Question 1 of 10

A fl oor polisher has a rotating disk that has a 15-cm radius. The disk rotates at a constant angular velocity of 1.4 rev/s and

N76 [4]

Answer:

59.4 meters

Explanation:

The correct question statement is :

A floor polisher has a rotating disk that has a 15-cm radius. The disk rotates at a constant angular velocity of 1.4 rev/s and is covered with a soft material that does the polishing. An operator holds the polisher in one place for 4.5 s, in order to buff an especially scuff ed area of the floor. How far (in meters) does a spot on the outer edge of the disk move during this time?

Solution:

We know for a circle of radius r and θ angle by an arc of length S at the center,

S=rθ

This gives

θ=S/r

also we know angular velocity

ω=θ/t where t is time

or

θ=ωt

and we know

1 revolution =2π radians

From this we have

angular velocity ω = 1.4 revolutions per sec = 1.4×2π radians /sec = 1.4×3.14×2×= 8.8 radians / sec

Putting values of ω and time t in

θ=ωt

we have

θ= 8.8 rad / sec × 4.5 sec

θ= 396 radians

We are given radius r = 15 cm = 15 ×0.01 m=0.15 m (because 1 m= 100 cm and hence, 1 cm = 0.01 m)

put this value of θ and r in

S=rθ

we have

S= 396 radians ×0.15 m=59.4 m

7 0

4 years ago

Read 2 more answers

Suppose your surface body temperature averaged 90 degrees F. How much radiant energy in W/m^2 would be emitted from your body?

Debora [2.8K]

$493 \; \text{W}\cdot \text{m}^{-2}$ .

<h3>Explanation</h3>

The Stefan-Boltzmann Law gives the energy radiation <em>per unit area</em> of a black body:

$\dfrac{P}{A} = \sigma \cdot T^{4}$

where,

$P$ the total power emitted,
$A$ the surface area of the body,
$\sigma$ the Stefan-Boltzmann Constant, and
$T$ the temperature of the body in degrees Kelvins.

$\sigma = 5.67 \times 10^{-8} \;\text{W}\cdot \text{m}^{-2} \cdot \text{K}^{-4}$ .

$T = 90 \; \textdegree{}\text{F} = (\dfrac{5}{9} \cdot (90-32) + 273.15) \; \text{K} = 305.372 \; \text{K}$ .

$\dfrac{P}{A} = \sigma \cdot T^{4} = 5.67 \times 10^{-8} \times 305.372^{4} = 493\; \text{W}\cdot \text{m}^{-2}$ .

Keep as many significant figures in $T$ as possible. The error will be large when $T$ is raised to the power of four. Also, the real value will be much smaller than $493\; \text{W}\cdot \text{m}^{-2}$ since the emittance of a human body is much smaller than assumed.