Hi..I'm new here..Can some one help mi with my school work plz

A cyclist makes the following trip along two vectors; he travels 9km to the north and then travels 6km to the east

Answer:

Final distance from the origin: 10.82 km. the vector points as shown in the attached image.

Angle with respect to the east: $56.31^o$

Explanation:

Please refer to the attached image. The cyclist's trip is indicated with the green arrows (9 km to the north followed by 6 km to the east.

So his final position is at the tip of this last vector, and indicated by the orange vector drawn form the point where the trip starts to the cyclist's final location.

We observe that this orange vector is in fact the hypotenuse of a right angle triangle, and we can estimate the distance from the origin by the Pythagorean theorem:

$d=\sqrt{9^2+6^2} \\d=\sqrt{81+36} \\d=\sqrt{117} \\d=10.82 \,\,km$

Notice that this is NOT the actual number of km that the cyclist pedaled to reach the final point.

Now, to find the value of the angle $\theta$ , we need to use trigonometry, and in particular the tangent function gives us the ratio between the side of the triangle "opposite" to the angle, divided the side "adjacent" to the angle:

$tan(\theta)=\frac{opp}{adj} \\tan(\theta)=\frac{9}{6}\\tan(\theta)=\frac{3}{2}\\$

Now we can find the value of the angle by using the arctan function:

$tan(\theta)=\frac{3}{2} \\\theta=arctan(\frac{3}{2} )\\\theta= 56.31^o$

6 0

3 years ago

Please can someone answer this question

EastWind [94]

B????????? I thinkkkkk

7 0

3 years ago

Read 2 more answers

outward from a wall just above floor level. A 1.5 kg box sliding across a frictionless floor hits the end of the spring and comp

sweet [91]

Answer:

v = 0.489 m/s

Explanation:

It is given that,

Mass of a box, m = 1.5 kg

The compression in the spring, x = 6.5 cm = 0.065 m

Let the spring constant of the spring is 85 N/m

We need to find the velocity of the box (v) when it hit the spring. It is based on the conservation of energy. The kinetic energy of spring before collision is equal to the spring energy after compression i.e.

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2$

$v=\sqrt{\dfrac{kx^2}{m}} \\\\v=\sqrt{\dfrac{85\times (0.065)^2}{1.5}} \\\\v=0.489\ m/s$

So, the speed of the box is 0.489 m/s.

3 0

4 years ago

Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P P and volume V V satisfy th

Verdich [7]

Answer:

the volume decreases at the rate of 500cm³ in 1 min

Explanation:

given

v = 1000cm³, p = 80kPa, Δp/t= 40kPa/min

PV=C

vΔp + pΔv = 0

differentiate with respect to time

v(Δp/t) + p(Δv/t) = 0

(1000cm³)(40kPa/min) + 80kPa(Δv/t) = 0

40000 + 80kPa(Δv/t) = 0

Δv/t = -40000/80

= -500cm³/min

the volume decreases at the rate of 500cm³ in 1 min

3 0

3 years ago

Which statement is true of a wave that’s propagating along the pavement and girders of a suspension bridge?

Readme [11.4K]

Had to look for the options and here is my answer.
The statement that is considered correct regarding a wave that is propagating along the pavement and girders of a suspension bridge is this: "t<span>he wave is mechanical, with particles vibrating in a direction that is perpendicular to that of the wave, forming compressions and rarefactions." Hope this helps.</span>

3 0

4 years ago

Hi..I'm new here..Can some one help mi with my school work plz​

Hi..I'm new here..Can some one help mi with my school work plz