Ask question

Ask question

All categories

3 years ago

6

Select the correct answer. An airplane is flying at a constant speed in a positive direction. It slows down when it approaches t

he airport where it's going to land. Which term describes the slowing of the plane?
A. stationaryB. positive velocityC. positive accelerationD. negative accelerationE. constant velocity

1 answer:

Triss [41]3 years ago

7 0

An airplane is flying at a constant speed in a positive direction. It slows down when it approaches the airport where it's going to land. this is an example of negative acceleration (D).

You might be interested in

What does the Nucleolus do?

SCORPION-xisa [38]

Answer:

C

Explanation:

6 0

3 years ago

Read 2 more answers

Identify the law, write the equation and calculate the answer to the problem below.

lyudmila [28]

Find refractive index first

$\\ \rm\Rrightarrow \mu=\dfrac{c}{v}$

$\\ \rm\Rrightarrow \mu=\dfrac{1.0003}{1.33}$

$\\ \rm\Rrightarrow \mu =0.75$

Now

$\\ \rm\Rrightarrow \dfrac{sini}{sinr}=\mu$

$\\ \rm\Rrightarrow \dfrac{sin45}{sinr}=0.75$

$\\ \rm\Rrightarrow \dfrac{sin45}{0.75}=sinr$

$\\ \rm\Rrightarrow sinr=0.94$

$\\ \rm\Rrightarrow r=sin^{-1}(0.94)$

$\\ \rm\Rrightarrow r=70^{\circ}$

5 0

2 years ago

The focal length for a spherical convex mirror is –20 cm. What is its radius of curvature?

Svetradugi [14.3K]

Answer:

40 cm

Explanation:

The focus of a spherical convex mirror is the point at which the rays of an object converge in the infinite, it is also the point at which an object must be placed so that its image is formed in the infinite. The distance from the focus to the origin is called the focal length and is called f. It is related to the radius of the mirror, R, according to:

$f=-\frac{R}{2}$

rewriting for R:

$R=-2f=-2*-20cm=40 cm$ .

4 0

3 years ago

A lighthouse is located on a small island, 3 km away from the nearest point on a straight shoreline, and its light makes four re

lbvjy [14]

Answer:

The beam of light is moving at the peed of:

$\frac{dy}{dt} = \frac{80\pi}{3}$ km/min

Given:

Distance from the isalnd, d = 3 km

No. of revolutions per minute, n = 4

Solution:

Angular velocity, $\omega = \frac{d\theta'}{dt} = 2\pi n = 2\pi \times 4 = 8\pi$ (1)

Now, in the right angle in the given fig.:

$tan\theta' = \frac{y}{3}$

Now, differentiating both the sides w.r.t t:

$\frac{dtan\theta'}{dt} = \frac{dy}{3dt}$

Applying chain rule:

$\frac{dtan\theta'}{d\theta'}.\frac{d\theta'}{dt} = \frac{dy}{3dt}$

$sec^{2}\theta'\frac{d\theta'}{dt} = \frac{dy}{3dt} = (1 + tan^{2}\theta')\frac{d\theta'}{dt}$

Now, using $tan\theta = \frac{1}{m}$ and y = 1 in the above eqn, we get:

$(1 + (\frac{1}{3})^{2})\frac{d\theta'}{dt} = \frac{dy}{3dt}$

Also, using eqn (1),

$8\pi\frac{10}{9})\theta' = \frac{dy}{3dt}$

$\frac{dy}{dt} = \frac{80\pi}{3}$

7 0

3 years ago

Qwertyuiop hhffhhhhf fiuefhasuhs afbauyhsbvzskhbdv

mars1129 [50]

Answer:

gfhvgtrtjrgvfjrrgfrfftuyrisnhdvfcgfridkjhsybvvtfvjfcgvwjfccegvghcvgrcgvrekgvrkgvkvvrvkvfgkerruuyti

4 0

3 years ago

Read 2 more answers