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allochka39001 [22]
3 years ago
10

Gave the examples of Civil law​

Physics
2 answers:
RUDIKE [14]3 years ago
8 0

Answer:

If someone slips on your store, you must pay for their medical bills because it is your fault.

KonstantinChe [14]3 years ago
6 0
Examples are murder, assault, theft,and drunken driving.
You might be interested in
The revolution of the earth around the sun demonstrate what motion?​
olga2289 [7]

Answer:

Anticlockwise directions

Please mark me Brainliest to help me

3 0
1 year ago
When the input voltages of a difference amplifier are 5.1 v and 6.4 v, the output voltage is 64.7 v. The inputs are changed to 4
daser333 [38]

Answer:

a) Differential mode gain = 48

b) Common mode gain = 0.4

c) CMRR = 120

Explanation:

The output of a difference amplifier is related to the input by the equation:

V_{0} = A_{1} V_{1} + A_{2} V_{2} \\

When V₁ = 6.4 V, V₂ = 5.1 V and V₀ = 64.7 V, the equation becomes

6.4 A₁ + 5.1 A₂ = 64.7.....................(1)

When V₁ = 5.6 V, V₂ = 4.9 V and V₀ = 35.7 V, the equation becomes

5.6 A₁ + 4.9 A₂ = 35.7.....................(2)

Multiply equation (1) by 5.6  and (2) by 6.4

35.84 A₁ + 28.56A₂ = 362.32.....................(3)

35.84 A₁ + 31.36 A₂ = 228.48....................................(4)

Subtract equation (3) from (4)

2.8 A₂ = -133.84

A₂ = -133.84/2.8

A₂ = -47.8

Put the value of  A₂ into equation (1)

6.4 A₁ + 5.1 (-47.8) = 64.7

6.4 A₁ = 64.7 + 243.78

A₁ = 308.48/6.4

A₁ = 48.2

a) Common mode gain = A₁ + A₂ = 48.2 + (-47.8)

Common mode gain = 0.4

b) Differential mode gain = (A₁ -A₂)/2

Differential mode gain = (48.2 - (-47.8))/2

Differential mode gain = 96/2

Differential mode gain = 48

c) Common Mode Rejection Ratio (CMRR)

CMRR = |\frac{Differential Mode Gain}{Common Mode Gain} |

CMRR = |\frac{48}{0.4} |\\CMRR = 120

4 0
3 years ago
A 100-kg spacecraft is in a circular orbit about Earth at a height h = 2RE .
maria [59]

To solve this problem it is necessary to apply the concepts related to the conservation of the Gravitational Force and the centripetal force by equilibrium,

F_g = F_c

\frac{GmM}{r^2} = \frac{mv^2}{r}

Where,

m = Mass of spacecraft

M = Mass of Earth

r = Radius (Orbit)

G = Gravitational Universal Music

v = Velocity

Re-arrange to find the velocity

\frac{GM}{r^2} = \frac{v^2}{r}

\frac{GM}{r} = v^2

v = \sqrt{\frac{GM}{r}}

PART A ) The radius of the spacecraft's orbit is 2 times the radius of the earth, that is, considering the center of the earth, the spacecraft is 3 times at that distance. Replacing then,

v = \sqrt{\frac{(6.67*10^{-11})(5.97*10^{24})}{3*(6.371*10^6)}}

v = 4564.42m/s

From the speed it is possible to use find the formula, so

T = \frac{2\pi r}{v}

T = \frac{2\pi (6.371*10^6)}{4564.42}

T = 8770.05s\approx 146min\approx 2.4hour

Therefore the orbital period of the spacecraft is 2 hours and 24 minutes.

PART B) To find the kinetic energy we simply apply the definition of kinetic energy on the ship, which is

KE = \frac{1}{2} mv^2

KE = \frac{1}{2} (100)(4564.42)^2

KE = 1.0416*10^9J

Therefore the kinetic energy of the Spacecraft is 1.04 Gigajules.

8 0
3 years ago
An insulated rigid tank contains 3 kg of H2O in the form of a saturated mixture of liquid and vapor at a pressure of 150 kPa and
Andre45 [30]

Answer:

change in entropy is 1.44 kJ/ K

Explanation:

from steam tables

At 150 kPa

specific volume

Vf = 0.001053 m^3/kg

vg = 1.1594 m^3/kg

specific entropy values are

Sf = 1.4337 kJ/kg K

Sfg = 5.789 kJ/kg

initial specific volume is calculated as

v_1 = vf + x(vg - vf)

      = 0.001053 + 0.25(1.1594 - 0.001053)

v_1 = 0.20964  m^3/kg

s_1 = Sf + x(Sfg)

     = 1.4337 + 0.25 \times 5.7894 = 2.88 kJ/kg K

FROM STEAM Table

at 200 kPa

specific volume

Vf = 0.001061 m^3/kg

vg = 0.88578 m^3/kg

specific entropy values are

Sf = 1.5302 kJ/kg K

Sfg = 5.5698 kJ/kg

constant volume  sov_1 -  v_2  = 0.29064 m^3/kg

v_2 = v_1 = vf + x(vg - vf)

       =0.29064 = x_2(0.88578 - 0.001061)

x_2 = 0.327

s_2 = 1.5302 + 0.32 \times 5.5968 = 3.36035 kJ/kg K

Change in entropy \Delta s = m(s_2 - s_1)

              =3( 3.36035 - 2.88) =  1.44 kJ/kg

8 0
3 years ago
An object, initially at rest, moves 250 m in 17 s. What is its acceleration?
mash [69]

Answer:

1.73 m/s²

Explanation:

Given:

Δx = 250 m

v₀ = 0 m/s

t = 17 s

Find: a

Δx = v₀ t + ½ at²

250 m = (0 m/s) (17 s) + ½ a (17 s)²

a = 1.73 m/s²

5 0
3 years ago
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