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Yuliya22 [10]
3 years ago
6

Two forces F1 and F2 are applied at the same point of an object. F1 has a magnitude of 20 N and is directed at an angle of 120 d

egrees with respect to the positive x axis. F2 has a magnitude of 5 N and is directed at an angle of -30 degrees with respect to the positive x axis. Determine net force acting on the object in i, j, k unit notation.
Physics
1 answer:
butalik [34]3 years ago
6 0

Answer:

The value is  \vec F_{n} =   -5.67 i +14.82 j  \ \  N

Explanation:

From the question we are told that

   The magnitude of the first force is  F_1 =  20 \  N

    The angle which it makes with the x-axis is  \theta_1 =  120 ^o

    The magnitude of the second force is  F_2 =  5 \  N

     The angle which it make with x-axis is  \theta_2 =   -30 ^o

Generally the x-component of the first force is  

       F__{1x}} =  20 cos (120 )'

=>     F__{1x}} =  -10 \  N

Generally the y-component of the first force is  

       F__{1y}} =  20 sin (120 )

=>     F__{1y}} =  17.32  \  N

Generally the vector representation of the first force is mathematically represented as

       \vec F_1 = [ -10 i  +  17.32 j \ ] \ N

Generally the x-component of the second force is  

       F__{2x}} =  20 cos (-30 )

=>     F__{2x}} =  4.33 \  N

Generally the y-component of the second force is  

       F__{2y}} =  20 sin (-30 )

=>     F__{2y}} =  -2.5  \  N

Generally the vector representation of the first force is mathematically represented as

       \vec F_2 = [ 4.33 i  - 2.5 j \ ] \ N

Generally the net force acting that point is mathematically represented as

       \vec F_{n} =   -10 i  +  17.32 j   + 4.33 i  - 2.5 j

=>    \vec F_{n} =   -5.67 i +14.82 j  \ \  N

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True since coulomb's law states that There is electric force between like charges or opposite charges. The negative sign only shows the nature of the force.

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coulombs formula is given by

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Now it states that if two charged particles are separated by the distance r and having same or opposite charge will attract or repel each other.

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The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371
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Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

q=m\times c\times (T_2-T_1)

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = 4.18J/g.K

T_1 = initial temperature of water = 293.0 K

T_2 = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

q=250g\times 4.18J/g.K\times (371.2-293.0)K

q=81719J

Now we have to calculate the enthalpy of combustion of octane.

\Delta H=\frac{q}{n}

where,

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Now put all the given values in the above formula, we get:

\Delta H=\frac{-81719J}{0.015mole}

\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

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