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3 years ago

Which vector has an x-component with a length of 3? А. С. B. d C. a D. b

Physics

2 answers:

zhannawk [14.2K]3 years ago

6 0

Answer:

D. b

Explanation:

the endpoint of vector b is at 3, which means the length is 3

worty [1.4K]3 years ago

4 0

Answer:

B.d

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2 years ago

A concave mirror has a focal length of 30.0 CM. an object is placed 15.0 CM from the mirror. what is the radius of curvature of

Nana76 [90]

Answer:

60 cm

Explanation:

We are given;

Focal length of a concave mirror as 30.0 cm
Object distance is 15.0 cm

We are required to determine the radius of curvature.

We need to know that the radius of a curvature is the radius of a circle from which the curved mirror is part.

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7 0

3 years ago

A particle of charge 2.0 x 10^-8C experiences an upward force of magnitude 4.0 x10^-6 when it is placed in a particular point in

koban [17]

Answer:

a) The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) The electric force is $2.0\times 10^{-6}$ newtons.

Explanation:

a) Let suppose that electric field is uniform, then the following electric field can be applied:

$E = \frac{F_{e}}{q}$ (1)

Where:

$E$ - Electric field, measured in newtons per coulomb.

$F_{e}$ - Electric force, measured in newtons.

$q$ - Electric charge, measured in coulombs.

If we know that $F_{e} = 4.0\times 10^{-6}\,N$ and $q = 2.0\times 10^{-8}\,C$ , then the electric field at that point is:

$E = \frac{4.0\times 10^{-6}\,N}{2.0\times 10^{-8}\,C}$

$E = 2.0\times 10^{2}\,\frac{N}{C}$

The electric field at that point is $2.0\times 10^{2}$ newtons per coulomb.

b) If we know that $E = 2.0\times 10^{2}\,\frac{N}{C}$ and $q = 1.0\times 10^{-8}\,C$ , then the electric force is:

$F_{e} = E\cdot q$

$F_{e} = \left(2.0\times 10^{2}\,\frac{N}{C} \right)\cdot (1.0\times 10^{-8}\,C)$

$F_{e} = 2.0\times 10^{-6}\,N$

The electric force is $2.0\times 10^{-6}$ newtons.

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3 years ago

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miss Akunina [59]

Explanation:

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8 0

3 years ago