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agasfer [191]
3 years ago
11

What are the coefficients in this expression?

Physics
2 answers:
skad [1K]3 years ago
7 0

6 and \frac{1}{3}

Explanation:

The expression:

     6a - \frac{3}{4}  + 7.5 - \frac{b}{3}

 A coefficient is a numerical constant usually placed before a variable in an expression.

They are used to multiply variables.

 Given equation is;

   6a - \frac{3}{4}  + 7.5 - \frac{b}{3}

There are two variables in this expression which are;

  a and  - b

  6 (a) - \frac{3}{4}  + 7.5 - b (\frac{1}{3} )

The coefficients in this expression are:

    6 and \frac{1}{3}

learn more:

Quadratic equation brainly.com/question/1357167

#learnwithBrainly

bekas [8.4K]3 years ago
5 0

Answer:

6 and Negative one-third    (D)

Explanation:

eDgE

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Two objects of same material are travelling near you. Object A is a 1.9 kg mass traveling 8 m/s; object B is a 2 kg mass traveli
Alex73 [517]

To determine the object which could give the greatest impact we will apply the concept of momentum. The object that has the highest momentum will be the object that will impact the strongest. Our values are

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Velocity of object A

v_A=8ms

Mass of object B

m_B=2 kg

Velocity of object B

v_B=5ms

The general formula for momentum is the product between mass and velocity, then

p = mv

For each object we have then,

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3 0
3 years ago
Charges of 4.0 μC and −6.0 μC are placed at two corners of an equilateral triangle with sides of 0.10 m. What is the magnitude o
jek_recluse [69]

Answer:

4.763 × 10⁶ N/C

Explanation:

Let E₁ be the electric field due to the 4.0 μC charge and E₂ be the electric field due to the -6.0 μC charge. At the third corner, E₁ points in the negative x direction and E₂ acts at an angle of 60 to the negative x - direction.

Resolving E₂ into horizontal and vertical components, we have

E₂cos60 as horizontal component and E₂sin60 as vertical component. E₁ has only horizontal component.

Summing the horizontal components we have

E₃ = -E₁ + (-E₂cos60) = -kq₁/r²- kq₂cos60/r²

= -k/r²(q₁ + q₂cos60)

= -k/r²(4 μC + (-6.0 μC)(1/2))

= -k/r²(4 μC - 3.0 μC)

= -k/r²(1 μC)

= -9 × 10⁹ Nm²/C²(1.0 × 10⁻⁶)/(0.10 m)²

=  -9 × 10⁵ N/C

Summing the vertical components, we have

E₄ = 0 + (-E₂sin60)

= -E₂sin60

= -kq₂sin60/r²

= -k(-6.0 μC)(0.8660)/(0.10 m)²

= -9 × 10⁹ Nm²/C²(-6.0 × 10⁻⁶)(0.8660)/(0.10 m)²

= 46.77 × 10⁵ N/C

The magnitude of the resultant electric field, E is thus

E = √(E₃² + E₄²) = √[(-9 × 10⁵ N/C)² + (46.77 10⁵ N/C)²) = (√226843.29) × 10⁴

= 476.28  × 10⁴ N/C

= 4.7628 × 10⁶ N/C

≅ 4.763 × 10⁶ N/C

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Answer:

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