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2 years ago

Frequency is measured in units called?

1 answer:

5 0

The frequency, f, of a wave is the number of waves passing a point in a certain time. We normally use a time of one second, so this gives frequency the unit hertz (Hz), since one hertz is equal to one wave per second.

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9. (02.04 LC)

ycow [4]

Answer:

Molecules speed up

Explanation:

This is caused because of the temperature increasing. The temperature increase is telling us that the thermal energy of the reaction is increasing. When the energy is increased molecules increase their speed, because they have more energy in them

7 0

3 years ago

The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me

Sav [38]

Answer:

(a): $\rm meter/ second^2.$

(b): $\rm meter/ second^3.$

(c): $\rm 2ct-3bt^2.$

(d): $\rm 2c-6bt.$

(e): $\rm t=\dfrac{2c}{3b}.$

Explanation:

Given, the position of the particle along the x axis is

$\rm x=ct^2-bt^3.$

The units of terms $\rm ct^2$ and $\rm bt^3$ should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of $\rm ct^2=meter$

Therefore, unit of $\rm c= meter/ second^2.$

(b):

Unit of $\rm bt^3=meter$

Therefore, unit of $\rm b= meter/ second^3.$

(c):

The velocity v and the position x of a particle are related as

$\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.$

(d):

The acceleration a and the velocity v of the particle is related as

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.$

(e):

The particle attains maximum x at, let's say, $\rm t_o$ , when the following two conditions are fulfilled:

$\rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.$
$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Applying both these conditions,

$\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.$

For $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c$

Since, c is a positive constant therefore, for $\rm t_o = 0$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0$

Thus, particle does not reach its maximum value at $\rm t = 0\ s.$

For $\rm t_o = \dfrac{2c}{3b}$ ,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.$

Here,

$\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}$

Thus, the particle reach its maximum x value at time $\rm t_o = \dfrac{2c}{3b}.$

7 0

3 years ago

How to calculate displacement?

Blizzard [7]

-- Take a straight ruler.

-- Lay it down with the 'zero' mark at the start point.

-- Rotate it around the start point until the end point is also touching the edge of the ruler.

-- From the marks on the ruler, read the straight-line distance from the start point to the end point.

-- Without moving the ruler, observe and write down the DIRECTION from the start point to the end point.

-- The Displacement is the straight-line distance and direction from the start point to the end point.

4 0

3 years ago

A car initially traveling at 24 m/s slams on the brakes and moves forward 196 m before coming to a complete halt. What was the m

Marrrta [24]

Answer:

$-1.47 m/s^2$

Explanation:

We can use the following SUVAT equation to solve the problem:

$v^2 - u^2 = 2ad$

where

v = 0 is the final velocity of the car

u = 24 m/s is the initial velocity

a is the acceleration

d = 196 m is the displacement of the car before coming to a stop

Solving the equation for a, we find the acceleration:

$a=\frac{v^2-u^2}{2d}=\frac{0-(24)^2}{2(196)}=-1.47 m/s^2$

4 0

2 years ago

PLEASE HELP 15 POINTS

zhuklara [117]

Answer & Explanation:

A magnifying glass is convex lens that forms a virtual image in your retina. A magnifying glass is curved or outward; meaning that it is convex. Please rank Brainliest if this helps. Thanks!

4 0

3 years ago