A net torque of magnitude 600 Nm is ex-

Kepler deduced this law of motion from observations of Mars. What information confirms his conclusion that the orbit of Mars is

Leno4ka [110]

Kepler noticed an imaginary line drawn from a planet to the Sun and this line swept out an equal area of space in equal times, If we then draw a triangle out from the Sun to a planet’s position at one point in time, it is notice that the area doesn't change even after the planet has left the original position say like after 2 to 3days or 2hours. So to have same area of triangle means that the the planet move faster when that are closer to the sun and slowly when they are far from the sun.

This led to Kepler's law of orbital motion.

First Law: Planetary orbits are elliptical with the sun at a focus.

Second Law: The radius vector from the sun to a planet sweeps equal areas in equal times.

Third Law: The ratio of the square of the period of revolution and the cube of the ellipse semi-major axis is the same for all planets.

It is this Kepler's law that makes Newton to come up with his own laws on how planet moves the way they do.

6 0

3 years ago

The electron dot diagram shows the arrangement of dots without identifying the element.

olasank [31]

Answer: tellurium (Te)

atomic number = 52 ,

Number of energy levels = 5;

First energy level = 2

Second energy level = 8

Third energy level = 18

Fourth energy level = 18

Fifth energy level = 6

In this electron configuration, 0uter most electrons are 6.

5 0

3 years ago

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A truck using a rope to tow a 2230-kg car accelerates from rest to 13.0 m/s in a time of 15.0s. How strong must the rope be? μk

Leokris [45]

Answer:

The rope must have a force of 10084,21 N

Explanation

Acceleration calculation

The car acceleration is equal to the acceleration of the truck

ac: car acceleration $\frac{m}{s^{2} }$

at: truck acceleration $\frac{m}{s^{2} }$ )

$ac = at= \frac{vf-vi}{t-ti}$ equation(1)

Known information:

vi = Initial speed = 0, ti = initial time = 0

vf = Final speed = 13 $\frac{m}{s}$ , t = final time =5 s

We replaced the known information in the equation(1):

$ac = at = \frac{13-0}{15-0}$

$ac=ac=\frac{13}{15} \frac{m}{s}$

Dynamic analysis

The forces acting on the car are the following:

Wc: Car weight

N: normal force, road force on the car

Ff: Friction force

T: Force of tension

Car weight calculation:

$Wc=mc*g$

mc = Car mass = 2230kg

g = Gravity acceleration=9.8 $\frac{m}{s^{2} }$

$Wc= 2230*9.8$

$Wc=21854 N$

Normal force calculation:

Newton's first law

$sum Fy= 0$

$N-W=0$

$N=W$

$N=21854 N$

Friction force calculation (Ff):

We have the formula to calculate the friction force:

Ff = μk * N Equation (3)

μk kinetic coefficient of friction

We know that μk = 0.373and N= 21854N ,then:

$Ff=0.373*21854$

$Ff=8151.54 N$

Calculation of the tension force in the rope (T):

Newton's Second law

$sum Fx= mc*ac$

$T-Ff=mc*ac$

$T=2230(\frac{13}{15}) + 8151.54$

T=10084,21 N

Answer: The rope must have a force of 10084,21 N

8 0

3 years ago

Is the color spectrum simply a small segment of the electromagnetic spectrum?

ch4aika [34]

Answer:

yup, u r correct

Explanation:

3 0

3 years ago

A 0.11 N m torque is applied to a fan that was initially at rest. The fan has moment of inertia of 0.034 kg m2. Determine the ki

Mars2501 [29]

Answer:

2.72*10-3 Joules

Explanation:

From Newton's second law of motion

F=ma

$\tau=I*\alpha$

given

$\tau= 0.11Nm\\\\I=0.034kgm^2\\\\t= 8s\\\\\alpha=?$

$\alpha =0.11/0.034\\\\\alpha=3.23 rad/s^2$

the angular velocity is

$\omega = \alpha/t\\\\\omega =3.23/8\\\\\omega =0.4 rad/s$

$KE= 1/2*I* \omega^2$

$KE=1/2*0.034*0.4^2\\\\KE=1/2*0.034*0.16\\\\KE=0.00272\\\\KE=2.72*10^-3J$

4 0

3 years ago