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3 years ago

How to know if a position time graph table is balanced or unbalanced

1 answer:

3 0

When balanced forces act on an object at rest, the object will not move. If you push against a wall, the wall pushes back with an equal but opposite force. Neither you nor the wall will move. Forces that cause a change in the motion of an object are unbalanced forces.

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A doorframe is twice as tall as it is wide. There is a positive charge on the top left corner and an equal but negative charge i

Andrei [34K]

Answer:

α = 141.5° (counterclockwise)

Explanation:

q₁ = +q

q₂ = -q

q₃ < 0

b = 2*a

We apply Coulomb's Law as follows

F₁₃ = K*q₁*q₃ / d₁₃² = + K*q*q₃ / (2*a)² = + K*q*q₃ / (4*a²)

F₂₃ = K*q₂*q₃ / d₂₃² = - K*q*q₃ / (5*a²)

(d₂₃² = a² + (2a)² = 5*a²)

Then

∅ = tan⁻¹(2a/a) = tan⁻¹(2) = 63.435°

we apply

F₃x = - F₂₃*Cos ∅ = - (K*q*q₃ / (5*a²))* Cos 63.435°

⇒ F₃x = - 0.0894*K*q*q₃ / a²

F₃y = - F₂₃*Sin ∅ + F₁₃

⇒ F₃y = - (K*q*q₃ / (5*a²))* Sin 63.435° + (K*q*q₃ / (4*a²))

⇒ F₃y = 0.0711*K*q*q₃ / a²

Now, we use the formula

α = tan⁻¹(F₃y / F₃x)

⇒ α = tan⁻¹((0.0711*K*q*q₃ / a²) / (- 0.0894*K*q*q₃ / a²)) = - 38.5°

The real angle is

α = 180° - 38.5° = 141.5° (counterclockwise)

4 0

4 years ago

A briefcase sits stationary in an elevator. The mass of the briefcase if 4.5 kg. The elevator then begins accelerating upwards a

Korvikt [17]

Answer:

D. 48.985 N

Explanation:

Newton's second law states that:

$\sum F = ma$

which means that the net force acting on an object is equal to the product between the object's mass and its acceleration.

The equation of the forces for the briefcase in the elevator therefore is given by:

$N-mg=ma$

where

N is the normal reaction exerted on the briefcase

(mg) is the weight of the briefcase, with

m = 4.5 kg being its mass

g = 9.8 m/s^2 is the acceleration of gravity

a = 1.10 m/s^2 is the acceleration

Here we chose upward as positive direction.

Solving for N, we find the normal force:

$N=mg+ma=m(g+a)=(4.5)(9.81+1.10)=49.095 N$

So the closest answer is

D. 48.985 N

3 0

3 years ago

A force in the +x-direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 8.90 kg box that is sitting on the horizontal,

Lady_Fox [76]

Answer:

6.875 m/s

Explanation:

The force is variable which is given by

F(x) = 18 - 0.53 x

mass of the box, m = 8.9 kg

initially it is at rest at x = 0

Let the velocity is v after travelling a distance of 15 m.

According to the work energy theorem, the work done by all the forces is equal to the change in kinetic energy of the body.

Work done = change in kinetic energy

$\int \overrightarrow{F}.d\overrightarrow{x}=\Delta K.E$

$\int_{0}^{15} \left ( 18-0.53 x \right )dx=\frac{1}{2}\times m \left ( v^{2}-u^{2} \right )$

$\left ( 18x-0.265x^{2} \right )_{0}^{15}=\frac{1}{2}\times 8.9\times \left ( v^{2}-0^{2} \right )$

18 x 15 - 0.265 x 15 x 15 = 4.45 x v²

270 - 59.625 = 4.45 v²

v² = 47.275

v = 6.875 m/s

Thus, the final velocity of the box is 6.875 m/s.

4 0

3 years ago

A particle moves with a velocity    1 v 5j 3j 6k ms      under influence of a contact force  F 10i 10j 20k N.    

katrin2010 [14]

Answer:

P = 200 W

Explanation:

The expression for the power is

P = W / t

the job

W = F d

we substitute

P = F d / t

P = F. v

We apply this equation to our case where the velocity is

v = (5 i + 3j + 6k) m / s

and the force is

F = (10i + 10j +20 k) N

we substitute in the power equation, remember that the scalar product of the unit vectors is

i.i = j.j = k.k = 1 and the other products are zero

P = 10 5 + 10 3 + 20 6

P = 200 W

6 0

3 years ago

A charged particle enters into a uniform magnetic field such that its velocity vector is perpendicular to the magnetic field vec

Anit [1.1K]

Answer:

The charged particle will follow a circular path.

Explanation:

The magnetic force exerted on the charged particle due to the magnetic field is given by:

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the particle

B is the magnetic field

$\theta$ is the angle between v and B

In this problem, the velocity is perpendicular to the magnetic field, so $\theta = 90^{\circ}, sin \theta=1$ and the force is simply

$F=qvB$

Moreover, the force is perpendicular to both B and v, according to the right-hand rule. Therefore, we have:

- a force that is always perpendicular to the velocity, v

- a force which is constant in magnitude (because the magnitude of v or B does not change)

--> this means that the force acts as a centripetal force, so it will keep the charged particle in a uniform circular motion. So, the correct answer is

The charged particle will follow a circular path.

8 0

4 years ago