Copernicus incorrectly assumed that the planets?

1 answer:

6 0

Nicholas Copernicus correctly assumed that the planets revolved around the sun but he incorrectly assumed that the planets followed a perfect circle orbit around the sun. It was later on discovered by Johannes Kepler that the planets moved around the sun following an elliptical orbit.

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If you weigh 660 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun an

Arada [10]

Answer:

8.93*10^13 N.

Explanation:

Assuming that in this case, the weight is just the the force exerted on you by the mass of the star, due to gravity, we can apply the Universal Law of Gravitation:

$F_{g}= \frac{G*m_{1}*m_{s}}{r_{s}^{2} }$

where, m1 = mass of the man = 660 N / 9.81 m/s^2 = 67.3 kg, ms = mass of the star = 1.99*10^30 kg, G= Universal Constant of Gravitation, and rs= radius of the star = 10.0 km. = 10^4 m.
Replacing by the values, we get:

$F_{g}= \frac{6.67e-11Nm^2/kg^2*1.99e30 kg*67.3 kg}{10e4m^2} = 8.93e13 N$

Fg = 8.93*10^13 N.

6 0

3 years ago

The doctor writes a prescription for na heparin 20,000 units in 500 ml n.s. infuse over 8 hours. what is the flow rate in ml/hr?

Ne4ueva [31]

I think this type of equation could be conducted in simple division equation since it does not involve drop rate.

we know that there is 500 ml of substance and should be infused within 8 hours period.

So the flow rate in ml/hr would be:

500/8 = 62.5 ml/hr

8 0

3 years ago

Help I don’t understand this question

7nadin3 [17]

Metals
3/4 of the periodic table
Good conductors of heat and electricity
Malleable

Nonmetals
1/4 of the periodic table
Bad conductors of heat and electricity
Not bendable

5 0

3 years ago

A 1300 kg steel beam is supported by two ropes. (Figure

Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

the net horizontal force acting on the beam is

$R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0$

where $R_1,R_2$ are the magnitudes of the tensions in ropes 1 and 2, respectively;

the net vertical force acting on the beam is

$R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0$

where $m=1300\,\rm kg$ and $g=9.8\frac{\rm m}{\mathrm s^2}$ .

Eliminating $R_2$ , we have

$\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)$