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nata0808 [166]
3 years ago
9

A(n) ___ keeps a reaction from occurring by increasing the activation energy required for a reaction to occur. A. inhibitor B. c

atalyst C. stopper D. synthesis
Physics
2 answers:
skelet666 [1.2K]3 years ago
8 0
Due to requirements i am typing this but the ans is a
Katena32 [7]3 years ago
8 0
A. inhibitor

think about what inhibit means, it means to keep something from happening.
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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
What is the minimum v (in km/s) the rocket engines must provide to allow the craft to escape from the Earth?
coldgirl [10]

Answer:

11 kilometers (7 miles) per second, or over 40,000 kilometers per hour (25,000 miles per hour)

Explanation:

3 0
3 years ago
Consider a block of mass m attached to two springs, one on the left with spring constant k1 and one on the right with spring con
Andrew [12]

Answer:

T= 2\pi\times sqrt(m/(k1+k2))

Explanation:

When the block is displaced by x units

F= spring force

two springs are connected parallel

F =-k_1x - k_2x

Writing Newtons second law, F = ma

-k_1x - k_2x =ma

-k_1x - k_2x = mx''

a= x" ( differentiating x w.r.t time twice)

x''+(k_1/m + k_2/m) x=0

this the standard form of equation of oscillation spring mass system

This is the differential equation, x'' means that double differentiation of x , i.e, x'' is acceleration

since,  Period T=2\pi\sqrt{\frac{m}{K_{eq.}} }

therefore,

T= 2\pi\times sqrt(m/(k1+k2))

3 0
3 years ago
Liquid can flow but solid cannot give reason ​
Airida [17]

<em>Answer:</em>

<em>well..</em>

<em>Explana</em><em>tion</em><em>:</em>

<em>L</em><em>iquid</em><em> can flow but solid cannot because of differences in their properties</em>

<em>property of liquid which lets it flow:</em>

  • <em>i</em><em>nter-particular</em><em> space is large</em>
  • <em>inter-particular attraction is small</em><em> </em><em>t</em><em>hese</em><em> properties tend to make the molecules of liquid free to flow</em><em> </em>

<em>property</em><em> </em><em>of</em><em> </em><em>solid</em><em> </em><em>which</em><em> </em><em>tends</em><em> </em><em>to</em><em> </em><em>obstruct</em><em> </em><em>flow</em><em>:</em>

  • <em>inter-particular</em><em> </em><em>spa</em><em>c</em><em>e</em><em> </em><em>is</em><em> </em><em>small</em><em> </em><em>and</em><em> </em><em>so</em><em> </em><em>it's </em><em>compac</em><em>t</em>
  • <em>inter-molecular</em><em> </em><em>attra</em><em>ction</em><em> </em><em>is</em><em> </em><em>strong</em><em> </em><em>hence</em><em> </em><em>no</em><em> </em><em>tenden</em><em>cy</em><em> </em><em>to</em><em> </em><em>flow</em>

<em>H</em><em>o</em><em>p</em><em>e</em><em> </em><em>this</em><em> </em><em>helps</em><em>!</em>

6 0
3 years ago
What would be the best design for an experiment that tests how much water expands when frozen?
galben [10]

B. Purchase a small plastic container and mark 1-ounce increments on the outside to determine volume. Pour 5 ounces of water into the container, and place in the freezer for 8 hours. Compare the frozen or ending volume with the liquid or beginning volume.

<h3>How much water expands when frozen?</h3>

Ice is less denser than the liquid form. Water is the only known non-metallic substance that expands when it freezes because it is the unique property of water. Water density decreases and it expands approximately  about 9% by volume. For calculating the expansion of water, plastic container is the best option. We know that water expands when the water freezes because it is a unique property of water which allows the survival of aquatic organisms.

So we can conclude that option B is the right answer.

Learn more about water here: brainly.com/question/1313076

#SPJ1

5 0
1 year ago
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