The wavelength of the sound is 0.750m. What is the frequency?

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

1 year ago

In the absence of air resistance, at what other angle will a thrown ball go the same distance as one thrown at an angle of 75 de

snow_tiger [21]

As we know that range of the projectile motion is given by

$R = \frac{v^2 sin(2\theta)}{g}$

here we know that range will be same for two different angles

so here we can say the two angle must be complementary angles

so the two angles must be

$\theta, 90 - \theta$

so it is given that one of the projection angle is 75 degree

so other angle for same range must be 90 - 75 = 15 degree

so other projection angle must be 15 degree

5 0

1 year ago

The force diagram represents a girl pulling a sled with a mass of 6.0 kg to the left with a force of 10.0 N at a 30.0 degree ang

STatiana [176]

the correct answers are 54N and -1,2m/s^2

6 0

1 year ago

Read 2 more answers

: The truck is to be towed using two ropes. Determine the magnitudes of forces FA and FB acting on each rope in order to develop

Sholpan [36]

Answer:

Fa=774 N

Fb=346 N

Explanation:

We will solve this problem by equating forces on each axis.

On x-axis let forces in positive x-direction be positive and forces in negative x-direction be negative
On y-axis let forces in positive y-direction be positive and forces in negative y-direction be negative

While towing we know that car is mot moving in y-direction so net force in y-axis must be zero

⇒∑Fy=0

⇒ $Fa*sin(50)-Fb*sin(20)=0$

⇒ $Fa*sin(50)=Fb*sin(20)$

⇒ $Fa=2.24Fb$

Given that resultant force on car is 950N in positive x-direction

⇒∑Fx=950

⇒ $Fa*cos(20)+Fb*cos(50)=950$

⇒ $2.24*Fb*cos(20)+Fb(50)=950$

⇒ $Fb*(2.24*cos(20)+cos(50))=950$

⇒ $Fb=\frac{950}{2.24*cos(20)+cos(50)}$

⇒ $Fb=\frac{950}{2.24*0.94+0.64}$

⇒ $Fb=\frac{950}{2.75}=345.5$

⇒ $Fa=2.24*Fb$

$=2.24*345.5$

$=773.93$

Therefore approximately, Fa=774 N and Fb=346 N

5 0

1 year ago

Submarine a travels horizontally at 11.0 m/s through ocean water. it emits a sonar signal of frequency f 5 5.27 3 103 hz in the

xeze [42]

Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ

6 0

2 years ago