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MariettaO [177]
2 years ago
9

AGREE OR DISAGREE: All light can be seen.

Physics
1 answer:
Tema [17]2 years ago
3 0

Answer:

Disagree

Explanation:

We cannot see infared lights.

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Which of the following is an example of kinetic energy
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Running because kinetic energey is the engery created by motion
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A 120-V rms voltage at 60.0 Hz is applied across an inductor, a capacitor, and a resistor in series. If the rms value of the cur
klemol [59]

Answer:

The impedance of this circuit is 200 ohm.

Explanation:

Given that,

rms voltage = 120 v

Frequency = 60.0 Hz

rms current = 0.600 A

We need to calculate the impedance

Using formula of impedance

Z=\dfrac{V_{rms}}{I_{rms}}

Where, V_{rms} = rms voltage

I_{rms} = rms current

Z= impedance

Put the value into the formula

Z=\dfrac{120}{0.600}

Z=200\ Omega

Hence, The impedance of this circuit is 200 ohm.

8 0
3 years ago
Two objects, one of mass m and the other of mass 2m, are dropped from the top of a building. when they hit the ground
Setler79 [48]
They will both hit the ground at the same time because gravitational acceleration for all objects is the same.
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Calculate the mass of 1.35 moles of sodium chloride (NaCl).
Harlamova29_29 [7]

Answer:d

Explanation:

4 0
2 years ago
A well-insulated bucket of negligible heat capacity contains 129 g of ice at 0°C.
Luba_88 [7]

Answer:

The final equilibrium temperature of the system is T = 12.48^oC

For the ice it would melt completely the mass that would remain is Zero

Explanation:

In the following question we are provided with

Mass of the ice M_{i} = 129 g = 0.129 kg

Mass of the steam M_s = 19 g = 0.019 kg

Initial temperature is  T_i = 0°C

Temperature of  steam  T_s = 100°C

Following the change of state of water in the question

 The energy required by ice to change to water is mathematically given as

          Q_A = M_iL_f

Where L_f is a constant known as heat of fusion  and the value is 334*10^3 J/kg

           Q_A = 0.129 *334 *10^3  = 43086 J

The energy been released when the steam changes to water is mathematically given as

            Q_B = M_s * L_v

           Where L_v is a constant known as heat of vaporization and the value is 2256*10^3J/kg

           Q_B = 0.019 * 2256*10^3 = 42864J

         The energy released when the temperature of water decrease from 100°C to 0°C is

                 Q_C = M_s *C_water (100°C)

Where C_{water} is the specific heat of water which has a value 4186J/kg \cdot K

                  Q_C = 0.019 *4186*100 = 7953.4

Looking at the values we obtained we noticed that ]

             Q_B + Q_C > Q_A

What this means is that the ice will melt

bearing in mind the conservation of energy

     looking the way at which water at different temperature were mixed according to the question

     Heat lossed by the vapor   = heat gained by ice

        Q_B + M_s *C_{water}(100-T) = Q_A + M_i C_{water} T

                                               T = \frac{Q_B+M_s *C_{water}(100^oC)-Q_A}{(M_s *C_{water})+(M_i*C_{water})}

                                               T = \frac{42864+7953.4-43086}{(0.019+0.129)(4186)}

                                              T = 12.48^oC

       

3 0
3 years ago
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