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3 years ago

Where would a car traveling on a roller coaster have the most kinetic energy ? and why?

Physics

2 answers:

goldenfox [79]3 years ago

4 0

Answer:

As the car travels up the coaster it is gaining potential energy.

Explanation:

Because It has the greatest in amount of potential energy at the top of the coaster. when the car travels down the roller coaster it obtains speed and kinetic energy.

Soloha48 [4]3 years ago

3 0

Most potential energy at the top of a hill/loop and the most kinetic energy at the bottom of a hill/loop

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If you stretch a rubber hose and pluck it, you can observe a pulse traveling up and down the hose.(i) What happens to the speed

Monica [59]

The thing that happens to the speed of the pulse when you stretch the hose more tightly is that it increases.

<h3>What is wage speed?</h3>

It should be noted that wave speed simply means the distance that a wave travels during a particular time.

It should be noted that higher tension leads to an increase in the speed of the wave.

Therefore, the thing that happens to the speed of the pulse when you stretch the hose more tightly is that it increases.

Learn more about speed on:

brainly.com/question/13943409

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1 year ago

Separating the electron from the proton in a hydrogen atom takes 2.18 ✕ 10−18 j of work. through what electric potential differe

My name is Ann [436]

Electric potential = work done/charge of electron = 2.18×10⁻¹⁸/1.6×10⁻¹⁹
= 13.625 V

6 0

3 years ago

At a certain location, wind is blowing steadily at 9 m/s. Determine the mechanical energy of air per unit mass and the power gen

Misha Larkins [42]

Answer:

The specific mechanical energy of the air in the specific location is 40.5 J/kg.
The power generation potential of the wind turbine at such place is of 2290 kW
The actual electric power generation is 687 kW

Explanation:

The mechanical energy of the air per unit mass is the specific kinetic energy of the air that is calculated using: $\frac{1}{2} V^2$ where V is the velocity of the air.
The specific kinetic energy would be: $\frac{1}{2}(9\frac{m}{s})^2=40.5\frac{m^2}{s^2}=40.5\frac{m^2 }{s^2}\frac{kg}{kg}=40.5\frac{N*m }{kg}=40.5\frac{J}{kg}$ .
The power generation of the wind turbine would be obtained from the product of the mechanical energy of the air times the mass flow that moves the turbine.
To calculate mass flow it is required first to calculate the volumetric flow. To calculate the volumetric flow the next expression would be: $\frac{V\pi D_{blade}^2}{4} =\frac{9\frac{m}{s}\pi(80m)^2}{4} =45238.9\frac{m^3}{s}$
Then the mass flow is obtain from the volumetric flow times the density of the air: $m_{flow}=1.25\frac{kg}{m^3}45238.9\frac{m^3}{s}=56548.7\frac{kg}{s}$
Then, the Power generation potential is: $40.5\frac{J}{kg} 56548.7\frac{kg}{s} =2290221W=2290.2kW$
The actual electric power generation is calculated using the definition of efficiency: $\eta=\frac{E_P}{E_I}}$ , where η is the efficiency, $E_P$ is the energy actually produced and, $E_I$ is the energy input. Then solving for the energy produced: $E_P=\eta*E_I=0.30*2290kW=687kW$

6 0

3 years ago

List 5 signs of a chemical change?

iris [78.8K]

Answer:

Color, formation, gas, odor,

Explanation:

change are a change in color
formation of a precipitate
formation of a gas
odor change
temperature change

4 0

3 years ago

n the Bohr model of the hydrogen atom (see Section 39.3), in the lowest energy state the electron orbits the proton at a speed o

Ivahew [28]

Answer:

(a) $T=1.5*10^{-6}s$

(b) $I=1.1*10^{-3}A$

Explanation:

(a) The orbital period is the time that the electron spend to travel the orbit of the atom. Thus, it is given by the length of the circular orbit divided by its velocity:

$T=\frac{2\pi r}{v}\\T=\frac{2\pi(5.3*10^{-11}m)}{2.2*10^{6}\frac{m}{s}}\\T=1.5*10^{-6}s$

(b) Current means charge over time, So, in this case is charge over period:

$I=\frac{q}{t}\\I=\frac{e}{T}\\I=\frac{1.6*10^{-19}C}{1.5*10^{-6}s}\\\\I=1.1*10^{-3}A$

$\mu=IA$

Here A is the area of the orbit.

$\mu=I\pi r^2\\\mu=(1.1*10^{-3}A)\pi(5.3*10^{-11}m)^2\\\mu=9.71*10^{-24}A\cdot m^2$

4 0

3 years ago