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krek1111 [17]
3 years ago
6

Vector A is 3 units length and points along the positive x-axis vector B is 4 units in length and points along negative y-axis .

find magnitude and direction of the vector (a) A+B (b)A-B
Physics
1 answer:
iogann1982 [59]3 years ago
5 0

Answer:

a) < 3 , -4 >

b) < 3 , -4 >

Explanation:

a) If you can imagine this, adding vectors is like putting them "tip to tail", where you put the beginning point of vector B to the end point of vector A (or vice versa).  Your new vector (A+B) would be from the "tip" of vector A to the "tail" of vector B.

Mathematically, this is the same as adding the x-components of each vector together as well as the y-components.

Vector A: 3 units along the positive x-axis: < 3 , 0 >

Vector B: 4 units along the negative y-axis: < 0 , -4 >

A+B = < 3 , 0 > + < 0 , -4 > = < (3+0) , (0+(-4)) > = < 3 , -4 >

b) Subtracting is like adding a negative, so you could use the same "tip to tail" visual by adding the negative of vector B instead (which is B in the opposite direction).

Vector A: < 3 , 0 >

Vector B: < 0 , -4 >

Vector -B: < 0 , 4 >

A-B = A+(-B) = < 3 , 0 > + < 0 , 4 > = < (3+0) , (0+4) > = < 3 , 4 >

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3 years ago
A solid sphere of uniform density has a mass of 3.0 × 104 kg and a radius of 1.0 m. What is the magnitude of the gravitational f
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Answer:

a) Fg = 9.495x10⁻⁶N

b) Fg = 3.908x10⁻⁶N

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Explanation:

Given:

m₁ = mass = 3x10⁴kg

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Questions:

a) What is the magnitude of the gravitational force due to the sphere located at R = 1.4 m, Fg = ?

b) What is the magnitude of the gravitational force due to the sphere located at R= 0.21 m, Fg = ?

c) Write a general expression for the magnitude of the gravitational force on the particle at a distance r ≤ 1.0 m from the center of the sphere.

a) Since R > r, the equation for the gravitational force is:

F_{g} =\frac{Gm_{1}m_{2}  }{R^{2} }

Here,

G = gravitational constant = 6.67x10⁻¹¹m³/s² kg

Substituting values:

F_{g} =\frac{6.67x10^{-11}*3x10^{4}*9.3  }{1.4^{2} } =9.495x10^{-6} N

b) Since R < r, the equation for the gravitational force is:

F_{g} =\frac{Gm_{1}m_{2}R  }{r^{3} } =\frac{6.67x10^{-11}*3x10^{4}*9.3*0.21  }{1^{3} } =3.908x10^{-6} N

c) The general expression for the magnitude of the gravitational force on the particle at a distance r ≤ 1.0 is the same to b)

F_{g} =\frac{Gm_{1}m_{2}  R}{r^{3} }

8 0
3 years ago
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