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3 years ago

Jennifer and katie stand and lean on each other. Jennifer weighs 150 pounds and katie weighs 120 pounds.

Physics

1 answer:

Alinara [238K]3 years ago

7 0

Answer:

Jennifer

Explanation:

she has more mass which means she is using more force

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an unknown charge exerts an attractive force of 4.23 E 13 N. If these charges are separated by 8cm, what is the magnitude of the

mash [69]

Answer:

Since there is attraction force between two charges so the other charge must be - 3C

Explanation:

As we know that the force between two charges is given by formula

$F = \frac{kq_1q_2}{r^2}$

here we know that

$F = 4.23 \times 10^{13}N$

also we know that

$q_1 = 10C$

r = 8 cm

now we have

$4.23 \times 10^{13} = \frac{(9\times 10^9)(10)q}{0.08^2}$

so we have

$q = -3C$

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3 years ago

Of orbital radius, orbital velocity, orbital circumference, or radial area swept through, which of these stays constant accordin

kow [346]

Answer:

dyow

Explanation:

sorrrey

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3 years ago

Scientists might make a computer model of volcanic eruptions. What is the

Anna [14]

Answer:

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1 year ago

A tennis ball is traveling at 50 m/s and has a kinetic energy of 75J. Calculate the mass of the tennis ball.

Vitek1552 [10]

Couldn’t fit all of it , but here luv

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3 years ago

As you know, a common example of a harmonic oscillator is a mass attached to a spring. In this problem, we will consider a horiz

Eddi Din [679]

a)E= U + K = $\frac{1}{2}$ kx² + $\frac{1}{2}$ mv²

The total energy of the system at any point in the motion is equal to the sum of the elastic potential energy of the spring, U, and of the kinetic energy of the mass, K:

E= U + K = $\frac{1}{2}$ kx² + $\frac{1}{2}$ mv²

where

'k' represents the spring constant

'x' is the compression/stretching of the spring with respect to its equilibrium position

'm' is the mass of the block attached to the spring

and 'v' is the speed of the block

b) A= $\sqrt{\frac{2E}{k}}$

The amplitude of the motion compares to the most extreme displacement of the mass-spring system. The displacement of the system, x(t), at time t, for a simple harmonic oscillator is given by,

x= Asin(ωt+∅)

where

amplitude is 'A'

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency of the motion

t is the time

$\phi$ is the phase (we can take $\phi$ =0 )

The amplitude of the motion occurs when the displacement of the motion is maximum: x=A. Regarding energy, the mass-spring system is at its maximum displacement (x=A) when all the mechanical energy of the framework is elastic potential energy, so when the kinetic energy is zero:

K= $\frac{1}{2}mv^2$ =0

E= $\frac{1}{2}kA^2\\$ -->(1)

A= $\sqrt{\frac{2E}{k}}$

c) $v_{max}=\omega A$

When the elastic potential energy is zero, the maximum speed of the system occurs i.e U=0 and the kinetic energy is maximum, so:

U=0

E= $\frac{1}{2}mv_{max}^2$

According to the law of conservation of the mechanical energy, this energy must be equal to the energy of the system at its maximum displacement (1), so we can write

$\frac{1}{2}kA^2=\frac{1}{2}mv_{max}^2$

and solving for $v_{max}$ we find an expression for the maximum speed:

$v_{max}=\sqrt{\frac{kA^2}{m}}=\sqrt{\frac{k}{m}}A=\omega A$

<h2> $v_{max}=\omega A$ </h2>

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3 years ago