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2 years ago

When a moving object collides with an object that isn’t moving , what happens to the kinetic energy of each object?

Physics

1 answer:

torisob [31]2 years ago

7 0

Answer:

The moving object transfers kinetic energy with the object that isn't moving.

Explanation:

Kinetic energy is a form of energy that a object or particle has by the reason of motion. Potential energy is a form of energy that a object or particle has by the reason of no movement. The object that is moving knocks down or collides with an object that isn't moving, and the object that is moving transfers kinetic energy to potential energy, and the object that wasn't moving transfers potential energy to kinetic energy.

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4. The blades on a fan have a frequency of 15 Hz.

vichka [17]

Answer:

a) 4500 cycles b) 0.0667s c) 6.67s

Explanation:

a) 15 Hz= 15 cycles/ s

5 mins= 300s

15 cycles/s * 300s= 4500 cycles

b) Period= 1/ frequency

Period= 1/ 15 cycles/s

Period= 0.0667s

c) Period * number of revolutions= time

0.0667 * 100= 6.67s

6 0

3 years ago

A circuit component with a high resistance will have a low electric current.

Zepler [3.9K]

You haven't said what 'high' resistance or 'low' current means, so there's way not enough info to nail the statement as true or false. The most precise answer is "certainly could be but not necessarily". Anyway, the current in the circuit depends on BOTH the resistance AND the voltage. So without knowing the voltage too, you can't say anything about the current.

6 0

3 years ago

If the distance between the Earth and Moon were half what it is now, by what factor would the force of gravity between them be c

hichkok12 [17]

Answer:

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

$m_1$ = Mass of Earth

$m_2$ = Mass of Moon

r = Distance between Earth and Moon

Old gravitational force

$F_o=\dfrac{Gm_1m_2}{r^2}$

New gravitational force

$F_n=\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}$

Dividing the equations

$\dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{\dfrac{1}{4}r^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=4$

The ratio is $\dfrac{F_n}{F_o}=4$

The new force would be 4 times the old force

7 0

2 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: $E_P=\eta*E_I=0.30*2290kW=687kW$