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3 years ago

How can you tell that toast is not warmed up by conduction?

Physics

1 answer:

solmaris [256]3 years ago

8 0

Answer:

Toasting in a toaster is usually considered by (infra red) radiation. But the hot coils touch the toast so an element of heating by conduction occurs as well.

Explanation:

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The graph illustrates the activity level of three common digestive enzymes, across a range of pH values. Which enzyme is likely

vaieri [72.5K]

Answer:

(A) Pepsin

Explanation:

From the graph it is clear that pepsin is the only enzyme which works in highly acidic condintion in the digestive system.

less than 7 the liquid is acidic
above 7 the liquid is basic
at 7 the liquid is neutral

It has an optimum pH of about 1.5 at which its activity level is 8.5 as shown in graph.

7 0

3 years ago

State the meaning of specific heat capacity of copper is 380 J/kg°C.

Arisa [49]

Answer:

Specific heat capacity means that in order to raise copper by 1C per kg, it needs 380J of heat energy.

7 0

3 years ago

Practical steam engines utilize 450ºC steam, which is later exhausted at 270ºC.

Naily [24]

(a) 0.249 (24.9 %)

The maximum efficiency of a heat engine is given by

$\eta = 1-\frac{T_C}{T_H}$

where

Tc is the low-temperature reservoir

Th is the high-temperature reservoir

For the engine in this problem,

$T_C = 270^{\circ}C+273=543 K$

$T_H = 450^{\circ}C+273=723 K$

Therefore the maximum efficiency is

$\eta = 1-\frac{T_C}{T_H}=1-\frac{543}{723}=0.249$

(b-c) 0.221 (22.1 %)

The second steam engine operates using the exhaust of the first. So we have:

$T_H = 270^{\circ}C+273=543 K$ is the high-temperature reservoir

$T_C = 150^{\circ}C+273=423 K$ is the low-temperature reservoir

If we apply again the formula of the efficiency

$\eta = 1-\frac{T_C}{T_H}$

The maximum efficiency of the second engine is

$\eta = 1-\frac{T_C}{T_H}=1-\frac{423}{543}=0.221$

8 0

3 years ago

A 100-N force causes an object to<br> accelerate at 2 m/s/s. What is the<br> mass of the object?

Kitty [74]

Answer:

$50\; \rm kg$ .

Explanation:

By Newton's Second Law, the acceleration $a$ of an object is proportional to the net force $\sum F$ on it. In particular, if the mass of the object is $m$ , then

$\sum F = m \cdot a$ .

Rewrite this equation to obtain:

$\displaystyle m = \frac{\sum F}{a}$ .

In this case, the assumption is that the $100\; \rm N$ force is the only force that is acting on the object. Hence, the net force $\sum F$ on the object would also be

Make sure that all values are in their standard units. Forces should be in Newtons (same as $\rm kg \cdot m \cdot s^{-2}$ , and the acceleration of the object should be in meters-per-second-squared ( $\rm m \cdot s^{-2}$ ). Apply the equation $\displaystyle m = \frac{\sum F}{a}$ to find the mass of the object.

$\displaystyle m = \frac{100\; \rm N}{2\; \rm m \cdot s^{-2}} = 50\; \rm kg$ .

4 0

3 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