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lubasha [3.4K]
3 years ago
15

Which one of the following statements is true concerning the magnitude of the electric field at a point in space? It is a measur

e of the electric force on any charged object. It is a measure of the ratio of the charge on an object to its mass. It is a measure of the electric force per unit mass on a test charge. It is a measure of the electric force per unit charge on a test charge. It is a measure of the total charge on the object.
Physics
1 answer:
IceJOKER [234]3 years ago
5 0

Answer:

It is a measure of the electric force per unit charge on a test charge.

Explanation:

The magnitude of the electric field is defined as the force per charge on the test charge.

Since we define electric field as the force per charge, it will have the units of  force divided by the unit of charge. This implies that the SI unit of electric field is given as Newton/Coulomb (N/C).

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Explanation:

It is given that,

Mass of cardinal,m_c=4.2\times 10^{-2}\ kg

Mass of baseball, m_b=0.146\ kg

Both cardinal and baseball have same kinetic energy. We need to find the ratio of the cardinal's magnitude p_c of momentum to the magnitude p_c of the baseball's momentum.

E_c=E_b

Kinetic energy is given by, E=\dfrac{p^2}{2m}

\dfrac{p_c^2}{2m_c}=\dfrac{p_b^2}{2m_b}

(\dfrac{p_c}{p_b})^2=\dfrac{m_c}{m_b}

\dfrac{p_c}{p_b}=\sqrt{\dfrac{m_c}{m_b}}

\dfrac{p_c}{p_b}=\sqrt{\dfrac{4.2\times 10^{-2}}{0.146}}

So, \dfrac{p_c}{p_b}=\dfrac{53}{100}

So, the ratio of cardinal's magnitude of momentum to the magnitude of the baseball's momentum is 53 : 100

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Answer:

3.53 V

Explanation:

Electric charge: The is the rate of flow of electric charge along a conductor.

The S.I unit of electric charge is C.

Mathematically it is expressed as,

Q = It ............................ Equation 1

Where Q = electric charge, I = current, t = time.

I = Q/t.......................... Equation 2

From the question, charge flows through the conductor at the rate of 420 C/mim

Which means in 1 min, 420 C of charge flows through the conductor.

Hence,

Q = 420 C, t = 1 min = 60 seconds

Substitute into equation 2

I = 420/60

I =7 A

Also

P = VI......................... Equation 3

Where P = power, V = potential drop, I = current.

V = P/I................... Equation 4

Note: Power = Energy/time

From the question, P = 742/30 = 24.733 W. and I = 7 A.

Substitute these values into equation 4

V = 24.733/7

V = 3.53 V

Hence the potential drop across the conductor =  3.53 V

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