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3 years ago

The total amount of energy contained in an object

1 answer:

7 0

The total amount of energy stored in the particles of an object is called its internal energy. The internal energy of an object is made up of the kinetic energy due to the random motion of the particles and the potential energy due to the interactive forces among the particles.

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The unit of energy is derived unit. ( give reasons)

m_a_m_a [10]

Answer:

Because it is made by two different unit force (F) and displacement(s)

3 0

3 years ago

An engine has a hot-reservoir temperature of 980 K and a cold-reservoir temperature of 570 K. The engine operates at three-fifth

hram777 [196]

Answer:

efficiency of engine is 0.21

Explanation:

given:

Th=980 K

Tc=570 K

since efficiency of engine is given as:

n=3/5*n_m

=3/5(1-Tc/Th)

=0.21

efficiency of engine is 0.21

7 0

3 years ago

Read 2 more answers

Given three vectors A = 24i + 33j, B = 55i - 12j and C = 2i + 43j (a) Find the magnitude of each vector. (b) Write an expression

slega [8]

Answer:

(a) , . and .

(b) $\vec A - \vec C=22 \hat i -10 \hat j$ .

Explanation:

Given that,

and

$\vec {C}=2 \hat i +43 \hat j$

(a) The magnitude of a vector is the square root of the sum of the square of all the components of the vector, i.e. for a ,.

So, the magnitude of the is

$|\vec A|=\sqrt {24^2+ 33^2}$

$\Rightarrow |\vec A|=\sqrt {1665}$

The magnitude of the is

$|\vec B|=\sqrt {55^2+ (-12)^2}$

$\Rightarrow |\vec B|=\sqrt {3169}$

And, the magnitude of the is

$|\vec C|=\sqrt {2^2+ 43^2}$

$\Rightarrow |\vec C|=\sqrt {1853}$

(b) The difference between the two vectors is the difference between the corresponding components of the vectors. So, the required expression of is

$\vec A - \vec C=(24 \hat i +33 \hat j) - (2 \hat i +43 \hat j)$

$\Rightarrow \vec A - \vec C=24 \hat i +33 \hat j - 2 \hat i -43 \hat j$

$\Rightarrow \vec A - \vec C=22 \hat i -10 \hat j$

$\vec A - \vec N=(24 \hat i +33 \hat j) - (55 \hat i -12 \hat j)$

$\Rightarrow \vec A - \vec B=24 \hat i +33 \hat j - 55\hat i +12 \hat j$

$\Rightarrow \vec A - \vec B=-31 \hat i +45 \hat j\;\cdots (i)$

The magnitude of is

$|\vec A - \vec B|=\sqrt {(-31)^2+55^2}$

$\Rightarrow |\vec A - \vec B|=\sqrt {3986}$

$\Rightarrow |\vec A - \vec B|=63.13$

Now, if a vector $\vec V= -\alpha \hat i +\beta \hat j$ in 3rd quadrant having direction $\theta$ with respect to $\hat i$ direction, than

in the anti-clockwise direction.

Here, from equation (i), for the vector $\vec A - \vec C$ , $\alpha=31$ and $\beta=45$ .

$\Rightarrow \theta = \pi-\tan ^{-1}\left(\frac {45}{31}\right)$

180°-55.44° [as \pi radian= 180°]

124.56° in the anti-clockwise direction.

(d) Vector diagrams for $\vec A +\vec B$ and $\vec A - \vec B$ has been shown

in the figure(b) and figure(c) recpectively.

Vector $\vec A - \vec B$ is in 3rd quadrant as calculated in part (c).

While Vector $\vec A +\vec B=(24 \hat i +33 \hat j)+(55 \hat i -12 \hat j)$

$\Rightarrow \vec A +\vec B=79 \hat i +21 \hat j$ , which is in 1st quadrant as both the components are position has been shown in figure(b).

6 0

3 years ago

How much heat energy must be added to 52kg Of water at 68°F to raise the temperature to 212°F? The specific heat capacity for wa

Anna [14]

Answer:

The amount of energy added to rise the temperature Q = 17413.76 KJ

Explanation:

Mass of water = 52 kg

Initial temperature $T_{1}$ = 68 °F = 20° c

Final temperature $T_{2}$ = 212 °F = 100° c

Specific heat of water $C = 4.186 \frac{KJ}{kg c}$

Now heat transfer Q = m × C × ( $T_{2}$ - $T_{1}$ )

⇒ Q = 52 × 4.186 × ( 100 - 20 )

⇒ Q = 17413.76 KJ

This is the amount of energy added to rise the temperature.

4 0

3 years ago

There are 120 degrees in a straight angle. true false

mixas84 [53]

That would be false.
Straight angle= 180 degrees
Obtuse angle= Above 90, and lower than 180
Acute angle= Below 90
Hope this helps! (:

6 0

3 years ago

Read 2 more answers