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mezya [45]
3 years ago
9

Investigate and explain that when a force is applied to an object but it does not move, it is because another opposing force is

being applied by something in the environment so that the forces are balanced.
The diagram below shows a box lying on a wedge. Two opposite forces are acting on the box. (1 point)

Force 1 acting downward and Force 2 acting upward on a square box lying on a wedge.

If the box is stationary, which of the following is true about force 1 and force 2?
1. Force 1 Force 2 Comparison of Forces
Gravity Friction Force 1 is stronger than Force 2
2. Force 1 Force 2 Comparison of Forces
Friction Gravity Force 2 is stronger than Force 1
3. Force 1 Force 2 Comparison of Forces
Gravity Friction Force 1 is equal to Force 2
4. Force 1 Force 2 Comparison of Forces
Friction Gravity Force 1 is equal to Force 2
Chemistry
2 answers:
N76 [4]3 years ago
5 0

Answer:

4. Force 1 Force 2 Comparison of Forces

Friction Gravity Force 1 is equal to Force 2

Explanation:

oksian1 [2.3K]3 years ago
4 0
The answer is number 4
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This is the answer to question 19. If it’s not clear, send me a message.

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4 years ago
A light source of wavelength, l, illuminates a metal and ejects photoelectrons with a maximum kinetic energy of 1 eV. A second l
madreJ [45]

Explanation:

From first source, kinetic energy (K.E_{1}) ejected is 1 eV and wavelength of light is \lambda.

From second source, kinetic energy (K.E_{2}) ejected is 4 eV and wavelength of light is \frac{\lambda}{2}.

Relation between work function, wavelength, and kinetic energy is as follows.

                   K.E = \frac{hc}{\lambda} - \phi

where,        h = Plank's constant = 6.63 \times 10^{-34} J.s

                   c = speed of light = 3 \times 10^{8} m/s

Also, it is known that 1 eV = 1.6 \times 10^{-19} J

Therefore, substituting the values in the above formula as follows.

  • From first source,

                      K.E_{1} = \frac{hc}{\lambda} - \phi  

            1 eV = 1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\lambda} - \phi    

    1.6 \times 10^{-19} J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \phi        ........... (1)

  • From second source,

                  K.E_{2} = \frac{hc}{\lambda} - \phi  

          4 \times 1.6 \times 10^{-19} J = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{\frac{\lambda}{2}} - \phi        

                 6.4 \times 10^{-19} J = \frac{2 \times 1.98 \times 10^{-25} J.m}{\lambda} - \phi        ........... (2)        

Now, divide equation (2) by 2. Therefore, it will become

       {6.4 \times 10^{-19}J}{2} = \frac{2 \times 1.98 \times 10^{-25} J.m}{2\lambda} - \frac{\phi}{2}

                3.2 \times 10^{-19}J = \frac{1.98 \times 10^{-25} J.m}{\lambda} - \frac{\phi}{2}   ......... (3)

Now, subtract equation (3) from equation (1), we get the following.

                 1.6 \times 10^{-19} = \frac{\phi}{2}

                    \phi = 3.2 \times 10^{-19}

                          = 2 eV

Thus, we can conclude that work function of the metal is 2 eV.

3 0
4 years ago
How many moles of NO2 are present in 138.01 g
Sophie [7]

Answer:

3.0002 moles

Explanation:

Data Given:

mass of NO₂ = 138.01 g

No. of moles NO₂ = ?

Solution

To find mass of sample mole formula will be used

mole formula

          no. of mole = mass in grams / molar mass

Molar mass of NO₂ = 14 + 2(16)

Molar mass of NO₂ = 14 + 32 = 46 g/mol

Put values in above equation

               no. of mole = 138.01 g / 46 g/mol

               no. of mole = 3.0002 moles

no. of mole of NO₂ = 3.0002 moles

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Please number 12 and 13
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12.) 15 : 10 Times Each One By 0.3

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C) Metal is a conductor and the electricity will easily flow through it.
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