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3 years ago

What concerns people in regards to the federal government?

Physics

1 answer:

inysia [295]3 years ago

3 0

Answer:

Mulitple Changes.

Explanation:

Most people want something to be done, and never change again. But some others do, since they have there own beliefs. Alot of people are concerned about tax rates, taxes, nationial polices, own beliefs. People have there own beliefs and when someone is elected into the federal Government, they have there own Beliefs (Mostly Republican Or Democrate). People go against on what they dont think and go with what they think. At the end whoever sways the people on who to believe and vote for, Is elected.

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What is the hidden gene called

vagabundo [1.1K]

The hidden gene is called recessive

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3 years ago

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A 100 kg roller coaster comes over the first hill at 2 m/sec (vo). The height of the first hill (h) is 20 meters. See roller dia

aleksandr82 [10.1K]

For the 100 kg roller coaster that comes over the first hill of height 20 meters at 2 m/s, we have:

1) The total energy for the roller coaster at the initial point is 19820 J

2) The potential energy at point A is 19620 J

3) The kinetic energy at point B is 10010 J

4) The potential energy at point C is zero

5) The kinetic energy at point C is 19820 J

6) The velocity of the roller coaster at point C is 19.91 m/s

1) The total energy for the roller coaster at the initial point can be found as follows:

$E_{t} = KE_{i} + PE_{i}$

Where:

KE: is the kinetic energy = (1/2)mv₀²

m: is the mass of the roller coaster = 100 kg

v₀: is the initial velocity = 2 m/s

PE: is the potential energy = mgh

g: is the acceleration due to gravity = 9.81 m/s²

h: is the height = 20 m

The total energy is:

$E_{t} = KE_{i} + PE_{i} = \frac{1}{2}mv_{0}^{2} + mgh = \frac{1}{2}*100 kg*(2 m/s)^{2} + 100 kg*9.81 m/s^{2}*20 m = 19820 J$

Hence, the total energy for the roller coaster at the initial point is 19820 J.

2) The potential energy at point A is:

$PE_{A} = mgh_{A} = 100 kg*9.81 m/s^{2}*20 m = 19620 J$

Then, the potential energy at point A is 19620 J.

3) The kinetic energy at point B is the following:

$KE_{A} + PE_{A} = KE_{B} + PE_{B}$

$KE_{B} = KE_{A} + PE_{A} - PE_{B}$

Since

$KE_{A} + PE_{A} = KE_{i} + PE_{i}$

we have:

$KE_{B} = KE_{i} + PE_{i} - PE_{B} = 19820 J - mgh_{B} = 19820 J - 100kg*9.81m/s^{2}*10 m = 10010 J$

Hence, the kinetic energy at point B is 10010 J.

4) The potential energy at point C is zero because h = 0 meters.

$PE_{C} = mgh = 100 kg*9.81 m/s^{2}*0 m = 0 J$

5) The kinetic energy of the roller coaster at point C is:

$KE_{i} + PE_{i} = KE_{C} + PE_{C}$

$KE_{C} = KE_{i} + PE_{i} = 19820 J$

Therefore, the kinetic energy at point C is 19820 J.

6) The velocity of the roller coaster at point C is given by:

$KE_{C} = \frac{1}{2}mv_{C}^{2}$

$v_{C} = \sqrt{\frac{2KE_{C}}{m}} = \sqrt{\frac{2*19820 J}{100 kg}} = 19.91 m/s$

Hence, the velocity of the roller coaster at point C is 19.91 m/s.

I hope it helps you!

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3 years ago

A woman in a sprint race accelerates from rest to 8.7 m/s in 2.7 s. What is her displacement?

Trava [24]

Answer:

23.49m

Explanation:

Distance = velocity x time

8.7 x 2.7 = 23.49m

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2 years ago

Which statement about gravity is always True? A. The force of gravity between two objects is greater when the objects have less

garri49 [273]

The answer is c because the farther apart they are the greater there gravity is

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3 years ago

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A 4.40-kilogram hoop starts from rest at a height 1.70 m above the base of an inclined plane and rolls down under the influence

Anestetic [448]

Answer:

The linear velocity is $v=4.08m/s$

Explanation:

According to the law of conservation of energy

The potential energy possessed by the hoop at the top of the inclined plane is converted to the kinetic energy at the foot of the inclined plane

The kinetic energy can be mathematically represented as

$KE = \frac{mv^2}{2} + \frac{Iw}{2}$

Where $I$ is the moment of inertia possessed by the hoop which is mathematically represented as

$I = mr^2$

Here R is the radius of the hoop

$w$ is the angular velocity which the hoop has at the bottom of the lower part of the inclined plane which is mathematically represented as

$w = \frac{v}{r}$

Where v linear speed of the hoop's center of mass just as the hoop leaves the incline and rolls onto a horizontal surface

Now expressing the above statement mathematically

$potential \ energy = \frac{mv^2}{y} + \frac{Iw^2}{2}$

$mgh = \frac{mv^2}{y} + \frac{Iw^2}{2}$

=> $mgh =\frac{mv^2}{2} + \frac{(mr^2)(\frac{v}{r})^2 }{2}$

=> $mgh = \frac{mv^2}{2} + \frac{mv^2}{2}$

=> $mgh = mv^2$

=> $v = \sqrt{gh}$

Substituting values

$v = \sqrt{9.81 * 1.7}$

$v=4.08m/s$

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3 years ago

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