As a killer whale swims toward a seal, it sends out sound waves to determine the direction the seal is moving. When the

A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed

mafiozo [28]

Answer:

0.6375 m/s

Explanation:

Let x be the distance of the man from the building

from the figure attached

initially the value of x=12

Given:

$\frac{dx}{dt}=-1.7m/s$

where the negative sign depicts that the distance of the man from the building is decreasing.

Now, Let The length of the shadow be = y

we have to calculate $\frac{dy}{dt}$ when x=4

from the similar triangles

we have,

$\frac{2}{12-x}=\frac{y}{12}$

or

$y=\frac{24}{12-x}$

Differentiating with respect to time 't' we get

$\frac{dy}{dt}=-\frac{24}{12-x}^2\frac{-dx}{dt}$

or

$\frac{dy}{dt}=\frac{24}{12-x}^2\frac{dx}{dt}$

Now for x = 4, and $\frac{dx}{dt}=-1.7m/s$ we have,

$\frac{dy}{dt}=\frac{24}{12-4}^2\times (-1.7)$

or

$\frac{dy}{dt}=-0.6375m/s$

<u>here, the negative sign depicts the decrease in length and in the question it is asked the decreasing rate thus, the answer is </u><u>0.6375m/s</u>

4 0

3 years ago

When discussing Newton's laws of motion,which terms do people most likely yse when talking about Newton's third law of motion?

mestny [16]

When discussing Newton's laws of motion, particularly Newton's third law of motion, the terms that almost everyone will use are "action" and "reaction".

You must not take this to mean that they understand what they're talking about.

4 0

3 years ago

An astronaut drops a rock on the surface of an asteroid.The rock is released from rest at a height of 0.86 m above the ground, a

SCORPION-xisa [38]

Answer:

$a_y=0.92m/s^2$

Explanation:

To solve this problem we use the formula for accelerated motion:

$y=y_0+v_{y0}t+\frac{a_yt^2}{2}$

We will take the initial position as our reference ( $y_0=0m$ ) and the downward direction as positive. Since the rock departs from rest we have:

$y=\frac{a_yt^2}{2}$

Which means our acceleration would be:

$a_y=\frac{2y}{t^2}$

Using our values:

$a_y=\frac{2(0.86m)}{(1.37s)^2}=0.92m/s^2$

5 0

3 years ago

I need the ans for this question QUICK PLEASE!!!

Basile [38]

Explanation:

A) 1.05

B) 1.33

C) 1.16

D) 0.62

All units in cm

3 0

3 years ago

It decomposes in a first-order reaction with a rate constant of 14 s–1. how long would it take for an initial concentration of 0

VikaD [51]

The rate constant of a reaction can be computed by the ratio of the changes in the concentration and time take taken for it to decompose. Thus, if the rate constant is given to be 14 M/s, we have

$rate = \frac{-(C_{new} - C_{old})}{t}$

where C are the concentration values and t is the time taken for it to decompose.

$14 = \frac{-(0.02 - 0.06)}{t}$
$t = 0.003 s$

Thus, it will take 0.003 s for it to decompose.
Answer: 0.003 s

4 0

3 years ago