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ivann1987 [24]
3 years ago
11

hich muscle fibers are best suited for activities that involve lifting large, heavy objects for a short period of time? cardiac

slow twitch intermediate fast twitch
Physics
1 answer:
Temka [501]3 years ago
6 0

Answer:

Dead lifting uses tho muscle fundamentals

Explanation:

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If a person weighs 600N on earth, how much does he weigh on the moon
Snezhnost [94]

Answer:

His weight would be 100 N

Explanation:

5 0
3 years ago
An equilibrium constant is not changed by a change in pressure <br> a. True<br> b. False
Umnica [9.8K]
Hi There! :)


An equilibrium constant is not changed by a change in pressurea. True
b. False

False! :P
7 0
3 years ago
Read 2 more answers
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
If 2N force is applied on 2 kg mass due east and same magnitude of force due west, thechange in velocity of the body in 2 sec is
klio [65]

Explanation:

F=m(v-u)/t

F=2N

m=2kg

t=2s

2=2(v-u)/2

cross multiply

2*2=2(v-u)

4=2(v-u)

4/2=v-u

v-u=2m/s

v-u is the change is velocity.

3 0
2 years ago
Can you please answer my question correctly ASAP​
lana66690 [7]

Answer:

1) C, 2) C, 3) B, 4) D, 5) A/B, 6) D, 7) D, 8) A, 9) B, 10) A

Explanation:

1) Seconds, minutes and days are time units. (Correct answer: C)

2) The displacement is shortest distance between the point of origin and the point of destination. (Correct answer: C)

3) The velocity is the rate of change of position in time. (Correct answer: B)

4) Graphs are a very useful ressource to infer characteristics inherent to a studied physical variable. (Correct answer: D)

5) There are two possible answers, since both milimeters and meters are units for distance. (Correct answers: A/B)

6) The distance is the arc length of the entire path that object travelled. (Answer: D)

7) Traditionally, in a position vs. time graph we plot time in the x-axis. (Answer: D)

8) In this case, we use a distance-time graph, where travelled distance is the dependent variable and time is the independent variable. (Answer: A)

9) The point of origin is the initial position of the object. (Answer: B)

10) The point of destination is the final position of the object. (Answer: A)

5 0
3 years ago
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