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rosijanka [135]
3 years ago
6

Drag each label to the correct location on the chart.

Physics
2 answers:
Soloha48 [4]3 years ago
7 0
Kinetic energy-flashlight, and guitar
potential-coal,and composed spring
slega [8]3 years ago
3 0

The answers would be:

POTENTIAL ENERGY

--- Energy of compressed spring

--- Energy of coal

KINETIC ENERGY

--- Energy of a guitar string when plucked.

--- Energy from a flashlight when turned on.

If you'd like to know why:

Energy can be divided into two broad categories, which are:

Potential and Kinetic energy. They differ from each other whereas <em>potential energy</em> is what we call <u>STORED </u>energy and <em>kinetic energy</em> is what we call energy <u>IN MOTION.</u>

Energy is then subdivided further into more specific forms of energy.

In your case, under potential energy you have the compressed spring and coal. A compressed spring's energy is stored through tension, which makes energy MECHANICAL energy. The energy in coal on the other hand comes in the form of CHEMICAL energy, where the energy is stored in the atomic bonds.

Under kinetic energy you have guitar string when plucked (the plucked part is the important clue) and the flashlight turned on. The guitar string creates sound when plucked, so the energy form in this case is what we call SOUND energy. The flashlight on the other hand creates light, which produces RADIANT energy.  The energy here is in motion and it is travelling in waves.

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A pulley system lifts a 500-lb. block 2.0 feet with an effort of only 50 lb. If the 50 lb. moves 30 feet, calculate the efficien
tatiyna
The answer to this is 67%
4 0
3 years ago
Read 2 more answers
An apparatus like the one Cavendish used to find G has large lead balls that are 5.2 kg in mass and small ones that are 0.046 kg.
Ber [7]

Answer:

The magnitude of gravitational force between two masses is 4.91\times 10^{-9}\ N.

Explanation:

Given that,

Mass of first lead ball, m_1=5.2\ kg

Mass of the other lead ball, m_2=0.046\ kg

The center of a large ball is separated by 0.057 m from the center of a small ball, r = 0.057 m

We need to find the magnitude of the gravitational force between the masses. It is given by the formula of the gravitational force. It is given by :

F=G\dfrac{m_1m_2}{r^2}\\\\F=6.67259\times 10^{-11}\times \dfrac{5.2\times 0.046}{(0.057)^2}\\\\F=4.91\times 10^{-9}\ N

So, the magnitude of gravitational force between two masses is 4.91\times 10^{-9}\ N. Hence, this is the required solution.

5 0
3 years ago
Over a time interval of 1.99 years, the velocity of a planet orbiting a distant star reverses direction, changing from +20.7 km/
madam [21]

Answer:

(a) - 42700 m/s

(b) - 6.8 x 10^-4 m/s^2

Explanation:

initial velocity of star, u = 20.7 km/s

Final velocity of star, v = - 22 km/s

time, t = 1.99 years

Convert velocities into m/s and time into second

So, u = 20700 m / s

v = - 22000 m/s

t = 1.99 x 365.25 x 24 x 3600 = 62799624 second

(a) Change in planet's velocity = final velocity - initial velocity

  = - 22000 - 20700 = - 42700 m/s

(b) Accelerate is defined as the rate of change of velocity.

Acceleration = change in velocity / time

                     = ( - 42700 ) / (62799624) = - 6.8 x 10^-4 m/s^2

8 0
3 years ago
Suppose that the current in the solenoid is i(t). The self-inductance L is related to the self-induced EMF E(t) by the equation
Artemon [7]

Answer:

L =   μ₀ n r / 2I

Explanation:

This exercise we must relate several equations, let's start writing the voltage in a coil

        E_{L} = - L dI / dt

 

Let's use Faraday's law

       E = - d Ф_B / dt

in the case of the coil this voltage is the same, so we can equal the two relationships

        - d Ф_B / dt = - L dI / dt

The magnetic flux is the sum of the flux in each turn, if there are n turns in the coil

        n d Ф_B = L dI

we can remove the differentials

      n Ф_B = L I

magnetic flux is defined by

     Ф_B = B . A

in this case the direction of the magnetic field is along the coil and the normal direction to the area as well, therefore the scalar product is reduced to the algebraic product

      n B A = L I

the loop area is

      A = π R²

     

we substitute

       n B π R² = L I                    (1)

To find the magnetic field in the coil let's use Ampere's law

        ∫ B. ds = μ₀ I

where B is the magnetic field and s is the current circulation, in the coil the current circulates along the length of the coil

           s = 2π R

we solve

              B 2ππ R =  μ₀ I

              B =  μ₀ I / 2πR

we substitute in

       n ( μ₀ I / 2πR) π R² = L I

       n  μ₀ R / 2 = L I

       L =   μ₀ n r / 2I

4 0
4 years ago
Two objects exert a gravitational force of 3 N on one another when they are 10 m apart. What would that force be if the distance
jeka94
Gravitational force = G ( m1 m2 ) / r²
3 = G ( m1 m2 ) / ( 10 )²x = G ( m1 m2 ) / ( 5 )²We shall divide those two equations:3 / x = 1/100 / 1/25 = 25 / 100 = 1 / 4x · 1 = 3 · 4x = 12Answer:C. 12 N
8 0
3 years ago
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