I think transfers is the answer
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
Answer: You do not specify what is being asked for. ∆E? ∆H?
∆E = (430 - 238) J = 192 J
∆H = 430 J
Explanation:
If asked for the value of ∆H the answer is simply the change in heat, and in the question, it states introduction of 430 J of heat is causing the system to expand.
Therefore ∆H = 430 J
If asked for ∆E, we know that ∆E = ±q (heat) + work (-P∆V) = ±q + w
The question states that 238 J of work are done AND the system expanded
(work is negative because expansion means work is done BY the system, releasing energy/heat... Conversely, if the system were compressed, work is done ON the system, absorbing heat/energy)
Therefore, ∆E = (430 - 238) J = 192 J
<span>electromagnetic.........</span>
