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Yanka [14]
2 years ago
8

Define cyber law and its types.​

Physics
1 answer:
butalik [34]2 years ago
5 0

Answer:

Cyber Law is a generic term referring to all the legal and regulatory aspects of the internet. Everything concerned with or related to or emanating from any legal aspects or concerning any activities of the citizens in the cyberspace comes within the ambit of cyber laws.

<h2>Types :-</h2>
  • Copyright
  • Patents
  • Trademarks or Service Marks
  • Trade Secrets
  • Domain Disputes
  • Contracts
  • Privacy
  • Employment
  • Defamation

Explanation:

<h2>HOPE IT HELPS YOU!! </h2>
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Objects with masses of 130 kg and a 430 kg are separated by 0.300 m.(a) Find the net gravitational force exerted by these object
fredd [130]

Answer

given,

mass of object

m₁ = 130 Kg

m₂ = 430 Kg

distance between them = 0.3 m

a) net force when 35 kg is place in between them

   F = \dfrac{GMm}{R^2}

now,

   F = - \dfrac{6.67 \times 10^{-11}\times 130 \times 30}{0.15^2} +\dfrac{6.67 \times 10^{-11}\times 430 \times 30}{0.15^2}

   F = 3.12 \times 10^{-5}\ N

   Direction of force will be toward the mass of 430 Kg

b) position where force will be zero

 F =-\dfrac{6.67 \times 10^{-11}\times 130 \times 30}{(0.3-x)^2} +\dfrac{6.67 \times 10^{-11}\times 430 \times 30}{x^2} = 0

 -\dfrac{130}{(0.3-x)^2} +\dfrac{430}{x^2} = 0

 x^2- 0.6 x + 0.09=\dfrac{130}{430}x^2

 0.7 x^2- 0.6 x + 0.09 =0

solving the above equation

  x = 0.1936 m

the distance of third mass will be at x = 0.1936 m from 430 Kg mass.

7 0
3 years ago
If one firecracker produces a sound intensity of 90db, what would be the intensity of a sound produced by 1000 firecrackers, all
Kryger [21]

Answer:

90db

Explanation:

The 1000, will produce sa.e intensity, since the firecrackers are made of same materials

5 0
4 years ago
Read 2 more answers
What causes melting of material under divergent plate boundaries?
Triss [41]
Melting: as mantle material rise toward the divergent plate boundary the pressure is reduced which causes melting
5 0
4 years ago
The pressure at the bottom of a full barrel of water is Poriginal . Determine what happens to the pressure when the radius or he
Sholpan [36]

Answer:

a)   P' = P_original, b)  P ’= P_original + ρ  g Δh

Explanation:

The expression for nanometric pressure is

          P = ρ g h

where ρ  is the density of the liquid and h is the height

a) we change the radius of the barrel, but keeping the same height

as the pressure does not depend on the radius it remains the same

        P' = P_original

b) We change the barrel height

         h ’≠ h

we substitute in the equation

      P ’= ρ  g h’

      h ’= h + Δh

      P ’= ρ  g (h + Δh)

      P ’= (ρ  g h) + ρ  g Δh

      P ’= P_original + ΔP

In this case, the pressure changes due to the new height,

*if it is higher than the initial one, the pressure increases

*if the height is less than the initial one, the pressure is less

3 0
3 years ago
A 50g ball is released from rest 1.0 above the bottom of thetrack
ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

y=\dfrac{1}{4}x^2

We need to calculate the velocity

Using conservation of energy

\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

\Delta U_{i}=\Delta K_{f}

Put the value into the formula

mgh=\dfrac{1}{2}mv^2

Put the value into the formula

50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2

v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}

v=\sqrt{19.6}

v=4.42\ m/s

We need to calculate the maximum height of the ball

Using again conservation of energy

\dfrac{1}{2}mv^2=mgh

Here, h = y highest point

Put the value into the formula

\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h

y=\dfrac{0.5\times(4.42)^2}{9.8}

y=0.996\ m

Put the value of y in the given equation

y=\dfrac{1}{4}x^2

x^2=4\times0.996

x=\sqrt{4\times0.996}

x=1.99\ m\ \approx 2 m

Hence, The maximum height of the ball is 2 m.

4 0
3 years ago
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