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2 years ago

KFC has engineered chickens that do not have beaks and double breast

1 answer:

6 0

I don't think so.

Before I can accept that rumor, I'll need to see the signed testimony of
two reputable adult witnesses, who have no connection to any business
that profits in any way from the production, sale, or consumption of food,
and who will swear in a public meeting, presided over by three respected
members of their community, that they personally saw more than one of
the alleged beakless double-breasted chickens in the same place at the
same time.

A photo is not acceptable ... not for me, and not for the two witnesses.

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A cyclist going downhill is accelerating at 1. 2 m/s2. If the final velocity of the cyclist is 16 m/s after 10 seconds, what is

mel-nik [20]

Answer:

$\boxed {\boxed {\sf v_i= 4 \ m/s}}$

Explanation:

We are asked to find the cyclist's initial velocity. We are given the acceleration, final velocity, and time, so we will use the following kinematic equation.

$v_f= v_i + at$

The cyclist is acceleration at 1.2 meters per second squared. After 10 seconds, the velocity is 16 meters per second.

$v_f$ = 16 m/s
a= 1.2 m/s²
t= 10 s

Substitute the values into the formula.

$16 \ m/s = v_i + (1.2 \ m/s^2)(10 \ s)$

Multiply.

$16 \ m/s = v_i + (1.2 \ m/s^2 * 10 \ s)$

$16 \ m/s = v_i + 12 \ m/s$

We are solving for the initial velocity, so we must isolate the variable $v_i$ . Subtract 12 meters per second from both sides of the equation.

$16 \ m/s - 12 \ m/s = v_i + 12 \ m/s -12 \ m/s$

$4 \ m/s = v_i$

The cyclist's initial velocity is <u>4 meters per second.</u>

6 0

2 years ago

Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .

Verdich [7]

According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit.

This Law is originally expressed as follows:

Where;

$G$ is the Gravitational Constant and its value is $6.674(10^{-11})\frac{m^{3}}{kgs^{2}}$

$M=1.9(10^{27})kg$ is the mass of Jupiter

$a=4.22(10^{5})km=4.22(10^{8})m$ is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}$

$T=\sqrt{\frac{2.966(10^{27})m^{3}}{1.268(10^{17})m^{3}/s^{2}}}$

$T=\sqrt{2.339(10^{10})s^{2}}$

Then:

Which is the same as:

Therefore, the answer is:

The orbital period of Io is 42.482 h

7 0

3 years ago

What is a Current (amp)

JulsSmile [24]

Answer:

Amp – an ampere a the unit for measuring electricity. The rate at which electricity flows is measured as an electric current. The electric current is measured in Amps.

Hope this helps:)

If correct, can I please have brainliest?

Thank you.

6 0

2 years ago

A glass of water sitting in direct sunlight evaporates over time. Explain this phase change in terms of the types of heat transf

olga_2 [115]

Well over time the temperature of the water increase causeing its moleclues to move alot which causes it's state change to a gas this is who we have rain.

6 0

3 years ago

Read 2 more answers

One gram of Uranium averages release 1.01 KJ (10^7) of energy. How much mass could be converted to energy to release this much e

frutty [35]

Answer:

The amount of mass that needs to be converted to release that amount of energy is $1.122 X 10^{-7} kg$