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3 years ago

What is the function of law

1 answer:

34kurt3 years ago

5 0

There are many functions and purposes of law, but here are some of the most important ones:
<span>establishing standards, maintaining order, resolving disputes, and protecting liberties and rights.</span>

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Please someone help me lol

Makovka662 [10]

From Largest to Smallest:

Earth > Biosphere > Biome > Ecosystem > Community > Population > Organism > Organelles

Hope this helps!

7 0

3 years ago

Substitution solve for x & y<br>-4x - 2y equals 14<br>-10x+7y=-25

den301095 [7]

Answer:

-4×-2y=14 (1)

-10×+7y=-25 (2)

multiplying eq 1 by 7 and eq 2 by 2 and add eq. 1 and 2

-28×-14y=98

-20×+14y=-50

___________

-28×=48

×=48/-28

×=-12/7

now

-4×-2y=14

-4*-12/7-2y=14

48/7-2y=14

-2y=14-48/7

-2y=(98-48)/7

-2y=50/7

y=-50/14

y=-25/7

8 0

3 years ago

In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it

bogdanovich [222]

Solution :

Given :

M = 0.35 kg

$m=\frac{M}{2}=0.175 \ kg$

Total mechanical energy = constant

or $K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$

But $K.E._{top} = 0$ and $P.E._{bottom} = 0$

Therefore, potential energy at the top = kinetic energy at the bottom

$\Rightarrow mgh = \frac{1}{2}mv^2$

$\Rightarrow v = \sqrt{2gh}$

$=\sqrt{2 \times 9.8 \times 0.35}$ (h = 35 cm = 0.35 m)

= 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$

here, $m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$

∴ $$v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}$

$=\frac{0.175}{0.525}+0$

= 0.33 m/s

Therefore, velocity after the collision of mass M = 0.33 m/s

3 0

3 years ago

An object located near the surface of Earth has a weight of a 245 N

gayaneshka [121]

Answer:

ثر أنواع التربة خصوبة التربحمراء .

ج- السوداء

Explanation:

ggfvbdgbvgbbbfgb

7 0

3 years ago

An electron moves with a speed of 8.0×106m/s along the -z-axis. It enters a region where there is a uniform magnetic field B = (

Crazy boy [7]

Answer:

Acceleration, $a=9.36\times 10^{18}\ m/s^2$

Explanation:

It is given that,

Speed of electron, $v=8\times 10^6\ m/s$

Charge on an electron, $q=1.6\times 10^{-19}\ C$

Mass of electron, $m=9.1\times 10^{-31}\ kg$

Magnetic field, $B=5.5i-3.7j$

Magnitude, $|B|=\sqrt{5.5^2+(-3.77)^2}=6.66\ T$

Magnetic force is given by :

$F=qvB$

Also, F = ma

$a=\dfrac{qvB}{m}$

$a=\dfrac{1.6\times 10^{-19}\times 8\times 10^6\times 6.66}{9.1\times 10^{-31}}$

$a=9.36\times 10^{18}\ m/s^2$

So, the acceleration of the electron is $9.36\times 10^{18}\ m/s^2$ . Hence, this is the required solution.

5 0

3 years ago