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4 years ago

What is the difference between a learner's license and a operator's licence

2 answers:

6 0

A learner's license, like the name suggest (learner), you are just beginning to drive and are "learning" the basics of the car.
For a operator's license, you can drive on your own, you have the knowledge and skills and have taken the required courses and passed them.

tamaranim1 [39]4 years ago

5 0

Also, with a learner's license/permit, you must have a licensed driver in the vehicle with you when you operate the vehicle.

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Si olvidas un trozo de chocolate cerca de la estufa encendida, al rato, observa tu chocolate derretido, ¿Cómo estarán las partíc

eduard

Answer:

partículas estarían separadas

Explanation:

ya que con el calor las partículas se alborotan y con frío se congelan

4 0

3 years ago

Hey Jatin .

snow_tiger [21]

Answer:

what? i cant understand you pls translate in English

6 0

4 years ago

A 1.80-kg monkey wrench is pivoted 0.250 m from its center of mass and allowed to swing as a physical pendulum. The period for s

Evgen [1.6K]

Answer:

(a) I (Moment of inertia)=0.0987 $kgm^{2}$

(b) W(Angular Speed)=2.66 $\frac{rad}{s}$

Explanation:

Given data

m (Monkey mass)= 1.80 kg

d=2.50 m

T (Time Period)=0.940 s

Angle= 0.400 rad

(a) I (Moment of Inertia)=?

(b) W (Angular Speed)=?

For part (a) I (Moment of Inertia)=?

Time Period Formula is given as

$T=2(3.14)\sqrt{\frac{I}{mgd} }$

After Simplifying we get

$I=\frac{mgdT^{2} }{4*(3.14)^{2}}$

$I=\frac{1.8*9.8*0.25*(0.94)^{2} }{4(3.14)^{2} }$

$I=0.0987 kgm^{2}$

For Part (b) Angular Speed

From Kinetic Energy we get

$KE=\frac{1}{2}IW^{2}$

Pontential Energy

$PE=mgd(1-Cosa)$

KE=PE

$\frac{1}{2}IW^{2}=mgd(1-Cosa)$

$W^{2}=\frac{2mgd(1-Cosa)}{I}$

$W=\sqrt{\frac{2*1.8*9.8*0.25*(1-Cos(0.4)rad)}{0.0987} }$

$W=2.66\frac{rad}{s}$

5 0

3 years ago

a hiker in africa discovers a skull that contains 43% of its original amount of c-14 Find the age of the skull to the nearest ye

11Alexandr11 [23.1K]

The amount of C-14 at time t is
$N(t) = N_{0} e^{-Kt}$
where
N₀ = inital amount of C-14
t = years
K = 0.0001 = 10⁻⁴

When the skull was discovered, the amount of C-14 contained 43% of N₀.
Therefore
$e^{-10^{-4}t} = 0.43 \\\\ -10^{-4} t = ln(0.43) \\\\ t = - \frac{ln(0.43)}{10^{-4}} =8439.7 \, years$

Answer:
The age of the skull s 8440 years (nearest integer)

8 0

4 years ago

Read 2 more answers

Rank the following compounds in order of decreasing vapor pressure.

liubo4ka [24]

Hello ? What compounds are you referring to ?

3 0

4 years ago