begin with vibrations of the eardrum. Sound waves cause the eardrum to vibrate. Cells in the cochlea detect the vibrations and send a message to the brain.

3 0

3 years ago

Read 2 more answers

A 0.473 kg ice puck, moving east with a speed of 2.76 m/s, has a head-on collision with a 0.819 kg puck initially at rest. Assum

Gekata [30.6K]

Answer:

The final speed of puck 1 is 0.739 m/s towards west and puck 2 is 2.02 m/s towards east .

Explanation:

Let us consider east as positive direction and west as negative direction .

Given

mass of puck 1 , $m_1= 0.473 kg$

mass of puck 2 , $m_2= 0.819 kg$

initial speed of puck 1 , $u_1=2.76\frac{m}{s}$

initial speed of puck 2 , $u_2=0.00\frac{m}{s}$

Final speed of puck 1 and puck 2 be $v_1\, and\, v_2$ respectively

Apply conservation of linear momentum

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

=> $0.473\times 2.76+0.0=0.473\times v_1+0.819\times v_2$

=> $1.594=0.5775\times v_1+ v_2$ -----(A)

Since collision is perfectly elastic , coefficient restitution e=1

$u_2-u_1=v_1-v_2$

=> $0-2.76=v_1-v_2$ ------(B)

From equation (A) and (B)

$v_1=-0.739\frac{m}{s}$

and $v_2=2.02\frac{m}{s}$

Thus the final speed of puck 1 is 0.739 m/s towards west and puck 2 is 2.02 m/s towards east .

3 0

3 years ago

The Cosmo Clock 21 Ferris wheel in Yokohama, Japan, has a diameter of 100 m. Its name comes from its 60 arms, each of which can

Sonja [21]

Answer:

Explanation:

a ) speed of passenger = circumference / time

= 2π R / Time

= 2 x 3.14 x 50 / 60

= 5.23 m /s

b )

centrifugal force = m v² /R

= (882 /9.8 ) x 5.23² / 50

= 77.47 N

Apparent weight at the highest point

real weight - centrifugal force

= 882 - 77.47

= 804.53 N

Apparent weight at the lowest point

real weight + centrifugal force

= 882 +77.47

= 959.47 N

c ) if the passenger’s apparent weight at the highest point were zero

centrifugal force = weight

mv² /R = mg

v² = gR

= 9.8 X 50

v = 22.13 m /s

d )

apparent weight

mg - mv² / R

= 882 - (882 / 9.8 )x 22.13²/50

= 882 + 882

= 1764 N

6 0

3 years ago

The weight of a block on the inclined plane is 500 N and the angle of incline is 30 degrees. What is the magnitude of the force

yanalaym [24]

Answer:

250 N

433 N

Explanation:

N = Normal force by the surface of the inclined plane

W = Weight of the block = 500 N

f = static frictional force acting on the block

Parallel to incline, force equation is given as

f = W Sin30

f = (500) Sin30

f = 250 N

Perpendicular to incline force equation is given

N = W Cos30

N = (500) Cos30

N = 433 N

3 0

3 years ago