How is a solution diluted?

A wind instrument that produces only odd-numbered standing wave modes has what configuration of its ends?

Luda [366]

Answer:

Open- closed

Explanation:

It has Open-closed configuration of its end

4 0

3 years ago

An object is placed 5.00 cm beyond the focal point of a convex lens whose focal length is 10.0 cm. If the object height is 3.0 c

Aleks04 [339]

Answer:

The height of the image is, h' = 6.0 cm

The image is erect.

Explanation:

Given data,

The object distance, u = -5 cm

The focal length of convex lens, f = 10 cm

The object height, h = 3 cm

The lens formula,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

$\frac{1}{10}=\frac{1}{v}-\frac{1}{-5}$

$\frac{1}{v}=\frac{1}{10}-\frac{1}{5}$

v = -10 cm

The magnification factor of lens

$m=\frac{-10}{-5}$

m = 2

$m=\frac{h'}{h}$

$h'=h\times m$

$h'=3\times 2$

h' = 6 cm

The height of the image is, h' = 6 cm

The image is erect.

4 0

3 years ago

A body with initial velocity 8.0 m/s moves along a straight line with constant acceleration and travels

Aleksandr [31]

Answer:

(a) The average velocity is 16 m/s

(b) The acceleration is 0.4 m/s^2

(c) The final velocity is 24 m/s

Explanation:

Constant Acceleration Motion

It's a type of motion in which the velocity (or the speed) of an object changes by an equal amount in every equal period of time.

Being a the constant acceleration, vo the initial speed, vf the final speed, and t the time, final speed is calculated as follows:

$v_f=v_o+at\qquad\qquad [1]$

The distance traveled by the object is given by:

$\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]$

(a) The average velocity is defined as the total distance traveled divided by the time taken to travel that distance.

We know the distance is x=640 m and the time taken t= 40 s, thus:

$\displaystyle \bar v=\frac{x}{t}=\frac{640}{40}=16$

The average velocity is 16 m/s

Using the equation [1] we can solve for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

(c) From [2] we can solve for a:

$\displaystyle a= 2\frac{x-v_ot}{t^2}$

Since vo=8 m/s, x=640 m, t=40 s:

$\displaystyle a= 2\frac{640-8\cdot 40}{40^2}=0.4$

The acceleration is 0.4 m/s^2

(b) The final velocity is calculated by [1]:

$v_f=8+0.4\cdot 40$

$v_f=8+16=24$

The final velocity is 24 m/s

3 0

3 years ago

If the 250 kg bumper car that you are riding in hits another bumper car that is sitting still while driving 3.5 m/s, how much fo

Vilka [71]

Answer:

875 N

Explanation:

From this question, you didn't state the time taken for the bumper car to move or to hit the other bumper car. In calculations of force, time is often needed, because

Force = mass * acceleration, while

Acceleration = velocity / time, basically

Force = mass * velocity / time.

We have our mass, we have our velocity, but we haven't time. So, for this calculation, I'd assume our time to be 1s.

Going by the formula I stated, we can then say that

Force = 250 * 3.5 / 1

Force = 875 N

This means the force my bumper car have while moving at 3.5 m/s for an estimated time of 1s is 875 N

3 0

3 years ago

A liquid is used to make a mercury-type barometer. The barometer is intended for space-faring astronauts. At the surface of the

Anarel [89]

Answer:

Density of liquid = 4730 kg/m³

Atmospheric pressure on planet X = 8401.7 N/m²

Explanation:

Pressure, P = ρgh where ρ = density of liquid, g =9.8 m/s² and h = height of column at earth's surface = 2185 mm. Since P = atmospheric pressure, for mercury, P = ρ₁gh₁ where ρ₁ = 13.6 g/cm³ and h₁ = 760 mm

So, ρgh = ρ₁gh₁

ρ = ρ₁h₁/h = 13.6 g/cm³ × 760/2185 = 4.73 g/cm³ = 4730 kg/m³

The atmospheric pressure on planet X

P = ρg₁h₃ g₁ = g/4 and h₃ = 725 mm = 0.725 m

on planet X

P = ρg₁h₃ = (4730 kg/m³ × 9.8 m/s² × 0.725 m)/4 = 8401.7 N/m²

6 0

3 years ago