PHYSICS COLLEGE

What best describes a societal law

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Answer 1

Answer:

Answer:

Societal laws are based on the behavior and conduct made by society or government.

hope it helps.

stay safe healthy and happy.

Related Questions

A person is pushing a lawnmower of mass m D 38 kg and with h D 0:75 m, d D 0:25 m, `A D 0:28 m, and `B D 0:36 m. Assuming that the force exerted on the lawnmower by the person is completely horizontal and that the mass center of the lawnmower is at G, and neglecting the rotational inertia of the wheels, determine the minimum value of this force that causes the rear wheels (labeled A) to lift off the ground. In addition, determine the corresponding acceleration of the mower.
A sphere has a charge of −84.0 nC and a radius of 5.00 cm. What is the magnitude of its electric field 3.90 cm from its surface?
a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How far from the wall should the observer stand to get the best view?
A turntable with a rotational inertia of 0.0120 kg∙m2 rotates freely at 2.00 rad/s. A circular disk of mass 200 g and radius 30.0 cm, and initially not rotating, slips down a spindle and lands on the turntable. (a) Find the new angular velocity. (b) What is the change in kinetic energy?
A smart phone charger delivers charge to the phone, in the form of electrons, at a rate of -0.75. How many electrons are delivered to the phone during 27 min of charging?

A rock is thrown from the top of a building 146 m high, with a speed of 14 m/s at an angle 43 degrees above the horizontal. When it hits the ground, what is the magnitude of its velocity (i.e. its speed).

Answers

Answer:

time is 32 s and speed is 304.3 m/s

Explanation:

Height, h = 146 m

speed, u = 14 m/s

Angle, A = 43 degree

Let it hits the ground after time t.

Use second equation of motion

$h = u t +0.5 at^2\n\n- 146 =14 sin 43 t - 4.9 t^2\n\n4.9 t^2 - 9.5 t - 146 =0 \n\nt =\frac{9.5\pm\sqrt {90.25 + 2861.6}}{9.8}\n\nt=(9.5\pm 54.3)/(9.8)\n\nt = 32.05 s, - 22.4 s$

Time cannot be negative so the time is t = 32 s .

The vertical velocity at the time of strike is

v' = u sin A - g t

v' = 14 sin 43 - 9.8 x 32 = 9.5 - 313.6 = - 304.1 m/s

horizontal velocity

v'' = 14 cos 43 =10.3 m/s

The resultant velocity at the time of strike is

$v=√(v'^2 + v''^2)\n\nv = √(304.1^2 +10.3^2 )\n\nv = 304.3 m/s$

A Lincoln Continental is twice as long as a VW Beetle, when they are at rest. On a 2-lane road, the Continental driver passes the VW Beetle. Unfortunately for the Continental driver, a stationary policeman has set up a speed trap, and the policeman observes that the Continental and the Beetle have the same length. The VW is going at half the speed of light. How fast is the Lincoln going ? (Express your answer as a multiple of c).

Answers

Answer:

V(t) = √13/4c

Explanation:

See attachment

A tightly wound solenoid is 15 cm long, has 350 turns, and carries a current of 4.0 A. If you ignore end effects, you will find that the value of app at the center of the solenoid when there is no core is approximately

Answers

Answer:

The magnetic field at the center of the solenoid is approximately 0.0117 T

Explanation:

Given;

length of the solenoid, L = 15 cm = 0.15 m

number of turns of the solenoid, N = 350 turns

current in the solenoid, I = 4.0 A

The magnetic field at the center of the solenoid is given by;

$B = \mu_o ((N)/(L) )I\n\nB = (4 \pi *10^(-7))((350)/(0.15) )(4.0)\n\nB = 0.0117 \ T$

Therefore, the magnetic field at the center of the solenoid is approximately 0.0117 T.

In order to work well, a square antenna must intercept a flux of at least 0.040 N⋅m2/C when it is perpendicular to a uniform electric field of magnitude 5.0 N/C.

Answers

Answer:

L > 0.08944 m or L > 8.9 cm

Explanation:

Given:

- Flux intercepted by antenna Ф = 0.04 N.m^2 / C

- The uniform electric field E = 5.0 N/C

Find:

- What is the minimum side length of the antenna L ?

Solution:

- We can apply Gauss Law on the antenna surface as follows:

Ф = $\int\limits^S {E} \, dA$

- Since electric field is constant we can pull it out of integral. The surface at hand is a square. Hence,

Ф = E.(L)^2

L = sqrt (Ф / E)

L > sqrt (0.04 / 5.0)

L > 0.08944 m

Final answer:

The area of a square antenna needed to intercept a flux of 0.040 N⋅m2/C in a uniform electric field of magnitude 5.0 N/C is 0.008 m². Consequently, each side of the antenna must be about 0.089 meters (or 8.9 cm) long.

Explanation:

The question pertains to the relationship between electric field and flux. The electric flux through an area is defined as the electric field multiplied by the area through which it passes, oriented perpendicularly to the field.

We are given that the electric field (E) is 5.0 N/C and the flux Φ must be 0.040 N⋅m2/C.

Hence, to intercept this amount of flux, the antenna must have an area (A) such that A = Φ / E.

That is, A = 0.040 N⋅m2/C / 5.0 N/C = 0.008 m².

Since the antenna is square, each side will have a length of √(0.008) ≈ 0.089 meters (or 8.9 cm).

Learn more about Electric Flux here:

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A meteoroid, heading straight for Earth, has a speed of 14.8 km/s relative to the center of Earth as it crosses our moon's orbit, a distance of 3.84 × 108 m from the earth's center. What is the meteroid's speed as it hits the earth

Answers

Answer:

The meteoroid's speed is 18.5 km/s

Explanation:

Given that,

Speed = 14.8 km/s

Distance $d= 3.84*10^(8)$

We need to calculate the meteoroid's speed

The total initial energy

$E_(i)=K_(i)+U_(i)$

$E_(i)=(1)/(2)mv_(i)^2-(GM_(e)m)/(r)$

Where, m = mass of meteoroid

G = gravitational constant

$M_(e)$ =mass of earth

r = distance from earth center

Now, The meteoroid hits the earth then the distance of meteoroid from the earth 's center will be equal to the radius of earth

The total final energy

$E_(f)=K_(f)+U_(f)$

$E_(f)=(1)/(2)mv_(f)^2-(GM_(e)m)/(r_(e))$

Where,

$r_(e)$ =radius of earth

Using conservation of energy

$E_(i)=E_(j)$

Put the value of initial and final energy

$(1)/(2)mv_(i)^2-(GM_(e)m)/(r)=(1)/(2)mv_(f)^2-(GM_(e)m)/(r_(e))$

$v_(f)^2=v_(i)^2+2GM_(e)((1)/(r_(e))-(1)/(r))$

Put the value in the equation

$v_(f)^2=(14.8*10^(3))^2+2*6.67*10^(-11)*5.97*10^(24)((1)/(6.37*10^(6))-(1)/(3.84*10^(8)))$

$v_(f)=\sqrt{(14.8*10^(3))^2+2*6.67*10^(-11)*5.97*10^(24)((1)/(6.37*10^(6))-(1)/(3.84*10^(8)))}$

$v_(f)=18492.95\ m/s$

$v_(f)=18.5\ km/s$

Hence, The meteoroid's speed is 18.5 km/s

Final answer:

To find the meteoroid's speed as it hits the Earth, we can use the principle of conservation of mechanical energy. The final velocity of the meteoroid is approximately 13.4 km/s.

Explanation:

To find the meteoroid's speed as it hits the Earth, we can use the principle of conservation of mechanical energy. Since there is no air friction, the mechanical energy of the meteoroid is conserved as it falls towards Earth. The initial kinetic energy of the meteoroid is equal to the final kinetic energy plus the gravitational potential energy.

First, we find the initial kinetic energy of the meteoroid using the formula KE = (1/2)mv^2, where m is the mass of the meteoroid and v is its initial velocity relative to the center of the Earth. Since the mass is not given, we can assume it cancels out in the equation.

Next, we calculate the gravitational potential energy of the meteoroid using the formula PE = mgh, where g is the acceleration due to gravity (approximately 9.8 m/s^2) and h is the height from which the meteoroid fell. The height can be calculated by subtracting the radius of the Earth from the distance from the center of the Earth to the moon's orbit (h = 3.84 × 10^8 m - 6.37 x 10^6 m).

Solving for the final velocity, we equate the initial kinetic energy and the sum of the final kinetic energy and gravitational potential energy. Rearranging the equation, we find that the final velocity is the square root of (initial velocity squared minus 2 times g times h).

Plugging in the given values, the final velocity of the meteoroid as it hits the Earth is approximately 13.4 km/s.

Learn more about conservation of mechanical energy here:

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A hemispherical surface (half of a spherical surface) of radius R is located in a uniform electric field of magnitude E that is parallel to the axis of the hemisphere. What is the magnitude of the electric flux through the hemisphere surface?

Answers

Answer:

π*R²*E

Explanation:

According to the definition of electric flux, it can be calculated integrating the product E*dA, across the surface.

As the electric field E is uniform and parallel to the hemisphere axis, and no charge is enclosed within it, the net flux will be zero, so, in magnitude, the flux across the opening defining the hemisphere, must be equal to the one across the surface.

The flux across the open surface can be expressed as follows:

$\int\ {E} \, dA = E*A = E*\pi *R^(2)$

As E is constant, and parallel to the surface vector dA at any point, can be taken out of the integral, which is just the area of the surface, π*R².

⇒Flux = E*π*R²

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