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andrezito [222]

3 years ago

5

As velocity _ kinetic energy and potential energy _____

1 answer:

quester [9]3 years ago

7 0

Answer:

Kinetic energy increases and potential energy decreases

Explanation:

Because kinetic energy is proportional to the square of the velocity, when velocity is doubled, kinetic energy is quadrupled. Kinetic energy K=mv^2/2. Kinetic energy is the energy of movement or change. Potential energy is the energy a system has as a result of being able to undergo some change. ... As an object falls its potential energy decreases, while its kinetic energy increases. The decrease in potential energy is exactly equal to the increase in kinetic energy.

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When the compasses are moved close to a wire with a current which will the compasses point

Sindrei [870]

<span>In this demonstration, there is a single compass and a piece of wire that is perpendicular to the plane that the compass sits on. The wire with current flowing through it can alter the direction the compass needle points.</span>

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4 years ago

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What is the energy equivalent of an object with a mass of 2.5 kg?

Bad White [126]

This is the answer on Edgenuity 2.25*10^17

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3 years ago

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If the CD rotates clockwise at 500 rpm (revolutions per minute) while the last song is playing, and then spins down to zero angu

gavmur [86]

Answer:

Explanation:

Given that,

Initial Angular velocity w=500rpm

Converting from rpm to rad/s

1rev =2πrad

1minutes =60secs

500rpm=500rev/mins

w = 500×2π/60

wi=52.36rad/s

The final angular velocity wf=0rad/s

Time to stop is t=2.6sec

We want to find angular acceleration α

Using the equation of angular motion

wf = wi + αt.

0 = 52.36 + 2.6α

-52.36=2.6α

α = -52.36/2.6

α = -20.14rad/s²

The angular acceleration is negative because it is decelerating.

Then, α=20.14rad/s²

8 0

3 years ago

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Two charges are located in the x–y plane. If q1 = -2.90 nC and is located at x = 0.00 m, y = 0.840 m and the second charge has m

Lunna [17]

Answer:

Epx= - 21.4N/C

Epy= 19.84N/C

Explanation:

Electric field theory

The electric field at a point P due to a point charge is calculated as follows:

E= k*q/r²

E= Electric field in N/C

q = charge in Newtons (N)

k= electric constant in N*m²/C²

r= distance from load q to point P in meters (m)

Equivalences

1nC= 10⁻⁹C

known data

q₁=-2.9nC=-2.9 *10⁻⁹C

q₂=5nC=5 *10⁻⁹C

r₁=0.840m

$r_{2} =\sqrt{1^{2} +0.8^{2} } =\sqrt{1.64}$

$sin\beta =\frac{0.8}{\sqrt{1.64} } =0.6246$

$cos\beta =\frac{1}{\sqrt{1.64} } =0.7808$

Calculation of the electric field at point P due to q1

Ep₁x=0

$Ep_{1y} =\frac{k*q_{1} }{r_{1}^{2} } =\frac{8.99*10^{9}*2.9*10^{-9} }{0.84^{2} } =36.95\frac{N}{C}$

Calculation of the electric field at point P due to q2

$Ep_{2x} =-\frac{k*q_{2} *cos\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.7808 }{(\sqrt{1.64})^{2} } =-21.4\frac{N}{C}$

$Ep_{2y} =-\frac{k*q_{2} *sin\beta }{r_{2}^{2} } =-\frac{8.99*10^{9}*5*10^{-9} *0.6242 }{(\sqrt{1.64})^{2} } =-17.11\frac{N}{C}$

Calculation of the electric field at point P(0,0) due to q1 and q2

Epx= Ep₁x+ Ep₂x==0 - 21.4N/C =- 21.4N/C

Epy= Ep₁y+ Ep₂y=36.95 N/C-17.11N =19.84N/C

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3 years ago

Tim observes that a force pulls him down when he jumps up in the air. Based on his observation, which conclusion is correct?

enyata [817]

Gravitational force is a non-contact force.

8 0

3 years ago