Two forces fa and fb are applied to an object whose mass is 5.42 kg. slader the larger force is fa. when both forces point due east, the object's acceleration has a magnitude of 0.651 m/s2. however, when fa points due east and fb points due west, the acceleration is 0.427 m/s2, due east. find (a) the magnitude of fa and (b) the magnitude of fb.

a)Fa=2.98 N

b)Fb=0.53  N

Explanation:

Given that

m= 5.42 kg

When force are acting in east direction:

So from second law of Newton's

Given that Fa is greater than Fb and acceleration is 0.651.

Fa+Fb= m a

Fa+  Fb =5.42 x 0.651

Fa+ Fb =3.52                -------------1

Now when both acting in apposite direction

Given that Fa is greater than Fb and acceleration is 0.427.

Fa-Fb= m a

Fa-  Fb =5.42 x 0.427

Fa- Fb =2.44                   -------------2

Now from equation 1 and 2

Add both equations

2 Fa = 5.96

Fa=2.98 N

From equation 1

Fb =3.52 -Fa

Fb=0.53  N

Fa = 2.92N

Fb = 0.61N

Explanation:

According to the Newton's second law,

Force = mass × acceleration

Given two forces Fa and Fb applied on an object,

∑F = ma

If both forces points due east (+x direction)

∑F = Fa+Fb = ma

Fa+Fb = 5.42kg(0.651 m/s²)

Fa+Fb = 3.53... (1)

If Fa points due east (+x drn) and Fb points due west (-ve x direction)

∑F = Fa-Fb = 5.42(0.427)

∑F = Fa-Fb = 2.31... (2)

Solving equation 1 and 2 simultaneously to get the magnitude of the force Fa and Fb

Fa+Fb = 3.53... (1)

Fa-Fb = 2.31... (2)

Adding both equation:

2Fa = 3.53+2.31

2Fa = 5.84

Fa = 5.84/2

Fa = 2.92N

Substituting Fa = 2.92N into equation 2

Fa - Fb = 2.31

2.92 - Fb = 2.31

-Fb = 2.31 - 2.92

-Fb = -0.61

Fb = 0.61N

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.

increases it