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Gnesinka [82]
3 years ago
10

What is the acceleration of an object that moves at constant velocity? What is the net force on the object in this case? What is

the acceleration of an object that moves at constant velocity? What is the net force on the object in this case? Acceleration is zero; net force is zero. Acceleration is zero; net force is greater than zero. Acceleration is greater than zero; net force is zero. Acceleration is greater than zero; net force is greater than zero.
Physics
1 answer:
Misha Larkins [42]3 years ago
3 0

Answer: What is the acceleration of an object that moves at constant velocity?  Acceleration is Zero

What is the net force on the object in this case? The ner force is also Zero

Explanation: If the velocity is constant means that the acceleration is zero, from Newton second law is clear that net force is also zero

Fnet= m* a as v is constant a=0 this Fnet is zero.

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Read 2 more answers
At an air show, a stunt pilot performs a vertical loop-the-loop in a circle of radius 3.63 x 103 m. During this performance the
san4es73 [151]

Answer:

189 m/s

Explanation:

The pilot will experience weightlessness when the centrifugal force, F equals his weight, W.

So, F = W

mv²/r = mg

v² = gr

v = √gr where  v = velocity, g = acceleration due to gravity = 9.8 m/s² and r = radius of loop = 3.63 × 10³ m

So, v = √gr

v = √(9.8 m/s² × 3.63 × 10³ m)

v = √(35.574 × 10³ m²/s²)

v = √(3.5574 × 10⁴ m²/s²)

v = 1.89 × 10² m/s

v = 189 m/s

5 0
3 years ago
Suppose two point charges, Q1 = -6.25 x 10-9 C and Q2 = -6.25 x 10-9 C, are separated by a distance d = 0.617 m.
n200080 [17]

Answer:

The answer to your question is the letter A) F =  9.23 x 10⁻⁷ N

Explanation:

Data

q₁ = -6.25 x 10⁻⁹ C

q₂ = -6.25 x 10⁻⁹ C

d = 0.617 m

k = 9 x 10⁹ Nm²/C²

F = ?

Formula

              F = k q₁q₂ /r²

-Substitution

              F = (9 x 10⁹)(-6.25 x 10⁻⁹)(-6.25 x 10⁻⁹) / (0.617)²

-Simplification

              F = 3.512 x 10⁻⁷ / 0.381

-Result

              F = 9.227 x 10⁻⁷ N  ≈ 9.23 x 10⁻⁷ N

8 0
3 years ago
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