When will an object accelerate in terms of force

Physics

1 answer:

GREYUIT [131]3 years ago

3 0

Answer:

when an external force applied on an object

Explanation:

there will be a change in velocity this means it is accelerating

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In a football game a kicker attempts a field goal. The ball remains in contact with the kicker's foot for 0.0580 s, during which

Fantom [35]

Answer:

V₀ₓ = 10.94 m/s

V₀y = 18.87 m/s

Explanation:

To find the launch velocity, we use 1st equation of motion.

Vf = Vi + at

where,

Vf = Final Velocity of Ball = Launch Speed = V₀ = ?

Vi = Initial Velocity = 0 m/s (Since ball was initially at rest)

a = acceleration = 376 m/s²

t = time = 0.058 s

Therefore,

V₀ = 0 m/s + (376 m/s²)(0.058 s)

V₀ = 21.81 m/s

Now, for x-component:

V₀ₓ = V₀ Cos θ

where,

V₀ₓ = x-component of launch velocity = ?

θ = Angle with horizontal = 59.9⁰

V₀ₓ = (21.81 m/s)(Cos 59.9°)

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for y-component:

V₀ₓ = V₀ Sin θ

where,

V₀y = y-component of launch velocity = ?

θ = Angle with horizontal = 59.9⁰

V₀y = (21.81 m/s)(Sin 59.9°)

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8 0

4 years ago

A 1.0 C charged object and a 2.0 C charged object are separated by 100 m. Where should a -1.0x10-3 C charged object be placed on

horsena [70]

Answer:

x = 41.2 m

Explanation:

The electric force is a vector magnitude, so it must be added as vectors, remember that the force for charges of the same sign is repulsive and for charges of different sign it is negative.

In this case the fixed charges (q₁ and q₂) are positive and separated by a distance (d = 100m), the charge (q₃ = -1.0 10⁻³ C)) is negative so the forces are attractive, such as loads q₃ must be placed between the other two forces subtract

F = F₁₃ - F₂₃

let's write the expression for each force, let's set a reference frame on the charge q1

F₁₃ = $k \frac{q_1 q_3}{x^2}$