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irakobra [83]
2 years ago
15

If an object has 25 newtons of force pushing from the left and 26 newtons of force pushing from the right the object will

Chemistry
1 answer:
IRISSAK [1]2 years ago
7 0

Answer:

C. move left

Explanation:

The object will move towards the left direction due to the unbalanced forces that are acting on it.

 The resultant force on the object will be 1N in the left direction

  • The resultant force on a body is that singular force that will have the same effect as the different forces that acts on a body
  • When forces acts in opposite directions, they are subtracted
  • The object will move in the direction of the one with the greater force

So;

  Resultant force = 26N - 25N  = 1N

The body moves 1N to the left

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What conclusion can be draw about the heat of fusion of water from the heating curve? A) The potential energy of fusion represen
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What would be the result on the Insect population if the Bats and Mice were removed?
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2 years ago
determine the frequency and wavelength (in nm) of the light emitted when the e- fell from n=4 and n=2
Lostsunrise [7]

Answer:

Frequency = 6.16 ×10¹⁴ Hz

λ = 4.87×10² nm

Explanation:

In case of hydrogen atom energy associated with nth state is,

En =  -13.6/n²

For n = 2

E₂ = -13.6 / 2²

E₂ = -13.6/4

E₂ = -3.4 ev

Kinetic energy of electron = -E₂ = 3.4 ev

For n = 4

E₄ = -13.6 / 4²

E₄ = -13.6/16

E₄ = -0.85 ev

Kinetic energy of electron = -E₄ = 0.85 ev

Wavelength of radiation emitted:

E = hc/λ = E₄ - E₂

hc/λ = E₄ - E₂

by putting values,

6.63×10⁻³⁴Js × 3×10⁸m/s / λ = -0.85ev   - (-3.4ev )

6.63×10⁻³⁴ Js× 3×10⁸m/s / λ = 2.55 ev

λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s  /2.55ev

λ = 6.63×10⁻³⁴ Js× 3×10⁸m/s  /2.55× 1.6×10⁻¹⁹ J

λ = 19.89 ×10⁻²⁶ Jm / 2.55× 1.6×10⁻¹⁹ J

λ = 19.89 ×10⁻²⁶ Jm / 4.08×10⁻¹⁹ J

λ = 4.87×10⁻⁷ m

m to nm:

4.87×10⁻⁷ m ×10⁹nm/1 m

4.87×10² nm

Frequency:

Frequency = speed of electron / wavelength

by putting values,

Frequency = 3×10⁸m/s /4.87×10⁻⁷ m

Frequency = 6.16 ×10¹⁴ s⁻¹

s⁻¹ = Hz

Frequency = 6.16 ×10¹⁴ Hz

3 0
3 years ago
Ideal He gas expanded at constant pressure of 3 atm until its volume was increased from 9 liters to 15 liters. During this proce
kkurt [141]

Explanation:

The given data is as follows.

             P = 3 atm

                = 3 atm \times \frac{1.01325 \times 10^{5} Pa}{1 atm}  

                 = 3.03975 \times 10^{5} Pa

    V_{1} = 9 L = 9 \times 10^{-3} m^{3}    (as 1 L = 0.001 m^{3}),  

        V_{2} = 15 L = 15 \times 10^{-3} m^{3}

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As relation between work, pressure and change in volume is as follows.

                  W = P \times \Delta V

or,                W = P \times (V_{2} - V_{1})

Therefore, putting the given values into the above formula as follows.

                  W = P \times (V_{2} - V_{1})

                      = 3.03975 \times 10^{5} Pa \times (15 \times 10^{-3} m^{3} - 9 \times 10^{-3} m^{3})

                      = 1823.85 Nm

or,                   = 1823.85 J

As internal energy of the gas \Delta E is as follows.

                     \Delta E = Q - W

                                  = 800 J - 1823.85 J

                                  = -1023.85 J

Thus, we can conclude that the internal energy change of the given gas is -1023.85 J.

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