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2 years ago

Did the kinetic frictional coefficient (for the wood/aluminum and felt/aluminum cases) vary with area of contact

Physics

1 answer:

lawyer [7]2 years ago

5 0

Answer:

Explanation:

Friction is defined as a force which acts at the surface of separation between two objects in contact and tends to oppose motion of one over the other.

While kinetic friction is the force that must be overcome so that a body can move with uniform speed over another.

Hence let consider one of the laws of friction which states that: '' Frictional force is independent of the area of the surfaces in contact.''

The value did not vary with area. This is because when calculating the kinetic fiction, the total contact area is not relevant and only the total weight of the system as well of as the block is put into consideration.

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The surface pressures at the bases of warm and cold columns of air are equal. air pressure in the warm column of air will ______

EastWind [94]

In atmospheric science, surface pressure is the atmospheric pressure at a location on Earth's surface. It is directly proportional to the mass of air over that location. For numerical reasons, atmospheric models such as general circulation models (GCMs) usually predict the nondimensional logarithm of surface pressure.

The answer is decrease more slowly

3 0

2 years ago

If no friction acts on a diver during a dive, then which of the following statements is true? A) The total mechanical energy of

EleoNora [17]

If no frictional work is considered, then the energy of the system (the driver at all positions is conserved.

Let
position 1 = initial height of the diver (h₁), together with the initial velocity (v₁).
position 2 = final height of the diver (h₂) and the final velocity (v₂).

The initial PE = mgh₁ and the initial KE = (1/2)mv₁²
where g = acceleration due to gravity,
m = mass of the diver.
Similarly, the final PE and KE are respectively mgh₂ and (1/2)mv₂².
PE in position 1 is converted into KE due to the loss in height from position 1 to position 2.

Therefore
(KE + PE) ₁ = (KE + PE)₂

Evaluate the given answers.
A) The total mechanical energy of the system increases.
FALSE

B) Potential energy can be converted into kinetic energy but not vice versa.
TRUE

C) (KE + PE)beginning = (KE + PE) end.
TRUE

D) All of the above.
FALSE

4 0

2 years ago

Read 2 more answers

Bradley gets an x-ray at a radiology clinic that employs its own technologists and radiologists. Would the coder at the clinic r

KengaRu [80]

Answer:

Explanation:

If Bradley examination was done and interpreted in the same facility, the radiologist code is used example- procedure code 72100- Radiologic examination, spine, lumbosacral, 2 or 3 views is reported.

if the X-ray was taken by Dr X but Dr X does not read or interpret the image but forward it to the radiologist for initial report, then a 26- modifier is used. E.g A reports by the technologist would be, procedure code 72050-Radiologic examination, spine, cervical, 2 or 3

views or 72050- TC in certain situations and the consulting radiologist would report 72050-26.

if Bradley’s x-ray were sent to an independent radiologist for interpretation, then the procedure code 76140 is used in reporting.

8 0

1 year ago

To a stationary observer, a man jogs east at 2.5 m/s and a woman jogs west at 1.5 m/s. from the woman's frame of reference, what

neonofarm [45]

T o a stationary observer, a man jogs east at 2.5 m/s and a woman jogs west at 1.5 m/s. from the woman's frame of reference, what is the man's velocity? it is 4m/s east

5 0

2 years ago

Read 2 more answers

A projectile is launched at an angle of 30° and lands 20 s later at the same height as it was launched. (a) What is the initial

Elina [12.6K]

Answer:

a)Initial speed of the projectile = 196.2 m/s

b)Maximum altitude = 490.5 m

c) Range of projectile = 3398.28 m

d) Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

Explanation:

Time of flight of a projectile is given by the expression,

$t=\frac{2usin\theta}{g}$

Here θ = 30° and t = 20 s

a) $t=\frac{2usin\theta}{g}\\\\20=\frac{2\times usin30}{9.81}\\\\u=196.2m/s$

Initial speed of the projectile = 196.2 m/s

b) Maximum altitude is given by

$H=\frac{u^2sin^2\theta}{2g}=\frac{196.2^2\times sin^230}{2\times 9.81}=490.5m$

Maximum altitude = 490.5 m

c) Range of projectile is given by

$R=\frac{u^2sin2\theta}{g}=\frac{196.2^2\times sin(2\times 30)}{9.81}=3398.28m$

Range of projectile = 3398.28 m

d) Horizontal velocity = ucosθ = 196.2 x cos 30 = 169.91 m/s

Vertical velocity = usinθ = 196.2 x sin 30 = 98.1 m/s

We have equation of motion s = ut + 0.5 at²

Horizontal motion

u = 169.91 m/s

a = 0 m/s²

t = 15 s

Substituting

s = 169.91 x 15 + 0.5 x 0 x 15² = 2548.71 m

Vertical motion

u = 98.1 m/s

a = -9.81 m/s²

t = 15 s

Substituting

s = 98.1 x 15 + 0.5 x -9.81 x 15² = 367.88 m

$\texttt{Total displacement =}\sqrt{2548.71^2+367.88^2}=2575.12m$

Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

7 0

2 years ago