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lawyer [7]
2 years ago
5

Newton's Principle states that the buoyant force on an object is equal to weight of the fluid displaced by the object.

Physics
1 answer:
Ilia_Sergeevich [38]2 years ago
3 0

Answer:

False it wasn't Newton who theorised it, it was Archimedes

Explanation:

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What is the frictional force between a box and the floor it is being pulled across if, the kinetic coefficient of friction is 0.
Artyom0805 [142]

If the pulling is done parallel to the floor with constant velocity, then the box is in equilibrium. In particular, the weight and normal force cancel, so that

<em>n</em> = 38 N

The friction force is proportional to the normal force by a factor of 0.27, so that

<em>f</em> = 0.27 (38 N) ≈ 10.3 N

and so the answer is D.

8 0
3 years ago
A puck of mass 0.110 kg slides across ice in the positive x-direction with a kinetic friction coefficient between the ice and pu
lara [203]

Answer:

a) Ffr = -0.18 N

b) a= -1.64 m/s2

c) t = 9.2 s

d) x = 68.7 m.

e) W= -12.4 J

f) Pavg = -1.35 W

g) Pinst = -0.72 W

Explanation:

a)

  • While the puck slides across ice, the only force acting in the horizontal direction, is the force of kinetic friction.
  • This force is the horizontal component of the contact force, and opposes to the relative movement between the puck and the ice surface, causing it to slow down until it finally comes to a complete stop.
  • So, this force can be written as follows, indicating with the (-) that opposes to the movement of the object.

       F_{frk} = -\mu_{k} * F_{n} (1)

       where μk is the kinetic friction coefficient, and Fn is the normal force.

  • Since the puck is not accelerated in the vertical direction, and there are only two forces acting on it vertically (the normal force Fn, upward, and  the weight Fg, downward), we conclude that both must be equal and opposite each other:

      F_{n} = F_{g} = m*g (2)

  • We can replace (2) in (1), and substituting μk by its value, to find the value of the kinetic friction force, as follows:

       F_{frk} = -\mu_{k} * F_{n} = -0.167*9.8m/s2*0.11kg = -0.18 N (3)

b)

  • According Newton's 2nd Law, the net force acting on the object is equal to its mass times the acceleration.
  • In this case, this net force is the friction force which we have already found in a).
  • Since mass is an scalar, the acceleration must have the same direction as the force, i.e., points to the left.
  • We can write the expression for a as follows:

        a= \frac{F_{frk}}{m} = \frac{-0.18N}{0.11kg} = -1.64 m/s2  (4)

c)

  • Applying the definition of acceleration, choosing t₀ =0, and that the puck comes to rest, so vf=0, we can write the following equation:

        a = \frac{-v_{o} }{t} (5)

  • Replacing by the values of v₀ = 15 m/s, and a = -1.64 m/s2, we can solve for t, as follows:

       t =\frac{-15m/s}{-1.64m/s2} = 9.2 s (6)

d)

  • From (1), (2), and (3) we can conclude that the friction force is constant, which it means that the acceleration is constant too.
  • So, we can use the following kinematic equation in order to find the displacement before coming to rest:

        v_{f} ^{2} - v_{o} ^{2} = 2*a*\Delta x  (7)

  • Since the puck comes to a stop, vf =0.
  • Replacing in (7) the values of v₀ = 15 m/s, and a= -1.64 m/s2, we can solve for the displacement Δx, as follows:

       \Delta x  = \frac{-v_{o}^{2}}{2*a} =\frac{-(15.0m/s)^{2}}{2*(-1.64m/s2} = 68.7 m  (8)

e)

  • The total work done by the friction force on the object , can be obtained in several ways.
  • One of them is just applying the work-energy theorem, that says that the net work done on the object is equal to the change in the kinetic energy of the same object.
  • Since the final kinetic energy is zero (the object stops), the total work done by friction (which is the only force that does work, because the weight and the normal force are perpendicular to the displacement) can be written as follows:

W_{frk} = \Delta K = K_{f} -K_{o} = 0 -\frac{1}{2}*m*v_{o}^{2} =-0.5*0.11*(15.0m/s)^{2}   = -12.4 J  (9)

f)

  • By definition, the average power is the rate of change of the energy delivered to an object (in J) with respect to time.
  • P_{Avg} = \frac{\Delta E}{\Delta t}  (10)
  • If we choose t₀=0, replacing (9) as ΔE, and (6) as Δt, and we can write the following equation:

       P_{Avg} = \frac{\Delta E}{\Delta t} = \frac{-12.4J}{9.2s} = -1.35 W (11)

g)

  • The instantaneous power can be deducted from (10) as W= F*Δx, so we can write P= F*(Δx/Δt) = F*v (dot product)
  • Since F is constant, the instantaneous power when v=4.0 m/s, can be written as follows:

       P_{inst} =- 0.18 N * 4.0m/s = -0.72 W (12)

7 0
3 years ago
How do you change matter into other phases of matter?
Vera_Pavlovna [14]
Hey there!

There's many ways to do it - like melting and evaporating.

For example, we'll use water. Plain old water in a water bottle. Right now, it's in its liquid state of matter, but say you put it in the freezer for an hour. That would change its state of matter to solid, since it would be solid ice. Now, if you were to put it out in the sun on a blazing hot day for a couple of hours, it would evaporate and become water vapor, a gas. Lastly, if you can cool that water vapor it becomes a liquid again.

Hope this helps!
3 0
3 years ago
Where does the name basketball come from?
umka21 [38]
Definitely ball and basket
7 0
3 years ago
What is the difference between an induced and a permanent magnet?
MissTica
INDUCTION MOTOR:-

Speed:-Less speed range than PMAC motors • Speed range is a function of the drive being used — to 1,000:1 with an encoder, 120:1 under field-oriented control


Reliability:-Waste heat is capable of degrading insulation essential to motor operation • Years of service common with proper operation

Power density:-Induction produced by squirrel cage rotor inherently limits power density

Accuracy:-Flux vector and field-oriented control allows for some of accuracy of servos

Cost:-Relatively modest initial cost; higher operating costs

PERMANENT MAGNET MORTOR:-

speed:-VFD-driven PMAC motors can be used in nearly all induction-motor and some servo applications • Typical servomotor application speed — to 10,000 rpm — is out of PMAC motor range

Reliability:-Lower operating temperatures reduces wear and tear, maintenance • Extends bearing and insulation life • Robust construction for years of trouble-free operation in harsh environments.

power density:-Rare-earth permanent magnets produce more flux (and resultant torque) for their physical size than induction types.

Accuracy:-Without feedback, can be difficult to locate and position to the pinpoint accuracy of servomotors

<span>Cost:-Exhibit higher efficiency, so their energy use is smaller and full return on their initial purchase cost is realized more quickly</span>
8 0
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