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Karolina [17]
3 years ago
13

Which of the following is NOT a function of the skeletal system?

Physics
2 answers:
puteri [66]3 years ago
4 0
D
Hope this helps
And
HACTEHA [7]3 years ago
4 0
I think the answer is D
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Which of the following would have the smallest coefficient of kinetic friction when slid along your hang.
Dmitriy789 [7]
A soap bar because the material of a carpet or sandpaper is more rough than a soapbar and thus has a higher coefficient of friction.
4 0
2 years ago
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A 3.0 kg mass is released from rest at point A. The mass slides along the curved surface to point B in 6.0 seconds. Point B is 2
timurjin [86]

Answer:

option d) -9 J

Explanation:

Given:

Mass, m = 3.0 kg

time, t = 6.0 seconds

Velocity of mass, v = 2.0 m/s

height, h = 2 m

Now, using the concept of work-Energy theorem

we have

Net work done = change in kinetic energy

or

Work done by gravity + work done by the friction = Final kinetic energy - Initial kinetic energy

mgh +W_f = \frac{1}{2}mv^2-0

on substituting the values in the above equation, we get

3 × 9.8 × 2 + W_f = \frac{1}{2}\times 3\times2^2

or

58.8 + W_f = 6

or

W_f = -52.8 J

here negative sign depicts that the work is done against the motion of the mass

also,

Power = (Work done)/time

or

Power = -52.8/6 = -8.8 W ≈ 9 J

Hence, option d) -9 J is correct

5 0
3 years ago
A 540 gram object is attached to a vertical spring, causing the spring’s length to change from 70 cm to 110 cm.
belka [17]

Answer:

Approximately 13\; {\rm N \cdot m^{-1}} (assuming that g = 9.81\; {\rm N \cdot kg^{-1}}.)

Explanation:

Let F_{\text{s}} denote the force that this spring exerts on the object. Let x denote the displacement of this spring from the equilibrium position.

By Hooke's Law, the spring constant k of this spring would ensure that F_\text{s} = -k\, x.

Note that the mass of the object attached to this spring is m = 540\; {\rm g} = 0.540\; {\rm kg}. Thus, the weight of this object would be m\, g = 0.540\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \approx 5.230\; {\rm N}.

Assuming that this object is not moving, the spring would need to exert an upward force of the same magnitude on the object. Thus, F_{\text{s}} = 5.230\; {\rm N}.

The spring in this question was stretched downward from its equilibrium by:

\begin{aligned} x &= (70\; {\rm cm} - 110\; {\rm cm}) \\ &= (-40)\; {\rm cm} \\ &= (-0.40) \; {\rm m}\end{aligned}.

(Note that x is negative since this displacement points downwards.)

Rearrange Hooke's Law to find k in terms of F_{\text{s}} and x:

\begin{aligned} k &= \frac{F_{\text{s}}}{-x} \\ &\approx \frac{5.230\; {\rm N}}{-(-0.40)\; {\rm m}} \\ &\approx 13\; {\rm N \cdot m^{-1}}\end{aligned}.

3 0
2 years ago
What do we call a change in the car’s speed or direction?<br><br><br> Help!!
zhuklara [117]
The answer for this question is acceleration
8 0
3 years ago
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What is the net force on this object?​
Sergio039 [100]
200N

Explanation:
600N-400N = 200N
6 0
3 years ago
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