All categories

3 years ago

An earth satellite in an elliptical orbit travels slowest when it is

1 answer:

8 0

The satellite travels slowest when it is at the maximum distance from the Earth.

We can verify this in two ways:

1) By using Kepler's second law: "A line segment joining a a satellite with the Earth covers equal areas during equal intervals of time". This means that the larger is the distance of the satellite from Earth, the slower it goes.

2) by looking at the forces acting on the satellite. There is only one force acting on it: the gravitational attraction exerted by Earth, and this force is the centripetal force that keeps the satellite in circular (elliptical, actually) motion. So we can write:
$G \frac{Mm}{r^2}=m \frac{v^2}{r}$
where on the left we wrote the formula of the gravitational force, while on the right the centripetal force. G is the gravitational constant, M the Earth's mass, m the satellite's mass, v its velocity and r the distance of the satellite from the center of Earth.
Simplifying, we get
$v= \sqrt{ \frac{GM}{r} }$
which is the speed of the satellite when it is at a distance r from Earth: the larger r, the smaller the speed v.

You might be interested in

Variations in the reactive properties of different organic molecules are most closely associated with _____.

azamat

Answer:

the presence or absence of functional groups

Explanation:

The functional group is the group of atoms that characterize a chemical function and that have well-defined characteristic properties.

In organic chemistry, the functional group is a set of submolecular structures, characterized by a specific elementary connectivity and composition that confers specific chemical reactivity to the molecule that contains them. These structures replace the hydrogen atoms lost by saturated hydrocarbon chains. Aliphatic, or open chain, groups are usually represented generically by R (alkyl radicals), while aromatic ones, or derivatives of benzene, are represented by Ar (aryl radicals).

7 0

2 years ago

Read 2 more answers

An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.80 kW. A receiving antenna 7

Alecsey [184]

Answer:0.1759 v

Explanation:

Intensity of wave at receiver end is

I= $\frac{P_{avg}}{A}$

I= $\frac{3.80\times 10^3}{4\times \pi \times \left ( 4\times 1609.34\right )^2}$

I= $7.296\times 10^{-6} W/m^2$

Amplitude of electric field at receiver end

$E_{max}=\sqrt{2I\mu _0c}$

Amplitude of induced emf

= $E_{max}d$

= $\sqrt{2\times 7.29\times 10-6\times 4\pi \times 3\times 10^8}\times 0.75$

= $17.591\times 10^{-2}=0.1759 v$

7 0

3 years ago

6 Fig. 6.1 is a full-scale diagram that represents a sound wave travelling in air

Oxana [17]

From the measured wavelength from diagram, the frequency of the sound is 6660 Hz.

<h3>What is the frequency of a wave?</h3>

The frequency of a wave is the number of complete oscillation per second completed by a wave.

Frequency is related to wavelength and speed by the following formula:

Frequency = velocity/wavelength

Velocity of sound in air = 330 m/s

The measured wavelength = 5.0 cm = 0.05 m

Frequency = 330/0.05 = 6660 Hz

Therefore, based on the measured wavelength from diagram, the frequency of the sound is 6660 Hz.

Learn more about frequency of sound at: https://brainly.in/question/15373132
#SPJ1

7 0

1 year ago

A weight lifter is trying to do a bicep curl with a weight of 300 N. At the "sticking point", the moment arm of this weight is 3

lesantik [10]

Answer:

The weight lifter would not get past this sticking point.

Explanation:

Generally torque applied on the weight is mathematically represented as

T = F z

To obtain Elbow torque we substitute 4000 N for F (the force ) and 2cm $= \frac{2}{100} = 0.02m$ for z the perpendicular distance

So Elbow Torque is $T_e= 4000 * 0.02$

$= 80Nm$

To obtain the torque required we substitute 300 N for F and 30cm $=\frac{30}{100} = 0.3 m$

So the Required Torque is $T_R = 300 *0.3$

$=90Nm$

Now since $T_e < T_R$ it mean that the weight lifter would not get past this sticking point

7 0

3 years ago

Two cars cover the same distance in a straight line. Car a covers the distance at a constant velocity. Car b starts from rest an

Artyom0805 [142]

a) For the motion of car with uniform velocity we have , $s = ut+\frac{1}{2}at^2$ , where s is the displacement, u is the initial velocity, t is the time taken a is the acceleration.