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Ludmilka [50]
3 years ago
7

PLEASE HELP IF U PASS TESTS

Physics
2 answers:
Alexxandr [17]3 years ago
6 0
Iron. Is the correct answer
bezimeni [28]3 years ago
5 0
I think iron? i’m not 100% sure
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Mass in physics is in kg and mass in chemistry is in grams. Therefore a person's mass is kg
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We know that an electric field contains energy. What causes the electric field itself?
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Electrical charges on one or more particles within the field cause the electric field

Each point in space has an electric field associated with it when a charge of any kind is present. The value of E, often known as the electric field strength, electric field intensity, or just the electric field, expresses the strength and direction of the electric field

A region of space surrounding an electrically charged particle or object known as an electric field is one in which an electric charge would experience force. A vector quantity called an electric field can be represented by arrows pointing in the direction of or away from charges. The force per unit charge exerted on a positive test charge that is at rest at a given position is the force per unit charge that is used to define the electric field analytically.

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1 year ago
A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated
Lelu [443]

Incomplete question as we have not told to find what quantity.The complete question is here

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated by vacuum and the inner radius of the outer shell is 5.00 cm.calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere.

Answer:

(a) C=16.7pF

(b) r_{a} =3.749cm

(c) E=2.24*10^{4} N/C

Explanation:

Given data

Q=3.50nC\\V=210V\\r_{b}=5.0cm

For part (a)

The Capacitance given by:

C=\frac{Q}{V}\\ C=\frac{3.50*10^{-9} C}{210V}\\C=1.6666*10^{-11}F\\or\\C=16.7pF

For part (b)

The Capacitance of coordinates is given as

C=\frac{4\pi e}{\frac{1}{r_{a} }-\frac{1}{r_{b} } }\\ So\\{\frac{1}{r_{a} }-\frac{1}{r_{b} } }=\frac{4\pi *8.85*10^{-12} }{1.666*10^{-11}}=6.672m^{-1} \\ \frac{1}{r_{a} }=6.672+(1 /0.05)\\\frac{1}{r_{a} }=26.672\\r_{a} =1/26.672\\r_{a} =0.0375m\\r_{a} =3.749cm

For part (c)

The electric field according to Gauss Law is given by:

EA=\frac{Q}{e}\\ E=\frac{Q}{4\pi er_{a}^{2}  }=\frac{kQ}{r_{a}^{2}}\\  E=\frac{9*10^{9}*3.50*10^{-9} }{(0.0375m)^{2} }\\ E=2.24*10^{4} N/C

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How is my Engineering Project?
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