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tia_tia [17]
3 years ago
8

What force would be required to accelerate a 1.100 kg car to 0.5 m/s?

Physics
2 answers:
sweet-ann [11.9K]3 years ago
7 0
I have no clue how to answer it
Anastaziya [24]3 years ago
6 0

Answer:

it’s 550N :))))

Explanation:

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2.
saul85 [17]

Answer:

<u>B</u>

Explanation:

Planets have different year lengths because it depends how far they revolve from a celestial body. Each planet has its own orbital period. Planets closer to the star will have a lower orbital period compared to the ones that lie far away from it.

4 0
2 years ago
Read 2 more answers
what is the magnitude of the electric force between charges of 0.25 C and 0.11 C at a separation of 0.88 m? if the separation be
leonid [27]
F = k \cdot \frac{q_1 q_2}{d} = 9 \cdot 10^{9} \cdot \frac{0.25 \cdot 0.11}{0.88} =  144 \cdot 5^{9} \ N.

If the separation between the charges is increased then the magnitude of the force will increase in fact how the distance is being used in that formula.
6 0
3 years ago
An electromagnet is a device in which moving electric charges (current) in a coil of wire create a magnet. What’s one advantage
yan [13]
I think the correct answer from the choices listed above is option D. One advantage of using electromagnets in devices would be that electromagnets can <span>easily be turned on and off. Hope this answers the question. Have a nice day.</span>
7 0
3 years ago
Read 2 more answers
A charge +Q is located at the origin and a second charge, +4Q, is at distance (d) on the x-axis.
tatiyna

Answer:

a)   x = ⅔ d , b) the charge must be negative, c) Q

Explanation:

a) In this exercise the force is electric between the charges, we are asked that the system of the three charges is in equilibrium, we use Newton's second law. Balance is on the third load that we are placing

         ∑ F = 0

        -F₁₂ + F₂₃ = 0

         F₁₂ = F₂₃

         

let's replace the values

        k Q Q / r₁₂² = k Q 4Q / r₂₃²

            Q² / r₁₂² = 4 Q² / r₂₃²

suppose charge 3 is placed at point x

        r₁₂ = x

        r₂₃ = d-x

             

we substitute

             1 / x² = 4 / (d-x) 2

             1 / x = 2 / (d-x)

             x = 2 (x-d)

             x = 2x -2d

            3x = 2d

              x = ⅔ d

b) The sign of the charge must be negative, to have an attractive charge on the two initial charges

c)  Q

5 0
4 years ago
Help me pls
Zina [86]

Answer:

Explanation:

the first one is D

8 0
4 years ago
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