The advantage of using the scientific method is that

A power station with an efficiency e generates W watts of electric power and dissipates D J of heat energy each second to the co

Andrews [41]

Answer: 13.94 tons/s

Explanation:

On adding heat energy to a substance, the temperature would be changed by a particular amount. This relationship between heat energy and temperature is often different for each material. The specific heat, is a value that describes how they relate.

Heat energy = mass flow rate * specific heat * Δ T

Q = MC (ΔΦ)

Heat energy, Q= 3.5*10^8J

Mass flow rate, M= ?

Specific heat, C= 4184j/KgC

Change in temperature, ΔΦ= 6°C

M = Q/CΔΦ

M = (3.5*10^8)/4184*6

M = 13942kg/s

M = 13.94 tons/s

3 0

3 years ago

Calculate the energy of the green light emitted, per photon, by a mercury lamp with a frequency of 5.49 × 1014 hz.

Tcecarenko [31]

The energy of a photon is given by
$E=hf$
where
$h=6.6 \cdot 10^{-34} Js$ is the Planck constant
f is the frequency of the photon

In our problem, the frequency of the light is
$f=5.49 \cdot 10^{14}Hz$
therefore we can use the previous equation to calculate the energy of each photon of the green light emitted by the lamp:
$E=hf=(6.6 \cdot 10^{-34}Js)(5.49 \cdot 10^{14} Hz)=3.62 \cdot 10^{-19} J$

8 0

3 years ago

PLEASE HELP WITH THIS QUESTION.

Sophie [7]

Answer:

#2 and #3 respectively

Explanation:

6 0

2 years ago

Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du

Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time $t$ is

$\theta=\dfrac\alpha2t^2$

It rotates 44.5 rad in this time, so we have

$44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}$

b. Since acceleration is constant, the average angular velocity is

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2$

where $\omega_f$ is the angular velocity achieved after 6.00 s. The velocity of the disk at time $t$ is

$\omega=\alpha t$

so we have

$\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}$

making the average velocity

$\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}$

Another way to find the average velocity is to compute it directly via

$\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}$

c. We already found this using the first method in part (b),

$\omega=14.8\dfrac{\rm rad}{\rm s}$

d. We already know

$\theta=\dfrac\alpha2t^2$

so this is just a matter of plugging in $t=12.0\,\mathrm s$ . We get

$\theta=179\,\mathrm{rad}$

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

$\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2$

Then for $t=6.00\,\rm s$ we would get the same $\theta=179\,\rm rad$ .

7 0

3 years ago

A 200g block on a 50cm long string swings in a circle, it's frictionless and 75rpm. What is its speed and tension on string

krek1111 [17]

Angular velocity = (75x2pie)/60
      =2.5pie ras^-1
linear velocity(or speed) at end of string, v = radius x angular velocity
       v= 0.5 x 2.5pie
   v=3.93 ms^-1

tension of string (I beleve is centeral force aplied by string), F= (mv^2)/r
      F= (0.2 x 3.93^2)/0.5
      F=6.18 N
(sorry if wrong)

3 0

2 years ago

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