Which part of the ear vibrates at the frequency of sound waves

A force of 5.0 N acts on a 15 kg body initially at rest. Compute the work done by the force in (a) the first, (b) the second, an

Leokris [45]

Answer:

(a) 0.833 j

(b) 2.497 j

(c) 4.1625 j

(d) 4.995 watt

Explanation:

We have given force F = 5 N

Mass of the body m = 15 kg

So acceleration $a=\frac{F}{m}=\frac{5}{15}=0.333m/sec^2$

As the body starts from rest so initial velocity u = 0 m/sec

(a) From second equation of motion $s=ut+\frac{1}{2}at^2$

For t = 1 sec

$s=0\times 1+\frac{1}{2}\times 0.333\times 1^2=0.1666m$

We know that work done W =force × distance = 5×0.1666 =0.833 j

(b) For t = 2 sec

$s=0\times 2+\frac{1}{2}\times 0.333\times 2^2=0.666m$

We know that work done W =force × distance = 5×0.666 =3.33 j

So work done in second second = 3.33-0.833 = 2.497 j

(c) For t = 3 sec

$s=0\times 3+\frac{1}{2}\times 0.333\times 3^2=1.4985m$

We know that work done W =force × distance = 5×1.4985 =7.4925 j

So work done in third second = 7.4925 - 2.497 -0.833 = 4.1625 j

(d) Velocity at the end of third second v = u+at

So v = 0+0.333×3 = 0.999 m /sec

We know that power P = force × velocity

So power = 5× 0.999 = 4.995 watt

4 0

4 years ago

How does temperature change as you go from the surface toward the center of the earth?

Nastasia [14]

Because lava is in the center and lavas hot

8 0

3 years ago

If the coefficient of friction is 0.3900 and the cylinder has a radius of 2.700 m, what is the minimum angular speed of the cyli

Aleks04 [339]

Answer:

w=3.05 rad/s or 29.88rpm

Explanation:

k = coefficient of friction = 0.3900

R = radius of the cylinder = 2.7m

V = linear speed of rotation of the cylinder

w = angular speed = V/R or to rewrite V = w*R

N = normal force to cylinder

N= $=\frac{m(V)^{2}}{R}=m*(w)^2*R$

$Friction force\\Ff = k*N\\Ff= k*m*w^2*R$

$Gravitational force \\Fg = m*g$

These must be balanced (the net force on the people will be 0) so set them equal to each other.

$Fg = Ff$

$m*g = k*m*w^2*R$

$g=k*w^{2}*R$

$w^2 =\frac{g}{k*R}$

$w=\sqrt{\frac{g}{k*R}} \\w =\sqrt{\frac{9.8\frac{m}{s^{2}}}{0.3900*2.7m}}\\ w=\sqrt{9.306}=3.05 \frac{rad}{s}$

There are 2*pi radians in 1 revolution so:

$RPM=\frac{w}{2\pi }*60\\RPM=\frac{3.05\frac{rad}{s}}{2\pi}*60\\RPM= 0.498*60\\RPM=29.88$

So you need about 30 RPM to keep people from falling out the bottom

7 0

4 years ago

CO2<br> NaCl<br> HCl<br> These may all be classified as

Nataly_w [17]

Gas because that’s what the periodic table says

6 0

3 years ago

Read 2 more answers

A machine that changes only the direction of a force has a mechanical advantage of

8090 [49]

Answer:

1

Explanation:

The word "only"= one!

5 0

3 years ago

Read 2 more answers