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3 years ago

Define the unit of time and unit of length

2 answers:

7 0

This is your perfect answer

The base unit for time is the second (the other SI units are: metre for length, kilogram for mass, ampere for electric current, kelvin for temperature, candela for luminous intensity, and mole for the amount of substance). The second can be abbreviated as s or sec.

djyliett [7]3 years ago

6 0

Answer:

hope you like

Explanation:

A unit of time is any particular time interval, used as a standard way of measuring or expressing duration.Its SI unit is second

A unit of length refers to any arbitrarily chosen and accepted reference standard for measurement of length. Its SI unit is meter

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What s physics?

Alex777 [14]

The answer is (B. The study of Matter and Energy) but technically you could consider physics all of these as engineering is based on physics and that would be the study of inventions, chemistry and biology were both discovered because of physics, and physics invokes more math than any other subject as it applies math to the entire Universe.

7 0

4 years ago

Read 2 more answers

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

A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. The ladder weigh

dem82 [27]

Answer:

Explanation:

a )

Reaction force of the ground

R = mg

= 160 N

Maximum friction force possible

= μ x R

= μ x 160

= .4 x 160

= 64 N .

b )

160 N will act at middle point . 740N will act at distance of 3 / 5 m from the wall ,

Taking moment about top point of ladder

160 x 1.5 + 740 x 3/5 + f x 4 = 900 x 3

240 + 444 + 4f = 2700

f = 504 N

c )

Let x be the required distance.

Taking moment about top point of ladder

160 x 1.5 + 740 x 3 x / 5 + .4 x 900 x 4 = 900 x 3 ( .4 x 900 is the maximum friction possible )

240 + 444 x + 1440 = 2700

x = 2.3 m

so man can go upto 2.3 at which maximum friction acts .

8 0

4 years ago

The heating of the filament is what causes the light production (photon emission), and heating is caused by the current in the l

alexira [117]

Answer:

explained

Explanation:

Yes, the heating of filament is what causes the light production (photon emission), and this heating is caused because of current in the light bulb

(H= i^2*R*t i=current, H= heat, t= time and R= resistance).But using constant current source is not a good idea because in constant current source resistance is very low that can cause short circuit and ultimately fusing it. Whereas in constant voltage source current adjusts itself and prevents fusing because of high resistance in the circuit.

8 0

4 years ago

A 5-turn square loop (10 cm along aside, resistance = 4.0 ) is placed in a magnetic field that makes an angle of30o with the pla

andreev551 [17]

Answer:

Explanation:

Area A of the coil = .1 x .1 = .01 m²

no of turns n = 5

magnetic field B = .5 t²

Flux Φ perpendicular to plane passing through it.= nBA sin30

rate of change of flux

dΦ/dt = nAdBsin30 / dt

= nA d/dt (.5t²x .5 )

= nA x 2 x .25 x t

At t = 4s

dΦ/dt = nA x 2

= 5x .01 x 2

= .1

current = induced emf / resistance

= .1 / 4

= .025 A

= 25 mA.

4 0

3 years ago

Define the unit of time and unit of length​

Define the unit of time and unit of length