6. Using trigonometry, solve for the x and y components of the following: a. 17.0 m/s @ 30.0o c. 13.0 m/s2 @ 120o b.

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6. Using trigonometry, solve for the x and y components of the following: a. 17.0 m/s @ 30.0o c. 13.0 m/s2 @ 120o b.

dn-cgi/l/email-protection" class="__cf_email__" data-cfemail="3b0a02150b567b090f0b54">[email protected] d. 22.0m/[email protected] 7. Usingthecomponentsbelow,sketchthevectorwithlabeledcomponentsand calculate its magnitude and direction. a. Ax =34.0m,Ay =-34.0m b. Bx =0.00km,By =150.0km c. Cx = -8.00 m/s2, Cy = 6.00 m/s2 8. A girl holds a stick at an angle with one end on the ground and the other end 0.750 m above the ground. The Sun shines straight downward on the stick, and the stick makes a shadow on the ground that is 1.25 m long. What is the length of the stick and the angle it makes with the ground? 9. A boat travels on a heading of 40.0o north of east for a distance of 300 km. How far north and how far east of its starting point does the boat sail? 10.A weather station releases a weather balloon. The balloon’s buoyancy accelerates it upward at 15 m/s2. At the same time, a wind accelerates it horizontally to the right at 6.5 m/s2. What is the resulting acceleration? 11.Using the analytical method of vector addition, sketch and solve the following: a. [email protected] [email protected] b. 22.0 m/s @ 225o + 40.0 m/s @ 315o c. 17.0 m/s2 @ 330o - 11.0 m/s2 @ 270o 12.A car travels along a curved section of road. Initially the car has velocity 30.0 m/s @ 90.0o. After 5.00 s the car has a velocity of 30.0 m/s @ 95.0o. a. Calculate the change in the car's velocity. b. Calculate the car's average acceleration.
I have all of the answers already, but I need to show work. If you could pls show the work for me that would be such a help!! Thanks!

Physics

1 answer:

Jet001 [13] · Answer 1 · 2022-08-18T12:25:11+03:00

6a) Vector: 17.0 m/s @ $30.0^{\circ}$

x-component: $a_x = (17.0) cos 30.0^{\circ}=14.7 m/s$

y-component: $a_y = (17.0) sin 30.0^{\circ}=8.5 m/s$

6b) Vector: 13.0 m/s^2 @ $120^{\circ}$

x-component: $b_x = (13.0) cos 120.0^{\circ}=-6.5 m/s^2$

y-component: $b_y = (13.0) sin 120.0^{\circ}=11.3 m/s^2$

6c) Vector: 19.0 m @ $240^{\circ}$

x-component: $c_x = (19.0) cos 240.0^{\circ}=-9.5 m$

y-component: $c_y = (19.0) sin 240.0^{\circ}=-16.4 m/s$

6d) Vector: 22.0 m/s @ $300^{\circ}$

x-component: $d_x = (22.0) cos 300.0^{\circ}=11 m/s$

y-component: $d_y = (22.0) sin 300.0^{\circ}=-19.0 m/s$

7a) Components: Ax =34.0m,Ay =-34.0m

Magnitude: $|a|= \sqrt{A_x^2 +A_y^2}=\sqrt{(34.0)^2+(-34.0)^2}=48.0 m$

Direction: $\theta=arctan(\frac{A_y}{A_x})=arctan(\frac{-34.0}{34.0})=arctan(-1)=-45^{\circ}$

7b) Components: Bx =0.00km,By =150.0km

Magnitude: $|b|= \sqrt{B_x^2 +B_y^2}=\sqrt{(0)^2+(150.0)^2}=150.0 km$

Direction: $\theta=90^{\circ}$ since it is in the y-direction (no component on x)

7c) Components: Cx = -8.00 m/s2, Cy = 6.00 m/s2

Magnitude: $|c|= \sqrt{C_x^2 +C_y^2}=\sqrt{(-8.0)^2+(6.0)^2}=10.0 m/s^2$

Direction: $\theta=arctan(\frac{C_y}{C_x})=arctan(\frac{6.0}{-8.0})=arctan(-0.75)=-36.8^{\circ}$

8) In this problem, 0.750 m corresponds to the vertical side and 1.25 m corresponds to the horizontal side of a right triangle. The length of the stick corresponds to the length of the hypothenuse of this triangle, that can be found using the Pytagorean's theorem:

$L=\sqrt{(0.750 m)^2+(1.25 m)^2}=1.458 m$

The angle the stick makes with the ground is given by:

$\theta=arctan (\frac{0.750 m}{1.25 m})=arctan(0.6)=31.0^{\circ}$

9) The displacement of the boat corresponds to a vector of length L=300 km and angle $\theta=40.0^{\circ}$ with respect to east. Therefore, the two components in the north and east directions are:

- north: $L_y = L sin 40^{\circ}=(300 km)sin 40^{\circ}=192.8 km$

- east: $L_x = L cos 40^{\circ}=(300 km) cos 40^{\circ} =229.8 km$

10) The two accelerations correspond to the two sides of a right triangle, therefore the resultant acceleration corresponds to the length of the hypothenuse of the triangle:

$a=\sqrt{a_x^2 +a_y^2 }=\sqrt{(6.5)^2+(15)^2}=16.3 m/s^2$

11a) [email protected] [email protected]

Let's resolve each vector in its components:

$v_{1x} = (9.0) cos 55^{\circ} = 5.2 m$

$v_{1y} = (9.0) sin 55^{\circ} = 7.4 m$

$v_{2x} = (6.0) cos 125^{\circ} = -3.4 m$

$v_{2y} = (6.0) sin 55^{\circ} = 4.9 m$

Now we sum the components in each direction:

$R_x = v_{1x}+v_{2x}=5.2-3.4 =1.8 m$

$R_y = v_{1y}+v_{2y}=7.4+4.9 =12.3 m$

So the magnitude of the resultant vector is:

$R=\sqrt{R_x^2+R_y^2}=\sqrt{(1.8)^2+(12.3)^2}=12.4 m$

And the direction is:

$\theta=arctan(\frac{R_y}{R_x})=arctan(\frac{12.3}{1.8})=81.7^{\circ}$

11b) 22.0 m/s @ 225o + 40.0 m/s @ 315o

Let's resolve each vector in its components:

$v_{1x} = (22.0) cos 225^{\circ} = -15.6 m/s$

$v_{1y} = (22.0) sin 225^{\circ} = -15.6 m/s$

$v_{2x} = (40.0) cos 315^{\circ} = 28.3 m/s$

$v_{2y} = (40.0) sin 315^{\circ} = -28.3 m/s$

Now we sum the components in each direction:

$R_x = v_{1x}+v_{2x}=15.6+28.3 =43.9 m/s$

$R_y = v_{1y}+v_{2y}=-15.6-28.3 =-43.9 m/s$

So the magnitude of the resultant vector is:

$R=\sqrt{R_x^2+R_y^2}=\sqrt{(43.9)^2+(-43.9)^2}=62.1 m/s$

And the direction is:

$\theta=arctan(\frac{R_y}{R_x})=arctan(\frac{-43.9}{43.9})=315^{\circ}$

11c) 17.0 m/s2 @ 330o - 11.0 m/s2 @ 270o

Let's resolve each vector in its components:

$v_{1x} = (17.0) cos 330^{\circ} = 14.7 m/s$

$v_{1y} = (17.0) sin 330^{\circ} = -8.5 m/s$

$v_{2x} = (11.0) cos 270^{\circ} = 0 m/s$

$v_{2y} = (11.0) sin 270^{\circ} = -11 m/s$

Now we calculate the difference between the components in each direction:

$R_x = v_{1x}+v_{2x}=14.7-0 =14.7 m/s$

$R_y = v_{1y}+v_{2y}=-8.5-(-11) =2.5 m/s$

So the magnitude of the resultant vector is:

$R=\sqrt{R_x^2+R_y^2}=\sqrt{(14.7)^2+(2.5)^2}=14.9 m/s$

And the direction is:

$\theta=arctan(\frac{R_y}{R_x})=arctan(\frac{2.5}{14.7})=9.7^{\circ}$

12a) The change in velocity is equal to the vector difference between the final velocity (vf) and the initial velocity (vi), so let's proceed as in the previous exercise:

$v_{1x} = (30.0) cos 90^{\circ} = 0 m/s$