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Papessa [141]
4 years ago
8

A source charge generates an electric field of 1236 N/C at a distance of 4 m. What is the magnitude of the source charge?

Physics
2 answers:
SashulF [63]4 years ago
8 0

Answer:

Explanation:

A region around a charged particle where another charged particle experiences a force is called electric field.

Electric field strength is defined as the force experienced by unit positive test cahrge when placed in an electric field.

The formula of electric field due to a point cahrge is given by

E = k q / r^2

Where k is a constant and it's value is

9 × 10^9 Nm^2/C^2.

1236 = 9 × 10^9 × q / 16

q = 1.48 × 10^-3 Coulomb

lara [203]4 years ago
6 0
The magnitude of the source is 2.2 C
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A 4.45 kg block of ice at 0.00 ?c falls into the ocean and melts. the average temperature of the ocean is 3.70 ?c, including all
rodikova [14]
The ocean does not change temperature but it does lose some entropy ( he gives heat to melt the ice and to warm it to 3.70° C ).
I ) For the ice:
1 ) For melting the ice:
Q = m · Lf = 4.45 kg · 334 · 10³ J/kg = 1,486,300 J
Δ S = Q / T = 1,486,300 J / 273.15 K = 5.441 · 10³ J/K.
2 ) To warm the melted ice to 3.70° C:
Q = m c Δ T = 4.45 kg · 4,190 J / kgK · 3.70 K = 68,988.35 J
Δ S = m c ln( T2/ T1 ) = 4.45 kg · 4,190 J/kgK · ln ( 276.85 / 273.15 ) =
= 18,645.5 · ln ( 1.01354 ) = 18.645.5 · 0.013454 = 250.8692 J/K
II ) For the ocean:
Δ S = Q / T = ( - 68,988.35  + 1,486,300 ) / 276.85 = - 5,617.8 J/K
The net entropy change:
Δ S = ΔS ice + ΔS ocean = 5,441.1 + 250.8692 - 5,617.8 = 74.1692 J/K
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4 0
3 years ago
Which component of the galaxy is shown in this image? Dust
Westkost [7]

Answer:

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3 0
3 years ago
If you weigh 665 NN on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun a
77julia77 [94]

Answer:

The weight on the neutron star is 99.69 × 10^{12} N.

Explanation:

Here,

weight on earth is 665 N. We can calculate the mass as :

                                                 W_{earth}= mg

                                           or, 665= m × 9.810

                                              ∴ m = \frac{665}{9.810} = 67.68 kg

Now weight of this mass on the surface of neutron star:

                            W_{neutron star} = \frac{GMm}{R^{2} }

                                                = \frac{(6.674*10^{-11})(1.99*10^{30})(67.68) }{9500^{2} }

                                                = 99.69 × 10^{12} N

The required weight is 99.69 × 10^{12} N.

7 0
3 years ago
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