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belka [17]
3 years ago
11

If water is leaking from a certain tank at a constant rate, is it leaking at a rate that is greater than 12 liters per hour?(1 l

iter = 1,000 milliliters)
Physics
1 answer:
Margarita [4]3 years ago
5 0

Answer:

Yes  it is possible because 3.33 is greater than 2.

Explanation:

Given that,

1 liter = 1000 milliliters

Suppose water is leaking from the tank at a rate that is greater than 2 milliliters per second.

We need to change the unit in ml/s

12 lit/hr=\dfrac{12000}{3600}

12 lit/hr=3.33\ ml/s

Hence, Yes  it is possible because 3.33 is greater than 2.

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labwork [276]

Answer:

The SI units for energy is Joules.

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3 years ago
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Explain how the plants of a tropical rain forest contribute to the characteristic hot, humid weather associated with the area. U
Talja [164]

Answer: The atmosphere of trophic rainforest is hot and humid due to high rate of transpiration.

Explanation:

The tropical rainforest is a biome which exhibit rich biodiversity of plants and animals. The average temperatures in this region remain high with warm summer. It remains frost free. The soil is nutrient deficient. Due to hot temperature the rate of transpiration remains high as a result the concentration of the water vapors remain high. This is responsible for increasing the humidity in the atmosphere also this region receives appreciable amount of rainfall annually. The average rainfall range is 200-450 centimeters.

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2 years ago
You have seen magnets sticking to the refrigerator door, or maybe in your science class room. They attract metal items, for exam
DENIUS [597]

Answer:

B) shrinks

Explanation:

The magnetic force is a force exerted between two magnets, or two magnetic materials, or also on an electric charge moving in a magnetic field.

If we talk about magnetic material, the magnetic field they generates can be represented using a dipole: essentially, they have a north pole (where the lines of the field go out) and a south pole (where the lines of the field go in).

Also, the lines spread apart as we move away from the magnet itself. This means that the strength of the field (and so, the intensity of the force) decreases as we move away from the magnet.

Using this description, we can now understand that when we move the paper clip further from the magnet, the force exerted on the clip decreases, as the magnetic field becomes weaker. So, the correct answer is B.

3 0
3 years ago
A 20 kg shopping cart moving at a velocity of 0.5 m/s collides with a store wall and
MAXImum [283]

Answer:

<h2>10 kg.m/s</h2>

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 20 × 0.5

We have the final answer as

<h3>10 kg.m/s</h3>

Hope this helps you

7 0
2 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
2 years ago
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