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Zanzabum
3 years ago
5

An object experiences an acceleration of -6.8 m/s​2.​ As a result, it accelerates from 54 m/s to a complete stop. How much dista

nce did it travel during that acceleration?
Physics
1 answer:
inessss [21]3 years ago
3 0

Answer:

The distance traveled during its acceleration, d = 214.38 m

Explanation:

Given,

The object's acceleration, a = -6.8 m/s²

The initial speed of the object, u = 54 m/s

The final speed of the object, v = 0

The acceleration of the object is given by the formula,

                                      a = (v - u) / t   m/s²

       ∴                              t = (v - u) / a

                                         = (0 - 54) / (-6.8)

                                         = 7.94 s

The average velocity of the object,

                                       V = (54 + 0)/2

                                           = 27 m/s

The displacement of the object,

                                 d = V x t   meter

                                    = 27 x 7.94

                                    = 214.38 m

Hence, the distance the object traveled during that acceleration is, a = 214.38 m

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If the kinetic energy of a given mass is to be doubled, its speed must be multiplied by
Tomtit [17]

Answer:

Either B or D. The answer itself is 2.

Explanation:

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When m is doubled, we can plug in 1 and 2 to compare our answers.

Plugging in 1 for mass would give us the answer 1/2*v^2.

Plugging in 2 for mass would give us v^2. This means that the velocity was multiplied by 2, meaning that the answer is it is multiplied by 2.

I am not sure which answer is correct since there seems to be two answer choices with 2 in it, but the answer is either B or D (I will call it ABCD because I do not want to cause confusion by saying 2 multiple times).

3 0
3 years ago
Three charges are located at a different position in a plane: q1= 10μC at →r1=(5,6)cm q2=−27μC at →r2=(−6,10)cm and q3=−12μC at
sasho [114]

Answer:

 E = (2.29 i ^ - 0.917 j ^) 10⁶ N / C

 E_{total} = 2,467 10⁶ N / A       θ = -21.8      

Explanation:

For this exercise we will use that the electric field is a vector quantity, so the total field is

        E_total = E₁₃ + E₂₃

bold font vectors .  We can work with the components of the electric field in each axis

X- axis

       E_ total x = E₁₃ₓ + E_{23x}

y-axis  

      E_{total y} = E_{13y} + E_{23y}

the expression for the electric field is

       E = k q / r²

where r is the distance between the charge and the positive test charge

       

in this exercise

Let's find the field created by charge 1

q₁ = 10 μC = 10 10⁻⁶ C

x₁ = 5 cm = 0.05 m

x₃ = 21 cm = 0.21 m

         E_{13x} = 9 10⁹ 10 10⁻⁶ / (0.21 -0.05)²

         E_{13x} = 3.516 10⁶ N / C

y₁ = 6 cm = 0.06 cm

y₃ = -12 cm = -0.12 m

        E_{13y} = 9 10⁹ 10 10⁻⁶ / (-0.12 - 0.06)²

        E_{13y} = 2,777 10⁶ N / C

let's find the field produced by charge 2

q₂ = -27 μC = - 27 10⁻⁶ C

x₂ = -6 cm = -0.06 m

x₃ = 0.21 m

        E_{23x} = 9 10⁹ 27 10⁻⁶ / (0.21 + 0.06)²

        E_{23x} = 1.23 10⁶ N / A

y₂ = 10 cm = 0.10 m

y₃ = -0.12 m

        E_{23y} = 9 10⁹ 27 10⁻⁶ / (-0.12 - 0.10)²

        E_{23y} = 1.86 10⁶ N / C

Taking the components we can calculate the total electric field, we must use that charge of the same sign repel and attract the opposite sign, remember that the test charge is always considered positive.

       E_{total x} = E_{13x} - E_{23x}

       E_{total x} = (3.516 - 1.23) 10⁶

       E_{total x} = 2.29 10⁶ N / A

       

       E_{total y} = -E_{13y} + E_{23y}

       E_{total y} = (-2.777 +1.86) 10⁶ N / A

       E_{total y} = -0.917 10⁶ N / A

we can give the result in two ways

         E = (2.29 i ^ - 0.917 j ^) 10⁶ N / C

or in the form of modulus and angle, let's use the Pythagorean theorem to find the modulus

                E_{total} = √ (E_{total x}^2 + E_{total y}^ 2)

                 E_{total} = √ (2.29² + 0.917²) 10⁶

                E_{total} = 2,467 10⁶ N / A

let's use trigonometry for the angle

                tan θ = E_total and / E_totalx

                θ = tan⁻¹ E_{total y} / E_{total x}

                θ = tan⁻¹ (-0.917 / 2.29)

                θ = -21.8

The negative sign indicates that the angle is measured with respect to the x-axis in a clockwise direction.

7 0
3 years ago
A 200 kg object is initially at rest on a horizontal frictionless surface. At time t = 0 , a horizontal force of 100 N applied t
s344n2d4d5 [400]

Answer:

b

Explanation:

5 0
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