Radiation transfers energy by moving matter. Please select the best answer from the choices provided T F

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

3 0

2 years ago

When an object is turning a corner what direction is the acceleration?

polet [3.4K]

If the corner is rounded and is perfectly circular, then the acceleration is centripetal and is always directed toward the center.

7 0

3 years ago

How can i find the acceleration?(rope and grinder have no weight) *** sorry for my english

Finger [1]

The main formula to be used here is

Force = (mass) x (acceleration).

We'll get to work in just a second. But first, I must confess to you that I see
two things happening here, and I only know how to handle one of them. So
my answer will be incomplete, but I believe it will be more reliable than the
first answer that was previously offered here.

On the right side ... where the 2 kg and the 3 kg are hanging over the same
pulley, those weights are not balanced, so the 3 kg will pull the 2kg down, with
some acceleration. I don't know what to do with that, because . . .

At the same time, both of those will be pulled up by the 10 kg on the other side
of the upper pulley.

I think I can handle the 10 kg, and work out the acceleration that IT has.

Let's look at only the forces on the 10 kg:

-- The force of gravity is pulling it down, with the whatever the weight of 10 kg is.

-- At the same time, the rope is pulling it UP, with whatever the weight of 5 kg is ...
that's the weight of the two smaller blocks on the other end of the rope.

So, the net force on the 10 kg is the weight of (10 - 5) = 5 kg, downward.

The weight of 5 kg is (mass) x (gravity) = (5 x 9.8) = 49 newtons.

The acceleration of 10 kg, with 49 newtons of force on it, is

Acceleration = (force) / (mass) = 49/10 = 4.9 meters per second²

7 0

3 years ago

A periodic wave with wavelength - 2 m has frequency f = 4 Hz. What is the wave's speed?

Sunny_sXe [5.5K]

Answer:

-8m/s

Explanation:

v=wavelength*f=-2*4=-8m/s

3 0

3 years ago

. Find the buoyant force exerted on a ball with radius of 2 meters, when the ball is entirely immersed in the water.

almond37 [142]

Explanation:

Buoyancy force is equal to the weight of the displaced fluid:

B = ρVg

where ρ is the density of the fluid,

V is the volume of the displaced fluid,

and g is the acceleration due to gravity.

The fluid is water, so ρ = 1000 kg/m³.

The volume displaced is that of a sphere with radius 2 m:

V = 4/3 π r³

V = 4/3 π (2 m)³

V ≈ 33.5 m³

The buoyancy force is therefore:

B = (1000 kg/m³) (33.5 m³) (9.8 m/s²)

B ≈ 328,400 N

Round as needed.

8 0

3 years ago