Select the correct answer.

72.0-kg person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface of the door. The doorknob is locat

expeople1 [14]

Answer: $I=2 kg-m^2$

Explanation:

Given

mass $m=72 kg$

Force $F=5 N$

door knob is located at a distance of r=0.8 m from axis

Angular acceleration of door $\alpha =2 rad/s^2$

Torque $T=I\alpha =F\times r$

where I=moment of inertia

$5\times 0.8=I\times 2$

$I=2 kg-m^2$

4 0

4 years ago

g A 4-foot spring is elongated167feet long after a mass weighing 16 pounds is attached to it. The medium throughwhich the mass m

marta [7]

Answer: hello question b is incomplete attached below is the missing question

a) attached below

b) V = 0.336 ft/s

Explanation:

Elongation ( Xo) = 16/ 7 feet

mass attached to 4-foot spring = 16 pounds

medium has 9/2 times instanteous velocity

<u>a) Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of 2 ft/s</u>

The motion is an underdamped motion because the value of β < Wo

Wo = 3.741 s^-1

attached below is a detailed solution of the question

3 0

3 years ago

If you take a deep breath of air, you can float more easily in water because you have _____.

Leni [432]

Answer:

the answer is c

6 0

3 years ago

Read 2 more answers

A disk 7.90 cm in radius rotates at a constant rate of 1 190 rev/min about its central axis. (a) Determine its angular speed. 12

Tanya [424]

Answer:

$124.62\ \text{rad/s}$

$3.71\ \text{m/s}$

$1.23\ \text{km/s}^2$

$20.28\ \text{m}$

Explanation:

r = Radius of disk = 7.9 cm

N = Number of revolution per minute = 1190 rev/minute

Angular speed is given by

$\omega=N\dfrac{2\pi}{60}\\\Rightarrow \omega=1190\times \dfrac{2\pi}{60}\\\Rightarrow \omega=124.62\ \text{rad/s}$

The angular speed is $124.62\ \text{rad/s}$

r = 2.98 cm

Tangential speed is given by

$v=r\omega\\\Rightarrow v=2.98\times 10^{-2}\times 124.62\\\Rightarrow v=3.71\ \text{m/s}$

Tangential speed at the required point is $3.71\ \text{m/s}$

Radial acceleration is given by

$a=\omega^2r\\\Rightarrow a=124.62^2\times 7.9\times 10^{-2}\\\Rightarrow a=1226.88\approx 1.23\ \text{km/s}^2$

The radial acceleration is $1.23\ \text{km/s}^2$ .

t = Time = 2.06 s

Distance traveled is given by

$d=vt\\\Rightarrow d=\omega rt\\\Rightarrow d=124.62\times 7.9\times 10^{-2}\times 2.06\\\Rightarrow d=20.28\ \text{m}$

The total distance a point on the rim moves in the required time is $20.28\ \text{m}$ .

8 0

3 years ago

Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light.

pogonyaev

Complete Question

Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcseconds in an arcminute).Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light

Answer:

The diameter is $D = 0.59 \ m$