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3 years ago

Suppose you walk 18.0 m straight west and then 25.0 m straight north, how far are you from your starting point? answer

Physics

1 answer:

KatRina [158]3 years ago

4 0

Solution:

Since the directions are perpendicular,

So we must use Pythagoras theorem:

d = √(18.0² + 25.0²) m = 30.8 m

Thus,

This the required distance from the starting point.

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If the position of a particle on the x-axis at time t is âˆ’5t2, then the average velocity of the particle for 0 â‰¤ t â‰¤ 3 is

Drupady [299]

Answer:

v = 15 m / s

Explanation:

In this exercise we are given the position function

x = 5 t²

and we are asked for the average velocity in an interval between t = 0 and t= 3 s, which is defined by the displacement between the time interval

$v= \frac{v_{f} - v_{o} }{t_{f} - t_{o} }$

let's look for the displacements

t = 0 x₀ = 0 m

t = 3 $x_{f}$ = 5 3 2

x_{f} = 45 m

we substitute

$v = \frac{45 -0}{3 - 0}$

v = 15 m / s

3 0

3 years ago

Charges q, q, and – q are placed on the x-axis at x = 0, x = 4 m, and x = 6 m, respectively. At which of the following points do

MrRissso [65]

Answer:

can you show a graph but if not i believe the answer is x=6m

Explanation:

7 0

3 years ago

Approximately how far is the sun from the center of the milky way galaxy?

enot [183]

The sun is approximately 27,000 light years away from the center of our galaxy.

8 0

3 years ago

Read 2 more answers

You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh

spayn [35]

To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

$\theta = \frac{1.22 \lambda }{D}$

Here,

D is diameter of the eye

$D = \frac{1.22 (539nm)}{5.11 mm}$

$D= 1.287*10^{-4}m$

The angle that relates the distance between the lights and the distance to the lamp is given by,

$Sin\theta = \frac{d}{L}$

For small angle, $sin\theta = \theta$

$sin \theta = \frac{d}{L}$

Here,

d = Distance between lights

L = Distance from eye to lamp

For small angle $sin \theta = \theta$

Therefore,

$L = \frac{d}{sin\theta}$

$L = \frac{0.691m}{1.287*10^{-4}}$

$L = 5367m$

Therefore the distance is 5.367km.

4 0

3 years ago

To calculate the change in kinetic energy, you must know the force as a function of _______. The work done by the force causes t

QveST [7]

Answer:

Explanation:

From the work-energy theorem, the workdone by a force on a body causes a change in kinetic energy of the body.

But, remember that the work done (W) by a force (F) on a body is the product of the force and the distance d, moved by the body caused by the force. i.e

W = F x d

This distance is a measure of the position of the body at a given instance.

Therefore, the work done is given by the force as a function of distance (or position).

3 0

3 years ago