The position of a particular particle as a function of is given by r=(8.5t i+5.6j-2tk )m (a) Determine the particle’s velocity and acceleration as a function of time.(b) Find the instantaneous velocity and acceleration of the particle at t=3.0 s.

Answer:

1.75 atm

Explanation:

Mass is conserved, so the mass flow before the constriction equals the mass flow after the constriction.

m₁ = m₂

ρQ₁ = ρQ₂

Q₁ = Q₂

v₁A₁ = v₂A₂

v₁ πd₁²/4 = v₂ πd₂²/4

v₁ d₁² = v₂ d₂²

Now use Bernoulli equation:

P₁ + ½ ρ v₁² + ρgh₁ = P₂ + ½ ρ v₂² + ρgh₂

Since h₁ = h₂:

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²

Writing v₂ in terms of v₁:

P₁ + ½ ρ v₁² = P₂ + ½ ρ (v₁ d₁²/d₂²)²

P₁ + ½ ρ v₁² = P₂ + ½ ρ v₁² (d₁/d₂)⁴

P₁ + ½ ρ v₁² (1 − (d₁/d₂)⁴) = P₂

Plugging in values:

P₂ = 2 atm + ½ (1000 kg/m³) (4.4 m/s)² (1 − (3.3 cm / 2.4 cm)⁴) (1 atm / 1.013×10⁵ Pa)

P₂ = 1.75 atm

Answer:

1a) The direction to throw the boot is directly away from the closest shore.

2b) The magnitude of the force that the thrown boot exerts on the engineer = 391 N

3c) Time taken to reach shore = 8.414 s

Explanation:

1a) Newton's third law of motion explains that for every action, there is an equal and opposite reaction.

The force generated by throwing the boot in one direction is exerted back on the engineer as recoil in the opposite direction.

Hence, the best direction to throw the boot is opposite the direction that the engineer intends to move towards.

2b) Just as explained in (1a) above, the force exerted in one direction always has a reaction of the same magnitude in the opposite direction.

Hence, the force exerted by the boot on the engineer is equal to the force exerted by the engineer on the boot = 391 N.

3c) For this part, we analyze the total motion of the engineer.

The force exerted by the boot on the engineer initially accelerates the engineer until the engineer reaches a constant velocity dictated the impulse of the initial force (since impulse is equal to change in momentum), this constant velocity then takes the engineer all the way to shore, since the ice surface is frictionless.

The weight of the engineer = W = 588 N

W = mg

Mass of the engineer = (W/g) = (588/9.8) = 60 kg

Force exerted on the engineer by the thrown boot = F = 391 N

F = ma

Initial acceleration of the engineer = (F/m) = (391/60) = 6.52 m/s²

We can then calculate the distance covered during this acceleration

X₁ = ut + ½at₁²

u = initial velocity of the engineer = 0 m/s (the engineer was initially at rest)

t₁ = time during which the force acts = 0.576 s

a = acceleration during this period = 6.52 m/s²

X₁ = 0 + 0.5×6.52×0.576² = 1.08 m

For the second part of the engineer's motion, the velocity becomes constant.

So, we first calculate this constant velocity

Impulse = Change in momentum

F×t = mv - mu

F = Force causing motion = 391 N

t = time during which the force acts = 0.576 s

m = mass of the engineer = 60 kg

v = final constant velocity of the engineer = ?

u = initial velocity of the engineer = 0 m/s

391 × 0.576 = 60v

v = (391×0.576/60) = 3.7536 m/s.

The distance from the engineer's initial position to shore is given as 30.5 m

The engineer covers 1.08 m during the time the force causing motion was acting.

The remaining distance = X₂ = 30.5 - 1.08 = 29.42 m

We can then calculate the time taken to cover the remaining distance, 29.42 m at constant velocity of 3.7536 m/s

X₂ = vt₂

t₂ = (X₂/v) = (29.42/3.7536) = 7.838 s

Time taken to reach shore = t₁ + t₂ = 0.576 + 7.838 = 8.414 s

Hope this Helps!!!

Answer:

Explanation:

Gauss' Law should be applied to find the E-field 3.9 cm from the surface of the sphere.

In order to apply Gauss' Law, an imaginary spherical shell (Gaussian surface) should be placed around the original sphere. The exact position of the shell must be 3.9 cm from the surface of the original sphere.

Gauss' Law states that

Here, the integral in the left-hand side is equal to the area of the imaginary surface. After all, the reason behind choosing the imaginary surface a spherical shell is to avoid this integral. The enclosed charge in the right-hand side is equal to the charge of the sphere, -84.0 nC. The radius of the imaginary surface must be 5 + 3.9 = 8.9 cm.

So,

Answer:

W = ½ m v²

Explanation:

In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation

We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved

initial instant. before separation

        p₀ = m v

final attempt. after separation

        = m /2  0 + m /2 v_{f}

       p₀ = p_{f}

       m v = m /2

       v_{f}= 2 v

this is the speed of the second part of the ship

now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body

initial energy

         K₀ = ½ m v²

final energy

        = ½ m/2  0 + ½ m/2 v_{f}²

        K_{f} = ¼ m (2v)²

        K_{f} = m v²

the expression for work is

         W = ΔK = K_{f} - K₀

         W = m v² - ½ m v²

         W = ½ m v²

Final answer:

The principle of conservation of momentum implies that no work is performed by the internal forces during the separation of the space vehicle. This is granted that external forces are ignored and the total momentum and kinetic energy of the closed system remain constant.

Explanation:

The subject you're asking about centers around the principle of conservation of momentum. In the case of this space vehicle, before separation, the momentum of the whole system is given by the product of the mass and velocity, mv. After separation, one piece is at rest, leaving the other piece with momentum mv. As there is no external force, the total momentum does not change, so no work is performed by the internal forces causing the separation.

In more detail, the principle of conservation of momentum states that the total linear momentum of a closed system remains constant, regardless of any interactions happening within the system. The system is 'closed' meaning that no external forces are acting upon it. In this case, the space vehicle and the two smaller pieces it separates into form a closed system. This is consistent with your question's stipulation to ignore external forces, such as gravitational forces.

This can also be understood from the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy. If we consider the vehicle before and after the separation, the kinetic energy of the system remains the same: initially all the energy is concentrated in the moving vehicle, and after the separation, all the kinetic energy is transferred to the moving piece while the at-rest piece has none. Therefore, the work done by the internal forces - which would change the kinetic energy - must be zero.

Learn more about Conservation of momentum here:

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Average velocity: 0 m/s. Average speed: 0.25 m/s. Greta returns to her starting point, so her displacement is 0m.

Greta's average velocity is 0 m/s because she ends up at the same point where she started. Her displacement is 0 meters, and since velocity is displacement divided by time, her average velocity is 0 / 240 = 0 m/s.

Her average speed, on the other hand, is calculated using the formula: Average Speed = TotalDistance / Total Time.

Initially, Greta walks 30 meters away from her truck, and then she returns 30 meters back to her starting point. So, the total distance she covers is 30 + 30 = 60 meters. Her total travel time is 240 seconds.

AverageSpeed = 60 meters / 240 seconds = 0.25 m/s.

In summary, Greta's average velocity is 0 m/s because her net displacement is 0 meters. Her average speed is 0.25 m/s because she covers a total distance of 60 meters in 240 seconds.

To learn more about Average velocity  here

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Answer:

0 | for Velocity

.25 | for speed

Explanation: