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wlad13 [49]
3 years ago
7

For elliptical obits: the direction of the velocity of the satellite is _______________________ (always, seldom, never) perpendi

cular to the net force acting upon the satellite. For circular obits: the direction of the velocity of the satellite is ___________________ (always, seldom, never) perpendicular to the net force acting upon the satellite.
Physics
1 answer:
alexandr1967 [171]3 years ago
7 0

Answer:

For elliptical orbits: seldom

For circular orbits: always

Explanation:

We start by analzying a circular orbit.

For an object moving in circular orbit, the direction of the acceleration (centripetal acceleration) is always perpendicular to the direction of motion of the object.

Since acceleration has the same direction of the force (according to Newton's second law of motion), this means that the direction of the force (the centripetal force) is always perpendicular to the velocity of the object.

So for a circular orbit,

the direction of the velocity of the satellite is always perpendicular to the net force acting upon the satellite.

Now we analyze an elliptical orbit.

An elliptical orbit correponds to a circular orbit "stretched". This means that there are only 4 points along the orbit in which the acceleration (and therefore, the net force) is perpendicular to the direction of motion (and so, to the velocity) of the satellite. These points are the 4 points corresponding to the intersections between the axes of the ellipse and the orbit itself.

Therefore, for an elliptical orbit,

the direction of the velocity of the satellite is seldom perpendicular to the net force acting upon the satellite.

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Two identical cylindrical vessels with their bases at the same level each contain a liquid of density 1.23 g/cm3. The area of ea
motikmotik

Explanation:

Work done by gravity is given by the formula,

           W = \rho A (h_{1} - h)g (h - h_{2}) ......... (1)

It is known that when levels are same then height of the liquid is as follows.

           h = \frac{h_{1} + h_{2}}{2} ......... (2)

Putting value of equation (2) in equation (1) the overall formula will be as follows.

       W = \frac{1}{4} \rho gA(h_{1} - h_{2})^{2})

           = \frac{1}{4} \times 1.23 g/cm^{3} \times 9.80 m/s^{2} \times 3.89 \times 10^{-4} m^{2}(1.76 m - 0.993 m)^{2})

           = 0.689 J

Thus, we can conclude that the work done by the gravitational force in equalizing the levels when the two vessels are connected is 0.689 J.

3 0
3 years ago
You are tasked with calibrating the springs for a pinball machine. Your method of testing the springs is by attaching masses ont
choli [55]

Answer:

490.5\ \text{N/m}

Explanation:

m = Mass attached to spring = 14 kg

g = Acceleration due to gravity = 9.81\ \text{m/s}^2

x = Displacement of spring = 78-50=28\ \text{cm}

k = Spring constant

The force balance of the system is given by

kx=mg\\\Rightarrow k=\dfrac{mg}{x}\\\Rightarrow k=\dfrac{14\times 9.81}{0.28}\\\Rightarrow k=490.5\ \text{N/m}

The spring constant for that spring is 490.5\ \text{N/m}.

6 0
3 years ago
Conductivities are often measured by comparing the resistance of a cell filled with the sample to its resistance when filled wit
Bezzdna [24]

Answer:

1200 Sm^2mol^-1

Explanation:

Given data :

conductivity of water ( kwater ) = 76 mS m^-1 = 0.076 Sm^-1

conductivity of kcl (aq)( Kkcl ) = 1.1639 Sm^-1

Kkcl = 1.1639 - 0.076 = 1.0879  Sm^-1

Resistance = 33.21 Ω

where conductivity can be expressed as = \frac{Cell constant}{Resistance }

hence cell constant = conductivity * Resistance

                                 = 1.0879 * 33.21 = 36.13m^-1

conductivity of  CH3COOH ( kCH3COOH ) =  36.13 / 300

                                                                       = 0.120 Sm^-1

<u>Determine the molar conductivity of acetic acid</u>

= ( kCH3COOH * 1000 ) / C

C = 0.1 mol dm

=  (0.120 * 1000) / 0.1  =  1200 Sm^2mol^-1

3 0
3 years ago
What happens to myosin and actin as sarcomeres relax?
Leokris [45]
The myosin heads pull on the actin, bringing them closer together

4 0
3 years ago
Assume the equation x 5 At3 1 Bt describes the motion of a particular object, with x having the dimension of length and t having
igomit [66]

Answer:

(a) A = m/s^3, B = m/s.

(b) dx/dt = m/s.

Explanation:

(a)

x = At^3 + Bt\\m = As^3 + Bs\\m = (\frac{m}{s^3})s^3 + (\frac{m}{s})s

Therefore, the dimension of A is m/s^3, and of B is m/s in order to satisfy the above equation.

(b) \frac{dx}{dt} = 3At^2 + B = 3(\frac{m}{s^3})s^2 + \frac{m}{s} = m/s

This makes sense, because the position function has a unit of 'm'. The derivative of the position function is velocity, and its unit is m/s.

6 0
3 years ago
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