I need help with #24 ASAP .. I have to get it done today

A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the

lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

$l^2 = x^2 + y^2-----(1)$

Differentiating with respect to t ( time ),

$0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}$ ( l = 26 feet = constant )

$\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}$

$\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$

We have,

$y = 10, \frac{dx}{dt}= -3\text{ feet per min}$

$\frac{dy}{dt}=\frac{3x}{10}-----(X)$

(i) From equation (1),

$26^2 = x^2 + 10^2$

$676=x^2 + 100$

$576 = x^2$

$\implies x = 24\text{ feet}$

From equation (X),

$\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}$

(ii) From equation (X),

$\frac{dy}{dt}\propto x$

Thus, for different value of x the value of $\frac{dy}{dt}$ would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

$\frac{dy}{dt}=0\text{ feet per min}$

3 0

2 years ago

Calculate the number of electrons passing a point in the wire in 1 min when the current is 1 A

Vera_Pavlovna [14]

Explanation:

When one coulomb charge passes through any cross section of the wire per second,the current passing is one ampere. Charge of electron ,e=1.6X10^-19C. n=1/(1.6X10^-19)=6.25X10^18.Sep 17, 2017

7 0

2 years ago

Choose the scenario in which the sound frequency of the waves is higher.

mrs_skeptik [129]

Answer:

B) the sound source moves towards you at 100 m/sec

Explanation:

The Dopper Effect is a phenomenon that occur when there is relative motion between an observer and a source of a wave. When this situation occurs, there is an apparent shift in frequency of the wave, as observed by the observer.

The apparent frequency observed by the observer is given by

$f'=\frac{v\pm v_o}{v\pm v_s}f$

where

f is the original frequency of the wave

f' is the apparent frequency

v is the speed of the wave

$v_o$ is the velocity of the observer (positive if moving towards the source of the wave, negative otherwise)

$v_s$ is the velocity of the source (negative if moving towards from the observer, positive otherwise)

In this problem, we want to find the scenario in which the sound frequency is higher.

We see that in all 4 scenarios, the sound source is moving: this means we have to find the scenario in which the denominator of the equation is smaller.

First of all, we notice the sound source moves towards the observer, $v_s$ is negative, so the denominator is higher: this means that the correct option must be either A or B.

Also, we notice that since $v_s$ is negative, a value larger in magnitude will mean a smaller denominator: therefore, the correct answer will be

B) the sound source moves towards you at 100 m/sec

Since this situation will make the denominator of the formula the smallest possible.

5 0

2 years ago

What would occur if the index of refraction of a material was less than n=1?

Bingel [31]

The correct answer is (A). The speed of light would increase to a speed larger than the maximum speed of light in vacuum.

The index of refraction is the ratio of speed of light in vacuum to the speed of light in a medium.

n=C/V

here, n is the index of refraction, c the speed of light in vacuum, v is speed of light in any medium.

Now if the value of index of refraction is less than one, than the value of speed of light would be greater than the speed of light in the vacuum.

3 0

3 years ago

Read 2 more answers

Explain how a clinical thermometer should be sterilized and why

Kisachek [45]

Answer:

Explain why it is not advisable to sterilise a clinical thermometer on boiling water at normal atmospheric temperature. A clinical thermometer has small temperature range. The glass will crack/burst due to excessive pressure created by expansion of mercury.

4 0

2 years ago