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Elena L [17]
3 years ago
10

Does Mike wosowski blink or wink

Physics
2 answers:
aev [14]3 years ago
7 0

Answer:

both

Explanation:

GrogVix [38]3 years ago
6 0
Definitely both because of the one eye situation
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A. What is the coefficient of kinetic friction?
Andrews [41]

Explanation:

Coefficient of kinetic friction is the resistive force that opposes the motion of a body as it moves and is in contact with another body.

It is found by dividing the frictional force by the normal force.

  • Friction is a force that opposes motion.
  • Static friction is for bodies that are not in motion
  • Kinetic friction is for moving bodies.
7 0
3 years ago
The way your brain interprets the intensity of a sound is the ______.
Naddik [55]
C i would think
it sounds best 
8 0
3 years ago
Why does a gymnast rub powder on his hands before exercising on a set of parallel bars? (I need a scientific answer)
dlinn [17]

<u>Answer</u>

To increase friction for a better grip.

<u>Answer</u>

Most human beings do sweat hands especially on the palm. When this happens the person will not have a good grip of heavy objects because they will slide/slip from the hand.

<em>By applying the powder, you are trying to make the hand dry hence increasing the friction for a better grip. </em>

If the gymnast doesn't do this the parallel bars may slip from the hands and injure himself or herself.

4 0
2 years ago
Read 2 more answers
A horizontal spring is lying on a frictionless surface. One end of the spring is attached to a wall, and the other end is connec
juin [17]

Answer:

Explanation:

Given that:

angular frequency = 11.3 rad/s

Spring constant (k) = = \omega^2  \times m

k = (11.3)² m

k = 127.7 m

where;

x_1 = 0.065 m

x_2  = 0.048 m

According to the conservation of energies;

E_1=E_2

∴

\Big(\dfrac{1}{2} \Big) kx_1^2 =\Big(\dfrac{1}{2} \Big) mv_2^2 + \Big(\dfrac{1}{2} \Big) kx_2^2

kx_1^2 = mv_2^2 + kx_2^2

(127.7 \ m) \times 0.065^2 = v_2^2 + (127.7 \ m) \times 0.048^2

0.5395325 = v_2^2 +0.2942208 \\ \\ 0.5395325  - 0.2942208 = v_2^2 \\ \\  v_2^2 = 0.2453117 \\ \\  v_2 = \sqrt{0.2453117} \\ \\ \mathbf{ v_2 \simeq0.50 \ m/s}

4 0
3 years ago
A 1.31 kg object is attached to a horizontal spring of force constant 2.70 N/cm and is started oscillating by pulling it 6.20 cm
goldfiish [28.3K]

Answer:

The answer is below

Explanation:

a) The change in energy is the difference between the final energy and the initial energy.

ΔE (energy change) = Ef (final energy) - Ei (initial energy)

\Delta E=\frac{1}{2}kA_f^2 -\frac{1}{2}kA_i^2\\\\k=force\ constant=2.7\ N/cm=270\ N/m, A_f=final\ dispalacment= 3.7\ cm=0.037\ m,\\ A_i=initial \ displacement = 6.2\ cm=0.062\ m\\\\Hence:\\\\\Delta E=\frac{1}{2}(270)(0.037)^2 -\frac{1}{2}(270)(0.062)^2\\\\\Delta E=-0.334 \ J

The negative sign shows that energy is lost to the environment. Hence 0.334 J is lost to the environment.

b) According to the law of conservation of energy, energy cannot be created or destroyed but transformed from one form to another.

The oscillating object loses energy due to wind resistance, friction between the spring and the object. Given that the air is frictionless, hence the energy loss is due to friction which is converted to heat.

6 0
3 years ago
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