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icang [17]
2 years ago
15

A compression spring with a spring constant of 200 N/m catches a ball that's falling with

Physics
1 answer:
mrs_skeptik [129]2 years ago
3 0

Answer:

distance of compression: 0.07071 m

Explanation:

\sf energy = \dfrac{1}{2} kx^2

\rightarrow \sf 0.5= \dfrac{1}{2} (200)x^2

\rightarrow \sf 0.5= (100)x^2

\rightarrow \sf 5\ *\ 10^{-3}= x^2

\rightarrow \sf x = \sqrt{5\ *\ 10^{-3}}

\rightarrow \sf x =0.07071 \ m

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A bomb, originally sitting at rest, explodes and during the
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Answer:

opposite

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2 years ago
A car is traveling at 20 km in 15 minutes. What is its average velocity
yarga [219]
The velocity is 4374.45 m/s.
I got the answer by using v=d/t.
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7 0
1 year ago
Suppose you could suddenly increase the speed of every molecule in a gas by a factor of 2. Would the gas pressure increase by a
Elena L [17]

<h2>The pressure will become double </h2>

Explanation:

The gas pressure is directly proportional to the mean root square velocity of the constituent molecules of gas .

P ∝ \sqrt{\frac{C_1^2+C_2^2+C_3^2----C_n^2}{n} }      I

Here C₁ , C₂ ------------ Cₙ is the velocities of molecules .

By making these velocities double

The pressure P₀ ∝ 2\sqrt{\frac{C_1^2+C_2^2+C_3^2----C_n^2}{n} }          II

By dividing II by I

P₀ = 2 P

Thus pressure will become double than its previous value

5 0
3 years ago
Water, initially saturated vapor at 4 bar, fills a closed, rigid container. The water is heated until its temperature is 360°C.
salantis [7]

Explanation:

Using table A-3, we will obtain the properties of saturated water as follows.

Hence, pressure is given as p = 4 bar.

u_{1} = u_{g} = 2553.6 kJ/kg

v_{1} = v_{g} = 0.4625 m^{3}/kg

At state 2, we will obtain the properties. In a closed rigid container, the specific volume will remain constant.

Also, the specific volume saturated vapor at state 1 and 2 becomes equal. So, v_{2} = v_{g} = 0.4625 m^{3}/kg

According to the table A-4, properties of superheated water vapor will obtain the internal energy for state 2 at v_{2} = v_{g} = 0.4625 m^{3}/kg and temperature T_{2} = 360^{o}C so that it will fall in between range of pressure p = 5.0 bar and p = 7.0 bar.

Now, using interpolation we will find the internal energy as follows.

 u_{2} = u_{\text{at 5 bar, 400^{o}C}} + (\frac{v_{2} - v_{\text{at 5 bar, 400^{o}C}}}{v_{\text{at 7 bar, 400^{o}C - v_{at 5 bar, 400^{o}C}}}})(u_{at 7 bar, 400^{o}C - u_{at 5 bar, 400^{o}C}})

     u_{2} = 2963.2 + (\frac{0.4625 - 0.6173}{0.4397 - 0.6173})(2960.9 - 2963.2)

                   = 2963.2 - 2.005

                   = 2961.195 kJ/kg

Now, we will calculate the heat transfer in the system by applying the equation of energy balance as follows.

      Q - W = \Delta U + \Delta K.E + \Delta P.E ......... (1)

Since, the container is rigid so work will be equal to zero and the effects of both kinetic energy and potential energy can be ignored.

            \Delta K.E = \Delta P.E = 0

Now, equation will be as follows.

           Q - W = \Delta U + \Delta K.E + \Delta P.E

           Q - 0 = \Delta U + 0 + 0

           Q = \Delta U

Now, we will obtain the heat transfer per unit mass as follows.

          \frac{Q}{m} = \Delta u

         \frac{Q}{m} = u_{2} - u_{1}

                      = (2961.195 - 2553.6)

                      = 407.595 kJ/kg

Thus, we can conclude that the heat transfer is 407.595 kJ/kg.

4 0
2 years ago
What type of seismic waves travel through Earth?
ycow [4]

Answer:d. Ground waves

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8 0
2 years ago
Read 2 more answers
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