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Montano1993 [528]
2 years ago
11

A closed system’s internal energy changes by 178 J as a result of being heated with 658 J of energy. The energy used to do work

by the system is J.
Physics
1 answer:
nika2105 [10]2 years ago
6 0

Answer:

480J

Explanation:

Using the formula:

Delta U = Q - W

Q:Heat (J)

Delta U: Changes in internal Energy (J)

W:Work (J)

We can plug in the give numbers, Q and W.

Delta U = 658J - 178J = 480J

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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
In a particular metal, the mobility of the mobile electrons is 0.0033 (m/s)/(N/C). At a particular moment, the electric field ev
Lapatulllka [165]

Answer:

the average drift speed of the mobile electrons in the metal is 1.089 x 10⁻⁴ m/s.

Explanation:

Given;

mobility of the mobile electrons in the metal, μ = 0.0033 (m/s)/(N/C)

the electric field strength inside the cube of the metal, E = 0.033 N/C

The average drift speed of the mobile electrons in the metal is calculated as;

v = μE

v =  0.0033 (m/s)/(N/C) x 0.033 N/C

v = 1.089 x 10⁻⁴ m/s.

Therefore, the average drift speed of the mobile electrons in the metal is 1.089 x 10⁻⁴ m/s.

6 0
3 years ago
Exercise will not help maintain the health of your endocrine system.
Mashcka [7]
A. True

Hope this helps :)
6 0
2 years ago
A yellow and green car traveled 400 miles to Dayton, OH. The green car made the trip in 10 hours. The yellow arrived in 8 hours.
elixir [45]

Answer: C)The yellow car was faster. Yellow traveled at a speed of 50 mph while green was traveling at an average of 40 mph.

Explanation:

The speed of each car is defined as:

v=\frac{d}{t}

where d is the distance traveled by the car and t is the time taken.

For the yellow car, d=400 mi and t=8 h, so its speed is

v=\frac{400 mi}{8 h}=50 mph

For the green car, d=400 mi and t=10 h, so its speed is

v=\frac{400 mi}{10 h}=40 mph

So, the correct choice is

C)The yellow car was faster. Yellow traveled at a speed of 50 mph while green was traveling at an average of 40 mph.

4 0
3 years ago
Read 2 more answers
Which statement describes how work and power are similar?
postnew [5]
The statement that describes how work and power are similar is D. you must know time and energy to calculate both.
I am not completely sure though, so I hope this helps. :)
6 0
3 years ago
Read 2 more answers
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