I can not find the answer to the one after potassium ?

please help! find magnitude and direction (the counterclockwise angle with the +x axis) of a vector that is equal to a + c

-BARSIC- [3]

Answer:

Option (2)

Explanation:

From the figure attached,

Horizontal component, $A_x=A\text{Sin}37$

$A_x=12[\text{Sin}(37)]$

= 7.22 m

Vertical component, $A_y=A[\text{Cos}(37)]$

= 9.58 m

Similarly, Horizontal component of vector C,

$C_x$ = C[Cos(60)]

= 6[Cos(60)]

= $\frac{6}{2}$

= 3 m

$C_y=6[\text{Sin}(60)]$

= 5.20 m

Resultant Horizontal component of the vectors A + C,

$R_x=7.22-3=4.22$ m

$R_y=9.58-5.20$ = 4.38 m

Now magnitude of the resultant will be,

From ΔOBC,

$R=\sqrt{(R_x)^{2}+(R_y)^2}$

= $\sqrt{(4.22)^2+(4.38)^2}$

= $\sqrt{17.81+19.18}$

= 6.1 m

Direction of the resultant will be towards vector A.

tan(∠COB) = $\frac{\text{CB}}{\text{OB}}$

= $\frac{R_y}{R_x}$

= $\frac{4.38}{4.22}$

m∠COB = $\text{tan}^{-1}(1.04)$

= 46°

Therefore, magnitude of the resultant vector will be 6.1 m and direction will be 46°.

Option (2) will be the answer.

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In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.

The equation that defines the linear moment is given by

$mV_i = (m-m_O)V_f - m_OV_O$

where,

m=Total mass

$m_O =$ Mass of Object

$V_i =$ Velocity before throwing

$V_f =$ Final Velocity

$V_O =$ Velocity of Object

Our values are:

$m_1=5.3kgm_2=7.9kg\\m_3=10.5kg\\m_A=75kg\\m_{Total}=m=98.7Kg$

Solving to find the final speed, after throwing the object we have

$V_f=\frac{mV_0+m_TV_O}{m-m_O}$

We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.

That way during each section the equations should be modified depending on the previous one, let's start:

A) $5.3Kg\rightarrow 15m/s$