Compound name: for Al2O3

Two car horns are sounded creating two sound waves with frequencies that differ by a factor of three. How does the speed of the

Andrews [41]

Answer:

Remains the same

Explanation:

The speed of waves of higher and lower frequency both will be same.

the speed of sound in a medium is constant and independent of it's frequency. Moreover, when the frequency changes wavelength changes accordingly, such that their product remains constant.

we know that

υ×λ = constant = velocity

υ= frequency

λ= wavelength.

7 0

3 years ago

You are 1.8 m tall and stand 2.8 m from a plane mirror that extends vertically upward from the floor. on the floor 1 m in front

velikii [3]

This problem must be solved using a sketch. I attached an illustration of the problem.
You must trace the ray that reflects from the top off the table to your eyes. This how eyesight works, light rays reflects off the objects into your eyes.
Law of reflection tells us that light ray reflects off the surface at the same angle in which it falls on it( i attached another illustration of this).
Now we can write tangens equations:
$tan(\theta)=\frac{h-0.8}{1}\\
tan(\theta)=\frac{1.8-h}{2.8}\\
\frac{h-0.8}{1}=\frac{1.8-h}{2.8}$
We solve for h:
$\frac{h-0.8}{1}=\frac{1.8-h}{2.8}\\
2.8h-0.8\cdot 2.8=1.8-h\\
3.8h=1.8+2.24\\
h=\frac{4.04}{3.8}\\
h=1.06m$

6 0

3 years ago

A hot air balloon is traveling vertically upward at a constant speed of 4.5 m/s. When it is 28 m above the ground, a package is

ella [17]

Answer:

1.97 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s²

$s=ut+\frac{1}{2}at^2\\\Rightarrow 21=4.5t+\frac{1}{2}\times -9.8\times t^2\\\Rightarrow 21=-4.5t-4.9t^2\\\Rightarrow 4.9t^2+4.5t-28=0\\\Rightarrow 49t^2+45t-280=0$

Solving the above equation we get

$t=\frac{-45+\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}, \frac{-45-\sqrt{45^2-4\cdot \:49\left(-280\right)}}{2\cdot \:49}\\\Rightarrow t=1.97, -2.89$

So, the time the package was in the air is 1.97 seconds

3 0

3 years ago

The small piston of a hydraulic lift has a cross-sectional of 3 00 cm2 and its large piston has a cross-sectional area of 200 cm

Nesterboy [21]

Q: The small piston of a hydraulic lift has a cross-sectional of 3.00 cm2 and its large piston has a cross-sectional area of 200 cm2. What downward force of magnitude must be applied to the small piston for the lift to raise a load whose weight is Fg = 15.0 kN?

Answer:

225 N

Explanation:

From Pascal's principle,

F/A = f/a ...................... Equation 1

Where F = Force exerted on the larger piston, f = force applied to the smaller piston, A = cross sectional area of the larger piston, a = cross sectional area of the smaller piston.

Making f the subject of the equation,

f = F(a)/A ..................... Equation 2

Given: F = 15.0 kN = 15000 N, A = 200 cm², a = 3.00 cm².

Substituting into equation 2

f = 15000(3/200)

f = 225 N.

Hence the downward force that must be applied to small piston = 225 N

8 0

3 years ago

The vector position of an object is given by r what is the torque acting on the object about the origin when a force f = (−12.5i

attashe74 [19]

Let the vector position of the object in the (x-y) plane be
$\vec{r} = x \hat{i} + y \hat{j}$

The applied force is
$\vec{f} = -12.5 \hat{i}
$

By definition, the applied torque is
$\vec{T} = \vec{r} \times \vec{f} = (x\hat{i} + y\hat{j}) \times (-12.5y \hat{i}) = 12.5\hat{k}$

Answer: $12.5y \, \hat{k}$

7 0

3 years ago