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3 years ago

당신을 결코 뒤지지 않는 사람을 사랑하는 방법과 당신에게 맞는 사람을 찾는 방법

Physics

1 answer:

Fiesta28 [93]3 years ago

7 0

Explanation:

내가 좋은 사람이 필요하고 내가 믿을 수 있는 사람이 필요하기 때문에 친구가 없다고 말하는 사람은 거의 없지만 대부분은 가짜이고 한국어를 모릅니다.

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A battery has an emf of 15.0 V. The terminal voltage of the battery is 12.2 V when it is delivering 14.0 W of power to an extern

solong [7]

Answer:

The value of R and the internal resistance of the battery are 10.6 ohm and 2.45 ohm

Explanation:

Given that,

Emf of battery = 15.0 V

Voltage = 12.2 V

Power = 14.0 W

(a). We need to calculate the value of R

Using formula of power

$P=IV$

$P=\dfrac{V^2}{R}$

$R=\dfrac{V^2}{P}$

Where, R = resistance

P = power

V = voltage

Put the value into the formula

$R =\dfrac{(12.2)^2}{14.0}$

$R=10.6\ \Omega$

(b). We need to calculate the internal resistance of the battery

Firstly we calculate the current

Using formula of current

$I=\dfrac{V}{R}=\dfrac{P}{V}$

Put the value of P and V into the formula

$I =\dfrac{14}{12.2}$

$I=1.14\ A$

We calculate the internal resistance

Using formula of emf

$e-V=Ir$

$r =\dfrac{e-V}{I}$

Put the value into the formula

$r=\dfrac{15.0-12.2}{1.14}$

$r = 2.45\ \Omega$

Hence, The value of R and the internal resistance of the battery are 10.6 ohm and 2.45 ohm

3 0

3 years ago

A 0.40 kg ball is suspended from a spring with spring constant 12 n/m . part a if the ball is pulled down 0.20 m from the equili

PolarNik [594]

The total mechanical energy of the system at any time t is the sum of the kinetic energy of motion of the ball and the elastic potential energy stored in the spring:
$E=K+U= \frac{1}{2}mv^2+ \frac{1}{2}kx^2$
where m is the mass of the ball, v its speed, k the spring constant and x the displacement of the spring with respect its rest position.

Since it is a harmonic motion, kinetic energy is continuously converted into elastic potential energy and vice-versa.

When the spring is at its maximum displacement, the elastic potential energy is maximum (because the displacement x is maximum) while the kinetic energy is zero (because the velocity of the ball is zero), so in this situation we have:
$E=U_{max}= \frac{1}{2}k(x_{max})^2$

Instead, when the spring crosses its rest position, the elastic potential energy is zero (because x=0) and therefore the kinetic energy is at maximum (and so, the ball is at its maximum speed):
$E=K_{max}= \frac{1}{2}m(v_{max})^2$

Since the total energy E is always conserved, the maximum elastic potential energy should be equal to the maximum kinetic energy, and so we can find the value of the maximum speed of the ball:
$U_{max}=K_{max}$
$\frac{1}{2}k(x_{max})^2 = \frac{1}{2}m(v_{max})^2$
$v_{max}= \sqrt{ \frac{k x_{max}^2}{m} }= \sqrt{ \frac{(12 N/m)(0.20 m)^2}{0.4 kg} }=1.1 m/s$