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3 years ago

In the context of energy transfers with hot and cold reservoirs, the sign convention is that _______________.

Physics

1 answer:

Likurg_2 [28]3 years ago

8 0

Answer:

B. QC > 0; QH < 0

Explanation:

Given that there are two reservoir of energy.

Sign convention for heat and work :

1.If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.

2. If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.

From hot reservoir heat is going out that is why it is taken as negative

$Q_H$

From cold reservoir heat is coming inside the reservoir that is why it is taken as positive

$Q_C>0$

That is why the answer will be

$Q_H$ , $Q_C>0$

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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.

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3 years ago

A 5 newton force and a 7 newton force act concurrently on a point. As the angle between the forces is increased from 0 to 180 th

Reika [66]

Answer:

The magnitude of the resultant decreases from A+B to A-B

Explanation:

The magnitude of the resultant of two vectors is given by

$R=\sqrt{A^2 +B^2 +2AB cos \theta}$

where

A is the magnitude of the first vector

B is the magnitude of the second vector

$\theta$ is the angle between the directions of the two vectors

In the formula, A and B are constant, so the behaviour depends only on the function $cos \theta$ . The value of $cos \theta$ are:

- 1 (maximum) when the angle is 0, so the magnitude of the resultant in this case is

$R=\sqrt{A^2 +B^2+2AB}=\sqrt{(A+B)^2}=A+B$

- then it decreases, until it becomes 0 when the angle is 90 degrees, where the magnitude of the resultant is

$R=\sqrt{A^2 +B^2+0}=\sqrt{A^2+B^2}$

- then it becomes negative, and continues to decrease, until it reaches a value of -1 when the angle is 180 degrees, and the magnitude of the resultant is

$R=\sqrt{A^2 +B^2-2AB}=\sqrt{(A-B)^2}=A-B$

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3 years ago

A gasoline tank has the shape of an inverted right circular cone with base radius 4 meters and height 5 meters. Gasoline is bein

RSB [31]

Answer:

h'=0.25m/s

Explanation:

In order to solve this problem, we need to start by drawing a diagram of the given situation. (See attached image).

So, the problem talks about an inverted circular cone with a given height and radius. The problem also tells us that water is being pumped into the tank at a rate of $8m^{3}/s$ . As you may see, the problem is talking about a rate of volume over time. So we need to relate the volume, with the height of the cone with its radius. This relation is found on the volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi r^{2}h$

notie the volume formula has two unknowns or variables, so we need to relate the radius with the height with an equation we can use to rewrite our volume formula in terms of either the radius or the height. Since in this case the problem wants us to find the rate of change over time of the height of the gasoline tank, we will need to rewrite our formula in terms of the height h.

If we take a look at a cross section of the cone, we can see that we can use similar triangles to find the equation we are looking for. When using similar triangles we get:

$\frac {r}{h}=\frac{4}{5}$

When solving for r, we get:

$r=\frac{4}{5}h$

so we can substitute this into our volume of a cone formula:

$V_{cone}=\frac{1}{3} \pi (\frac{4}{5}h)^{2}h$

which simplifies to:

$V_{cone}=\frac{1}{3} \pi (\frac{16}{25}h^{2})h$

$V_{cone}=\frac{16}{75} \pi h^{3}$

So now we can proceed and find the partial derivative over time of each of the sides of the equation, so we get:

$\frac{dV}{dt}= \frac{16}{75} \pi (3)h^{2} \frac{dh}{dt}$

Which simplifies to:

$\frac{dV}{dt}= \frac{16}{25} \pi h^{2} \frac{dh}{dt}$

So now I can solve the equation for dh/dt (the rate of height over time, the velocity at which height is increasing)

So we get:

$\frac{dh}{dt}= \frac{(dV/dt)(25)}{16 \pi h^{2}}$

Now we can substitute the provided values into our equation. So we get:

$\frac{dh}{dt}= \frac{(8m^{3}/s)(25)}{16 \pi (4m)^{2}}$

so:

$\frac{dh}{dt}=0.25m/s$

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2 years ago

Why do the graphs differ?

melisa1 [442]

Well first graph represents rectangular hyperbola

vu = c^2 ( c is constant)

AS 1/v + 1/u = 1/f

Take1/ f to be constant c

1/v = c - 1/u

it is of the form y = - x + k

Slope = -1 having intercept k as shown in fig 2

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3 years ago

You push a coin across a table. The coin stops. How does this motion relate to balanced and unbalanced forces?

Snezhnost [94]

If you are pushing the coin across the table at a constant rate, the friction of the table and the horizontal force of your hand pushing are equal, and the coin itself moves at a constant rate. If you push a coin and let it go, there is no horizontal force keeping the coin going. Friction slows the coin to a stop. In both cases, the gravitational downward pull of Earth is equally but oppositely resisted by the upward push of table on the coin.

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3 years ago