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The following table shows the duration of a year on some unknown planets of equal mass. Duration of Year Planet Duration of Year

Soloha48 [4]

Gravitational force between sun and planet is given by,

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

Thus, The force of gravitation is inversely proportional to the distance.

Kepler's law states that the square of period of planet is proportional to cube of semi major axis of its orbit.

The time period of planet is more when the gravitational force is less.

Thus, planet Y is closer to earth because sun exert greater gravitational force than planet Z.

8 0

4 years ago

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A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180 kg. The free end of the stri

Dmitrij [34]

Answer:

(a). The angular speed of the rotating is 33.8 rad/s.

(b). The speed of its center is 2.7 m/s.

Explanation:

Given that,

Radius = 8.00 cm

Mass = 0.180 kg

Height = 75.0 m

We need to calculate the angular speed of the rotating

Using conservation of energy

$\dfrac{1}{2}I\omega_{1}^2+\dfrac{1}{2}mv_{1}^{2}+mgh_{1}=\dfrac{1}{2}I\omega_{2}^2+\dfrac{1}{2}mv_{2}^{2}+mgh_{2}$

Here, initial velocity and angular velocity are equal to zero.

$mgh_{1}=\dfrac{1}{2}I\omega_{2}^2+\dfrac{1}{2}mv_{2}^{2}+mgh_{2}$

$mg(h_{1}-h_{2})=\dfrac{1}{2}I\omega_{2}^2+\dfrac{1}{2}mv_{2}^{2}$

$mgH=\dfrac{1}{2}mr^2\omega_{2}^2+\dfrac{1}{2}m(r\omega_{2})^2$

Here, $H = h_{1}-h_{2}$

$gH=r^2\omega_{2}^2$

$\omega_{2}^2=\dfrac{gH}{r^2}$

$\omega_{2}=\sqrt{\dfrac{9.8\times75.0\times10^{-2}}{(8.00\times10^{-2})^2}}$

$\omega_{2}=33.8\ rad/s$

The angular speed of the rotating is 33.8 rad/s.

(b). We need to calculate the speed of its center

Using formula of speed

$v=r\omega$

Put the value into the formula

$v=8.00\times10^{-2}\times33.8$

$v=2.7\ m/s$

Hence, (a). The angular speed of the rotating is 33.8 rad/s.

(b). The speed of its center is 2.7 m/s.

3 0

3 years ago

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An undamped spring-mass system contains a mass that weighs and a spring with spring constant . It is suddenly set in motion at b

balandron [24]

Answer:

Explanation:

When all other forces acting on the mass in a damped mass-spring system are grouped together into one term denoted by F(t), the differential equation describing

motion is

Mx''+ βx' + kx = F(t).

Note for an undamped system

β=0,

Then, the differential equation becomes

Mx'' + kx = F(t).

The force is in the form

F=Fo•Sinωo•t

Let solved for the homogeneous or complementary solution, I.e f(t) = 0

Using D operator

MD² + k = 0

MD²=-k

D²=-k/M

Then, D= ±√(-k/m)

D=±√(k/m) •i

So we have a complex root

Therefore, the solution is

x= C1•Cos[√(k/m)t] + C2•Sin[√(k/m)]

This is simple harmonic motion that once again we prefer to write in the form

x(t) = A•Sin[ √(k/M)t + φ]

Where A=√(C1²+C2²)

and angle φ is defined by the equations

sin φ = C1/A and cos φ = C2/A.

Quantity √(k/M), often denoted by ω, is called the angular frequency.

This is called the natural frequency (ωn) of the system

ωn=√(k/M)

ωn²= k/M

Now, for particular solution

Xp=DSinωo•t

Xp' = Dωo•Cosωo•t

Xp"=-Dωo²•Sinωo•t

Now substituting this into

Mx'' + kx = F(t).

M(-Dωo²•Sinωo•t) + k(DSinωo•t)=FoSinωo•t

Now, let solve for D

D(-Mωo²•Sinωo•t +kSinωo•t) = FoSinωo•t

D=Fo•Sinωo•t/(-Mωo²•Sinωo•t +kSinωo•t)

D=Fo•Sinωo•t / Sinωo•t(-Mωo²+k)

D=Fo / (-Mωo²+k)

D=Fo / (k-Mωo²)

Divide through by k

D=Fo/k ÷ (1 -Mωo²/k)

Note from above

ωn²= k/M

Therefore,

D=Fo/k ÷ (1-ωo²/ωn²)

D=Fo/k ÷ [1-(ωo/ωn)²]

Then,

Xp=DSinωo•t

Xp=(Fo/k ÷ [1-(ωo/ωn)²]) Sinωo•t

Then the general solution is the sum of the homogeneous solution and particular solution

Xg(t)=(Fo/k ÷ [1-(ωo/ωn)²]) Sinωo•t + A•Sin[ √(k/M)t + φ]

Check attachment for the graph of homogeneous, particular and general solution.

Also, check for better way of writing the equations.

8 0

4 years ago

An object begins at position x = 0 and moves one-dimensionally along the x-axis witļi a velocity v

Liula [17]

Answer:

The answer is "between 20 s and 30 s".

Explanation:

Calculating the value of positive displacement:

$\ (x_{+ve}) = \frac{1}{2} \times 15 \times 20 \\\\$

$= \frac{1}{2} \times 300 \\\\= 150 \\\\$

Calculating the value of negative displacement upon the time t:

$(x_{-ve}) = \frac{1}{2} \times 5 \times 20- 20(t-20) \\\\$

$= \frac{1}{2} \times 100- 20t+ 400 \\\\= 50- 20t+ 400 \\\\$

$\to X= X_{+ve} + X_{-ve} \\\\$

$\to 150 - 50 -20t+400 =0\\\\\to 100 -20t+400 =0 \\\\\to 500 -20t =0\\\\\to 20t =500 \\\\\to t=\frac{500}{20}\\\\\to t=\frac{50}{2}\\\\\to t= 25$

That's why its value lie in "between 20 s and 30 s".

6 0

4 years ago

What's the symbol for an ionized gallium atom?

babunello [35]

Hi there!

Gallium is chemical element 31, (meaning 31 protons, 31 electrons).

The symbol for Gallium is "Ga"

Hope this helps you!

~DoctorLex

6 0

3 years ago