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Marat540 [252]
3 years ago
6

How much energy is contained in matter what the mass of 1 gram (0.001 kilogram)

Physics
1 answer:
kaheart [24]3 years ago
6 0

Classically and Newtonianly, it's the sum of the chemical energy if any,
the electrical energy if any, the thermal energy if any, and the mechanical
energy consisting of potential and kinetic energy if any.

The mechanical energy, consisting of potential and kinetic energy if any, is

           0.001 x [ (acceleration of gravity x height) +  (1/2) (speed)² ] .

But I've got a sneaky hunch that you're not talking about any of these.
You want to know how much [ <em><u>mc</u>² </em>] there is in 1 gram of mass.  No prob.

           E = m c² = (0.001) x (3 x 10⁸)² = <em>9 x 10¹³ joules</em>

That's the energy that a 1,000-watt toaster uses
in  <em>2,852 years</em>  of continuous toasting.


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Therese plays lacrosse, but she has injured her ankle and has to take a few weeks off. She is finding that she has a lot of trou
vivado [14]

Answer: No endorphins

Explanation: cuz' she injured her ankle

6 0
2 years ago
The gauge pressure inside an alveolus with a 200 µm radius is 25 mmHg, while the blood pressure outside is only 10 mmHg. Assumin
alekssr [168]

Answer:

The surface tension is 0.0318 N/m and is sufficiently less than the surface tension of the water.

Solution:

As per the question:

Radius of an alveolus, R = 200\mu m = 200\times 10^{- 6}\ m

Gauge Pressure inside, P_{in} = 25\ mmHg

Blood Pressure outside, P_{o} = 10\ mmHg

Now,

Change in pressure, \Delta P = 25 - 10 = 15\ mmHg = 1.99\times 10^{3}\ Pa

Since the alveolus is considered to be a spherical shell

The surface tension can be calculated as:

\Delta P = \frac{4\pi T}{R}

T = \frac{1.99\times 10^{3}\times 200\times 10^{- 6}}{4\pi} = 0.0318\ N/m = 0.318\ mN/m

And we know that the surface tension of water is 72.8 mN/m

Thus the surface tension of the alveolus is much lesser as compared to the surface tension of water.

3 0
3 years ago
How is power defined? Question 4 options: the quantity of work accomplished the direction of the work the total distance an obje
GarryVolchara [31]

Power can be defined as the rate at which work is accomplished.

Option D is the correct answer.

<h3></h3><h3>Power </h3>

The work done by an object in a given time interval is called the power of that object.

Suppose an external force F is applied to any object for the time interval T seconds. Due to this external force, the object will perform some amount of work for the time T seconds. This work W done by the object for the time interval T seconds is called the power of that object.

Power can be defined in mathematical term which is given below.

P = \dfrac {W}{T} \;\rm Watts

Thus the power can also be defined as the work done by the object per unit time interval.

Hence we can conclude that option D is the correct answer.

To know more about power, follow the link given below.

brainly.com/question/1618040.

8 0
2 years ago
Which body is in equilibrium?
sergey [27]
"(1) a satellite moving around Earth in a circular <span>orbit" is the only option from the list that describes an object in equilibrium, since velocity and gravity are working together to keep the orbit constant. </span>
6 0
2 years ago
Read 2 more answers
A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s
aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is U_{s} = 0

And, initial potential gravitational potential energy of the rod is U_{g} = \frac{mgL}{2}.

It is given that,

       mass of the bar = 0.795 kg

            g = 9.8 m/s^{2}

           L = length of the rod = 0.2 m

Initial total energy T = \frac{mgL}{2}

Now, when the rod is in horizontal position then final total energy will be as follows.

            T = \frac{1}{2}kx^{2} + I \omega^{2}

where,    I = moment of inertia of the rod about the end = \frac{mL^{2}}{3}

Also,    \omega = \frac{\nu}{L}

where,    \nu = speed of the tip of the rod

              x = spring extension

The initial unstrained length is x_{o} = 0.1 m

Therefore, final length will be calculated as follows.

              x' = \sqrt{(0.2)^{2} + (0.1)^{2}} m

Then,  x = x' - x_{o}

          x = \sqrt{(0.2)^{2} + (0.1)^{2}} m - 0.1 m

             = 0.1236 m

       k = 25 N/m

So, according to the law of conservation of energy

       \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}

      \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

Putting the given values into the above formula as follows.

   \frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}

  \frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}

          v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

7 0
3 years ago
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