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aleksandr82 [10.1K]
3 years ago
8

The drawing shows a top view of a door that is free to rotate about an axis of rotation that is perpendicular to the screen. Fin

d the net torque (magnitude and direction) produced by the forces F1 and F2 about the axis.
Physics
1 answer:
dangina [55]3 years ago
4 0

Answer:

τ = 23.34 Nm

Direction = Anti-Clockwise (Positive)

Explanation:

The image of the question is attached in the answer.

First, we will calculate the torque due to force F₁.

τ₁ = F₁ d₁ Sin θ₁

where,

τ₁ = Torque due to F₁ = ?

F₁ = 20 N

d₁ = moment arm of F₁ = 0.5 m

θ₁ = Angle between F₁ and d₁ = 90°

Therefore,

τ₁ = (20 N)(0.5 m)Sin 90°

τ₁ = - 10 Nm

Negative sign due to clockwise direction.

Now, we will calculate the torque due to force F₂.

τ₂ = F₂ d₂ Sin θ₂

where,

τ₂ = Torque due to F₂ = ?

F₂ = 35 N

d₂ = moment arm of F₂ = 1.10 m

θ₂ = Angle between F₂ and d₂ = 120°

Therefore,

τ₁ = (35 N)(1.1 m)Sin 120°

τ₁ = 33.34 Nm

Positive sign due to anti-clockwise direction.

Now, we find net torque on door:

τ = τ₁ + τ₂

τ = - 10 Nm + 33.34 Nm

<u>τ = 23.34 Nm</u>

<u>Direction = Anti-Clockwise (Positive)</u>

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2 years ago
How much heat in joules is required to melt an ice cube with a mass of 18.6 g at 0 °C? The Lf for water is 333 J/g.
EastWind [94]

Answer:

The heat energy required, Q = 6193.8 J

Explanation:

Given,

The mass of ice cube, m = 18.6 g

The heat of fusion of ice, ΔHₓ = 333 J/g

The heat energy of a substance is equal to the product of the mass and heat of fusion of that substance. It is given by the equation,

                               <em> Q = m · ΔHₓ     joules</em>

Substituting the given values in the above equation

                                Q = 18.6 g x 333 J/g

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Hence, the heat required to melt the ice cube is, Q = 6193.8 J

5 0
3 years ago
The great limestones caverns were formed by dripping water. If water droplets of 10 ml fall from a height of 5 m at a rate of 10
loris [4]

The average force of the water droplets is the force given by the impact

per second of the droplets on the limestone floor.

  • The average force exerted on the limestone floor is approximately <u>1.6013 × 10⁻² N</u>

Reasons:

The given parameters are;

Volume of a droplet = 10 ml = 1 × 10⁻⁵ m³

Height from which the water falls, <em>h </em>= 5 meters

Rate at which the water falls = 10 per minute

Required:

The average force exerted on the floor by the water droplets.

Solution:

According to Newton's Second Law of motion, we have;

Force = Rate of change of momentum

Momentum = Mass × Velocity

Mass of a droplet of water = Volume × Density

Density of water = 997 kg/m³

Mass of a droplet = 1 × 10⁻⁵ m³ × 997 kg/m³ = 0.00997 kg

The velocity just before the droplet reaches the ground, v = √(2·g·h)

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Which gives;

v = √(2 × 9.81 m/s² × 5 m) ≈ 9.905 m/s

The rate of change in momentum per minute = 1

Therefore;

\displaystyle The \ rate \ of \ change \ in \ momentum = Average \ force = \mathbf{\frac{\Delta Momentum }{\Delta Time}}

ΔMomentum = Mass × ΔVelocity

Considering the 10 drops per minute, we have;

ΔMomentum = 10 × 0.0097 kg × 9.905 m/s = 0.960785 kg·m/s

ΔTime = 1 minute = 60 seconds

Therefore;

\displaystyle Average \ force, \, F_{ave}  \frac{0.960785 \, kg\cdot m/s }{60 \, s} \approx =\mathbf{1.6013 \times 10^{-2} \, N}

  • The average force exerted on the limestone floor by the droplets of water is F_{ave} ≈ <u>1.6013 × 10⁻² N</u>

Learn more about Newton's Second Law of motion and force exerted water here:

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The net force on an object is equal to its mass times its acceleration.

Newton's third law:

For every action, there is an opposite and equal reaction.

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Please help brainliest, rattings, thanks, ect. its answer choice
Svetllana [295]

D: a force equal to the force

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