What are 4 ways individuals can influence the government?

Answers

Answer 1

Answer: Voting, running, speaking in front of government

Related Questions

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You're carrying a 3.6-m-long, 21 kg pole to a construction site when you decide to stop for a rest. You place one end of the pole on a fence post and hold the other end of the pole 35 cm from its tip. For the steps and strategies involved in solving a similar problem, you may view a Video Tutor Solution. Part A Part complete How much force must you exert to keep the pole motionless in a horizontal position? Express your answer in newtons. F = 114 N Previous Answers
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A disk of radius 10 cm is pulled along a frictionless surface with a force of 16 N by a string wrapped around the edge. At the instant when 28 cm of string has unwound off the disk, what is the torque exerted about the center of the disk?

Answers

The torque exerted about the center of the disk is 0.2845Nm (rounded to 4 decimal places).

What do you mean by torque?

Torque is the rotational equivalent of linear force. It is also referred to as the moment, moment of force, rotational force or turning effect, depending on the field of study. It represents the capability of a force to produce change in the rotational motion of the body.

Moreover, torque is a twisting or turning force that tends to cause rotation around an axis, which might be a center of mass or a fixed point. Torque can also be thought of as the ability of something that is rotating, such as a gear or a shaft, to overcome turning resistance.

Therefore, torque is defined as Γ=r×F=rFsin(θ). In other words, torque is the cross product between the distance vector (the distance from the pivot point to the point where force is applied) and the force vector, 'a' being the angle between r and F.

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Answer:

t = 0.2845Nm (rounded to 4 decimal places)

Explanation:

The disk rotates at a distance of an arc length of 28cm

Arc length = radius × central angle × π/180

28cm = 10cm × central angle × π/180

Central angle = $(28)/(10)$ × 180/π ≈ 160.4°

Torque (t) = rFsin(central angle) , where F is the applied force

Radius in meters = 10/100 = 0.1m

t = 0.1m × 16N × sin160.4°

t = 0.2845Nm (rounded to 4 decimal places)

Listed following are locations and times atwhich different phases of the Moon are visible fromEarth’s....? Listed following are locations and times atwhich different phases of the Moon are visible from Earth’sNorthern Hemisphere. Match these to the appropriate moon phase.

The three given moon phases are:

Waxing Crescent Moon

Waning Crescent Moon

Full Moon

The things we need to match to the above three topicsare:

*visible near western horizon an hour after sunset
*rises about the same time the sun sets
*visible near eastern horizon just before sun rises
*occurs about 3 days before new moon (i know this is waningcrescent)
*visible due south at midnite
*occurs 14 days after new moon (i know this is full moon)

Answers

Answer:

Explanation:

*visible near western horizon an hour aftersunset WAXING CRESCENT

*rises about the same time the sun sets FULLMOON

*visible near eastern horizon just before sun rises WANING CRESCENT

*occurs about 3 days before new moon WANING CRESCENT

*visible due south at midnite FULL MOON

*occurs 14 days after new moon FULL MOON

Two point charges have a total electric potential energy of -24 J, and are separated by 29 cm.If the total charge of the two charges is 45 μC, what is the charge, in μC, on the positive one?
What is the charge, in μC, on the negative one?

Answers

Answer:

The charge of the negative one is 13.27 microcoulombs and the positive one has a charge of 58.27 microcoulombs.

Explanation:

Electric potential energy between two point charges is derived from concept of Work, Work-Energy Theorem and Coulomb's Law and described by the following formula:

$U_(e) = (k\cdot q_(A)\cdot q_(B))/(r)$ (1)

Where:

$U_(e)$ - Electric potential energy, measured in joules.

$q_(A)$ , $q_(B)$ - Electric charges, measured in coulombs.

$r$ - Distance between charges, measured in meters.

$k$ - Coulomb's constant, measured in kilogram-cubic meters per square second-square coulomb.

If we know that $U_(e) = -24\,J$ , $q_(A) = 45* 10^(-6)\,C+ q_(B)$ , $k = 9* 10^(9)\,(kg\cdot m^(3))/(s^(2)\cdot C^(2))$ and $r = 0.29\,m$ , then the electric charge is:

$-24\,J = -(\left(9* 10^(9)\,(kg\cdot m^(3))/(s^(2)\cdot C^(2)) \right)\cdot (45* 10^(-6)\,C+q_(B))\cdot q_(B))/(0.29\,m)$

$-6.96 = -405000\cdot q_(B)-9* 10^(9)\cdot q_(B)^(2)$

$9* 10^(9)\cdot q_(B)^(2)+405000\cdot q_(B) -6.96 = 0$ (2)

Roots of the polynomial are found by Quadratic Formula:

$q_(B,1) = 1.327* 10^(-5)\,C$ , $q_(B,2) \approx -5.827* 10^(-5)\,C$

Only the first roots offer a solution that is physically reasonable. The charge of the negative one is 13.27 microcoulombs and the positive one has a charge of 58.27 microcoulombs.

You have a battery marked " 6.00 V ." When you draw a current of 0.361 A from it, the potential difference between its terminals is 5.07 V . What is the potential difference when you draw 0.591 A

Answers

Answer:

Explanation:

Battery voltage is 6V

A current of 0.361A is draw the voltage reduces to, 5.07V

This shows that the appliances resistance that draws the currents is

Using KVL

The battery has an internal resistance r

V=Vr+Va

Vr is internal resistance voltage

Va is appliance voltage

6=5.07+Va

Va=6-5.07

Va=0.93

Using ohms law to the resistance of the appliance

Va=iR

R=Va/i

R=0.93/0.361

R=2.58ohms

Then if the circuit draws a current of 0.591A

Then the voltage across the load is

V=iR

Va=0.591×2.58

Va=1.52V

Then the voltage drop at the internal resistance is

V=Vr+Va

Vr=V-Va

Vr=6-1.52

Vr=4.48V

Answer:

V = 4.48 V

Explanation:

• As the potential difference between the battery terminals, is less than the rated value of the battery, this means that there is some loss in the internal resistance of the battery.

• We can calculate this loss, applying Ohm's law to the internal resistance, as follows:

$V_(rint) = I* r_(int)$

• The value of the potential difference between the terminals of the battery, is just the voltage of the battery, minus the loss in the internal resistance, as follows:

$V = V_(b) - V_(rint) = 5.07 V = 6.00 V - 0.361 A * r_(int)$

• We can solve for rint, as follows:

$r_(int) =(V_(b)- V_(rint) )/(I) = (6.00 V - 5.07 V)/(0.361A) = 2.58 \Omega$

• When the circuit draws from battery a current I of 0.591A, we can find the potential difference between the terminals of the battery, as follows:

$V = V_(b) - V_(rint) = 6.00 V - 0.591 A * 2.58 \Omega = 4.48 V$

• As the current draw is larger, the loss in the internal resistance will be larger too, so the potential difference between the terminals of the battery will be lower.

Those who support the alien theory believe that the ancient Egyptians where very primitive. What evidence about the ancient Egyptians suggests that claim is inaccurate?

Answers

Answer:

They had a well-developed agricultural system , they raised domesticated animals , they developed a writing system, and they used early forms of tools.

Final answer:

The claim that the ancient Egyptians were primitive and relied on aliens to build their monuments is inaccurate. Evidence from archaeology and history shows that the ancient Egyptians had advanced knowledge and skills in various fields. Their construction techniques and use of mathematics in building the pyramids are well-documented.

Explanation:

The claim that the ancient Egyptians were primitive and that their accomplishments, such as building the pyramids, were assisted by aliens is inaccurate. There is evidence from archaeology and history that ancient monuments were built by ancient people using their own ingenuity and capabilities. The ancient Egyptians had advanced knowledge in various fields including architecture, engineering, astronomy, and mathematics. For example, their construction techniques and use of mathematics in building the pyramids and other structures are well-documented.

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A toy car that is 0.12 m long is used to model the actions of an actual car thatis 6 m long. Which ratio shows the relationship between the sizes of the
model and the actual car?
A: 5:1
B: 1:50
C: 50:1
D: 1:5

Answers

Answer:

it is B 1:50

Explanation:

just did it on apex

Final answer:

The ratio that shows the relationship between the sizes of the model car and the actual car is 1:50. This is because the actual car is 50 times longer than the model car.

Explanation:

The relationship between the sizes of the model car and the actual car is represented by a ratio. To find this ratio, we can divide the length of the actual car by the length of the model car. So, 6 m / 0.12 m = 50. This means that the actual car is 50 times longer than the model car, or in other words, the model car is 1/50th the size of the actual car. Therefore, the correct ratio is 1:50.

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An object is projected with speed of 4ms at an angle of 60° to horizontal. Calculate the time of flight of the object. (g=10ms2)

A large aquarium has portholes of thin transparent plastic with a radius of curvature of 1.95 m and their convex sides facing into the water. A shark hovers in front of a porthole, sizing up the dinner prospects outside the tank.a) If one of the sharks teeth is exactly 46.5 cm from the plastic, how far from the plastic does it appear to be to observers outside the tank? (You can ignore refraction due to the plastic.)b) Does the shark appear to be right side up or upside down?c) If the tooth has an actual length of 5.00 cm, how long does it appear to the observers?

If a small child swallowed a safety pin, whywould an X-ray photograph clearly show thelocation of the pin?

A lead ball is dropped into a lake from a diving board 6.10 mm above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 4.50 ss after it is released. How deep is the lake?