Perform calculations using the circuit illustrated. Round

A vector has a magnitude of 46.0 m and points in a direction 20.0° below the positive x-axis. A second vector, , has a magnitude

irina1246 [14]

Answer with Step-by -step explanation:

We are given that

b. $\mid A\mid=46 m$

$\theta=20^{\circ}$ below the positive x-axis

Therefore, the angle made by vector A in counter clockwise direction when measure from positive x-axis= $x=360-20=340^{\circ}$

x-component of vector A= $A_x=\mid A\mid cosx=46cos 340=46\times 0.94=43.24$

y-Component of vector A= $A_y=\mid A\mid sinx=46sin340=46(-0.34)=-15.64$

Magnitude of vector B=86 m

The vector B makes angle with positive x- axis= $x'=42^{\circ}$

x-component of vector B= $B_x=86cos42=63.64$

y-Component of vector B= $B_y=86sin42=57.62$

Vector A= $A_xi+A_yj=43.24i-15.64j$

Vector B= $B_xi+B_yj=63.64i+57.62j$

Vector C=A+B

Substitute the values

$C=43.24i-15.64j+63.64i+57.62j$

$C=106.88i+41.98j$

c.Direction= $\theta=tan^{-1}(\frac{y}{x})=tan^{-1}(\frac{41.98}{106.88})=21.5^{\circ}$

The direction of the vector C=21.5 degree

6 0

2 years ago

A short-wave radio antenna is supported by two guy wires, 155 ft and 175 ft long. Each wire is attached to the top of the antenn

Vladimir79 [104]

Answer:

163.8 ft

Explanation:

In triangle ABD

$AB$ = 155 ft

$Cos63 = \frac{BD}{AB} = \frac{BD}{155}\\BD = 155 Cos63 \\BD = 70.4 ft$

$Sin63 = \frac{AD}{AB} = \frac{AD}{155} \\AD = 166 Sin63\\AD = 148 ft$

Using Pythagorean theorem in triangle ADC

$AC^{2} = AD^{2} + DC^{2} \\175^{2} = 148^{2} + DC^{2} \\DC = 93.4 ft$

$d$ = distance between the anchor points

distance between the anchor points is given as

$d = BD + CD = 70.4 + 93.4\\d = 163.8 ft$

5 0

2 years ago

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An astronaut finds herself in a predicament in which she has become untethered from her shuttle. She figures that she could get

Blizzard [7]

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) $5.3Kg\rightarrow 15m/s$