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3 years ago

8

Which one of the following is an example of a solution?

1 answer:

insens350 [35]3 years ago

6 0

Answer:

it is maybe A but i'm not 100% sure

Explanation:

its the only one that is mixed

You might be interested in

Which famous scientist is credited as the founder of the scientific method?

anygoal [31]

Aristotle created and it’s credited as the creator.

3 0

3 years ago

You place a 55.0 kg box on a track that makes an angle of 28.0 degrees with the horizontal. The coefficient of static friction b

Radda [10]

Answer:

$\theta=34 \textdegree$

Explanation:

From the question we are told that:

Mass $m=55kg$

Angle $\theta =28.0$

Coefficient of static friction $\alpha =0.680$

Generally, the equation for Newtons second Law is mathematically given by

For

$\sum_y=0$

$N=mgcos \theta$

for

$\sum_x=0$

$F_{s}=mgsin\theta$

Where

$F_{s}=\alpha*N\\\\F_{s}=\alpha*m*gcos \theta$

$F_{s}=0.68*55*9.8*cos 28$

$F_{s}=323.62N$

Therefore

$\alpha mgcos \theta=mg sin \theta$

$\theta=tan^{-1}(0.68)$

$\theta=34 \textdegree$

6 0

2 years ago

Sharon the ant (Aaron’s sister) sits at the edge of a turntable of radius R that is spinning with period T. As she makes one-hal

Dmitry_Shevchenko [17]

Answer:

$a = \dfrac{4\pi^2R}{T^2}$

Explanation:

The acceleration of a circular motion is given by

$a = \omega^2 R$

where $\omega$ is the angular velocity and $R$ is the radius.

Angular velocity is related to the period, T, by

$\omega=\dfrac{2\pi}{T}$

Substitute into the previous formula.

$a = (\dfrac{2\pi}{T})^2 R$

$a = \dfrac{4\pi^2R}{T^2}$

This acceleration does not depend on the linear or angular displacement. Hence, the amount of rotation does not change it.

6 0

3 years ago

A new planet is discovered beyond Pluto at a mean distance to the sun of 4004 million miles. Using Kepler's third law, determine

AVprozaik [17]

Answer:

103239.89 days

Explanation:

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

a³ / T² = 7.496 × 10⁻⁶ (a.u.³/days²)

where,

a is the distance of the semi-major axis in a.u

T is the orbit time in days

Converting the mean distance of the new planet to astronomical unit (a.u.)

1 a.u = 9.296 × 10⁷ miles

$\frac{4004 * 10^{6}}{9.296 * 10^{7}} = 43.07\ a.u.$

Substituting the values into Kepler's third law equation;

$\frac{(43.07)^{3}}{T^{2}} = 7.496 * 10^{-6}$

$T^{2} = \frac{(43.07)^{3}}{7.496 * 10^{-6}}$ (days)²

$T^{2} = \sqrt{\frac{(43.07)^{3}}{7.496 * 10^{-6}}}$

T = 103239.89 days

An estimate time T for the new planet to travel around the sun in an orbit is 103239.89 days

7 0

3 years ago

A phonograph record accelerates from rest to 28.0 rpm in 5.73 s.

Arturiano [62]

Answer:

a) $\alpha=0.5117\ rad.s^{-2}$

b) $\theta=8.4\ rad$

Explanation:

Given:

initial rotational speed of phonograph, $\omega_i=0\ rad.s^{-1}$

final rotational speed of phonograph, $N_f=28\ rpm \Rightarrow \omega_f=2\pi\times\frac{28}{60} =2.932\ rad.s^{-1}$

time taken for the acceleration, $t=5.73\ s$

a)

Now angular acceleration:

$\alpha=\frac{\omega_f-\omega_i}{t}$

$\alpha=\frac{2.932-0}{5.73}$

$\alpha=0.5117\ rad.s^{-2}$

b)

Using eq. of motion:

$\theta=\omega_i.t+\frac{1}{2} \alpha.t^2$

$\theta=0+\frac{1}{2}\times 0.5117\times 5.73^2$

$\theta=8.4\ rad$

5 0

3 years ago