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3 years ago

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13) A mass attached to the free end of a spring executes simple harmonic motion according to the equation y = (0.50 m) sin (18π

t) where y is in meters and t is seconds. What is the period of vibration?

1 answer:

Kitty [74]3 years ago

4 0

Hi there!

The period is given by:

$T = \frac{2\pi}{w}$

T = Period (sec)

w = angular frequency (rad/sec)

According to the equation for SHM in terms of position:

y(t) = Asin(ωt + φ)

A = Amplitude (m)

ω = angular frequency (rad/sec)

t = time (sec)

φ = phase angle

In this instance, the angular frequency is given as 18π.

Plug this value into the equation for T:

$T= \frac{2\pi}{18\pi} = \frac{1}{9} = \boxed{0.111 s}$

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Answer:

<em>D. The total force on the particle with charge q is perpendicular to the bottom of the triangle.</em>

Explanation:

The image is shown below.

The force on the particle with charge q due to each charge Q = $\frac{kQq}{r^{2} }$

we designate this force as N

Since the charges form an equilateral triangle, then, the forces due to each particle with charge Q on the particle with charge q act at an angle of 60° below the horizontal x-axis.

Resolving the forces on the particle, we have

for the x-component

$N_{x}$ = N cosine 60° + (-N cosine 60°) = 0

for the y-component

$N_{y}$ = -f sine 60° + (-f sine 60) = -2N sine 60° = -2N(0.866) = -1.732N

The above indicates that there is no resultant force in the x-axis, since it is equal to zero ( $N_{x}$ = 0).

The total force is seen to act only in the y-axis, since it only has a y-component equivalent to 1.732 times the force due to each of the Q particles on q.

<em>The total force on the particle with charge q is therefore perpendicular to the bottom of the triangle.</em>

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4 years ago

A girl of mass m1=60 kilograms springs from a trampoline with an initial upward velocity of v1=8.0 meters per second. At height

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a) 5.0 m/s

This first part of the problem can be solved by using the conservation of energy. In fact, the mechanical energy of the girl just after she jumps is equal to her kinetic energy:

$E_i=\frac{1}{2}m_1v_1^2$

where m1 = 60 kg is the girl's mass and v1 = 8.0 m/s is her initial velocity.

When she reaches the height of h = 2.0 m, her mechanical energy is sum of kinetic energy and potential energy:

$E_f = \frac{1}{2}m_1 v_2 ^2 + m_1 gh$

where v2 is the new speed of the girl (before grabbing the box), and h = 2.0m. Equalizing the two equations (because the mechanical energy is conserved), we find

$\frac{1}{2}m_1 v_1^2 = \frac{1}{2}m_1 v_2 ^2 + m_1 gh\\v_1^2 = v_2^2 +2gh\\v_2 = \sqrt{v_1^2 -2gh}=\sqrt{(8.0 m/s)^2-(2)(9.8 m/s^2)(2.0 m)}=5.0 m/s$

b) 4.0 m/s

After the girl grab the box, the total momentum of the system must be conserved. This means that the initial momentum of the girl must be equal to the total momentum of the girl+box after the girl catches the box:

$p_i = p_f\\m_1 v_2 = (m_1 + m_2) v_3$

where m2 = 15 kg is the mass of the box. Solving the equation for v3, the combined velocity of the girl+box, we find

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c) 2.8 m

We can use again the law of conservation of energy. The total mechanical energy of the girl after she catches the box is sum of kinetic energy and potential energy:

$E_i = \frac{1}{2}(m_1+m_2) v_3^2 + (m_1+m_2)gh=\frac{1}{2}(75 kg)(4 m/s)^2+(75 kg)(9.8 m/s^2)(2.0m)=2070 J$

While at the maximum height, the speed is zero, so all the mechanical energy is just potential energy:

$E_f = (m_1 +m_2)gh_{max}$

where h_max is the maximum height. Equalizing the two expressions (because the mechanical energy must be conserved) and solving for h_max, we find

$E_i = (m_1+m_2)gh_{max}\\h_{max}=\frac{E_i}{(m_1+m_2)g}=\frac{2070 J}{(75 kg)(9.8 m/s^2)}=2.8 m$

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3 years ago

If a body having mass 40kg started moving initially with rest and it takes a velocity of 20m/sec in time 4 seconds. Find the val

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${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Mass \ of \ the \ body \ (m) = 40 \ kg}$

$\:\:\:\:\bullet\:\:\:\sf{Final \ velocity \ of \ the \ body \ (v) = 20 \ m/s}$

$\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ of \ the \ body \ (u) = 0}$

$\\$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Force \ exerted \ by \ the \ body \ ( F)}$

$\\$

${\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\$

☯ <u>Using 1st equation of motion </u>

$\\$

$\dashrightarrow\:\: \sf{v = u + at}$

$\\$

$\dashrightarrow\:\: \sf{20 = 0 + a(4)}$

$\\$

$\dashrightarrow\:\: \sf{20 = 4a}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{\cancel{20}}{\cancel{4}} = a}$

$\\$

$\dashrightarrow\:\: \sf{a = 5}$

$\\$

☯ <u>Now, Finding the force exerted </u>

$\\$

$\dashrightarrow\:\: \sf{F = ma}$

$\\$

$\dashrightarrow\:\: \sf{F = 40 \times 5}$

$\\$

$\dashrightarrow\:\: \sf{F = 200 \ N}$

$\\$

☯ <u>Hence</u>, $\\$

$\:\:\:\:\star\:\:\:\sf{The \ force \ exerted \ by \ the \ body \ is \ 200N}$

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