of mass 3.76 X 10^(-25) kg and charge 3.5 X 10^(-19) C are sep
Flauer [41]

Answer:

a. 0.394 T b. 0.255 A c. 1.309 × 10⁸ J

Explanation:

Here is the complete question

A certain commercial mass spectrometer (Fig. 28-12) is used to separate uranium ions of mass 3.92 x 10-25 kg and charge 3.20 x 10-19 C from related species. The ions are accelerated through a potential difference of 109 kV and then pass into a uniform magnetic field, where they are bent in a path of radius 1.31 m. After traveling through 180° and passing through a slit of width 0.752 mm and height 0.991 cm, they are collected in a cup. (a) What is the magnitude of the (perpendicular) magnetic field in the separator? If the machine is used to separate out 1.12 mg of material per hour, calculate (b) the current (in A) of the desired ions in the machine and (c) the thermal energy (in J) produced in the cup in 1.31 h.

Solution

a. The magnitude of the (perpendicular) magnetic field in the separator

The kinetic energy of the uranium ions = electric potential energy

¹/₂mv² = qV

v = √(2qV/m) where v = speed of uranium ions, q = uranium ion charge = 3.2 × 10⁻¹⁹ C , m = mass of uranium ions = 3.92 × 10⁻²⁵ kg and V = 109 kV = 1.09 × 10⁵ V

v = √(2qV/m) = √(2 × 3.2 × 10⁻¹⁹ C × 1.09 × 10⁵ V/3.92 × 10⁻²⁵ kg)

v = 4.22 × 10⁵ m/s

Also, the magnetic force on the uranium ions equals the centripetal force when it passes through the magnetic field.

Bqv = mv²/r

B = mv/rq   where B = magnetic field strength and r = radius of circle = 1.31 m

B = m(√(2qV/m))/rq

B = √(2mV/q)/r

B = √(2 × 3.92 × 10⁻²⁵ kg × 1.09 × 10⁵ V/3.2 × 10⁻¹⁹ C)/1.31 m

B = √0.26705/1.31

B = 0.394 T

(b) the current (in A) of the desired ions in the machine

Since a mass m of 3.92 × 10⁻²⁵ kg of uranium ions carries a charge q of 3.2 × 10⁻¹⁹ C, then 1.12 mg per hour = 1.12 × 10⁻³ kg/h. In 1.31 h, our mass is M = 1.12 × 10⁻³ kg/h × 1.31 h = 1.47 × 10⁻³ kg carries a charge of Q of

m/q = M/Q

Q = Mq/m

Q = 1.47 × 10⁻³ kg × 3.2 × 10⁻¹⁹ C/3.92 × 10⁻²⁵ kg

Q = 1200 C

The current i = Q/t where t = time = 1.31 h = 1.31 × 60 × 60 s = 4716 s

i = 1200/4716

i = 0.2545 A

i ≅ 0.255 A

(c) the thermal energy (in J) produced in the cup in 1.31 h.

The thermal energy produced in the cup equals the kinetic energy lost by the uranium ions hitting the cup in 1.31 h.

E = ¹/₂Mv² = ¹/₂ × 1.47 × 10⁻³ kg × (4.22 × 10⁵ m/s)²

E = 1.309 × 10⁸ J

3 0
4 years ago
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uysha [10]

Answer:

a c d

Explanation:

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5 0
4 years ago
Current is directly proportional to resistance.<br> a. True<br> b. False
ikadub [295]

IF voltage remains constant, then current is
inversely proportional to resistance.

The correct response is "b).", signifying "false" as the choice.

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