Whose contributions to astronomy explained how planets were held in their orbits?

Why are some forces considered to be noncontact forces? A. Objects must be far apart in order to exert a force. B. Objects do no

kompoz [17]

Answer:

B. Objects do not have to touch each other to experience a force.

Explanation:

For example ..One of the noncontact forces is magnetic force whereby a magnetic object will be attracted to another magnetic object of oppsite charged particles, through waves called electromagnetic waves. On the other hand, the two magnetic objects of similar charged particles can repel through electromagnetic waves..

8 0

3 years ago

If you were to mix 3 liters of 20 degrees celsius water with 4 liters of 30 degrees water at 7 liters of water what temperature

Debora [2.8K]

Lets say final temperature is @ celsius
3c(@-20)= 4c([email protected])
[email protected]=180
@=180/7

3 0

3 years ago

What is the differential equation governing the growth of current in the circuit as a function of time after t=0? express the ri

Reika [66]

Answer:

$v_{b}=ir+L\frac{di}{dt}$

Explanation:

A differential equation that contain a term with di(t)/dt is in a RL circuit. Here we have

$v_{b}=v_{r}+v_{i}$

where vr is the voltage in the resistance, vi is the voltage in the inductance and vb is the source voltage. But also we have that

$v_{r}=ir\\v_{i}=L\frac{di}{dt}$

where L is the inductance of the circuit, r is the resistance an i is the current. By replacing we have the differential equation

$v_{b}=ir+L\frac{di}{dt}$

I hope this is useful for you

regards

3 0

3 years ago

The orbital period of a satellite is 2 × 106 s and its total radius is 2.5 × 1012 m. The tangential speed of the satellite, writ

LenaWriter [7]

The orbital period of the satellite[T] is given as $2*10^{6} S$ .

The radius of the satellite is given [R] $2.5*10^{12} m$ .

we are asked here to calculate the tangential speed of the satellite.

Before going to get the solution first we have understand the tangential speed.

The tangential speed of a satellite is given as the speed required to keep the satellite along the orbit. If satellite speed is less than tangential speed,there is the chance of it falling down towards earth. If it is more,then it will deviate from it orbit and can't stick to the orbit further.In a simple way the tangential speed is the linear speed of an object in a circular path.

Now we have to calculate the tangential speed [V].

Mathematically the tangential speed [V] written as -

V $=\frac{2\pi R}{T}$

where T is the time period of the satellite and R is the radius of the satellite.

$V=\frac{2*3.14*10^{12} }{2*10^{6} }$

$= 7.85*10^{6} m/s$

There is also another way through which we can get the solution as explained below-

We know that the tangential speed of a satellite V $=\sqrt{\frac{GM}{R^{2} } }$

where G is the gravitational constant and M is the mas of central object.

But we know that $g=\frac{GM}{R^{2} }$

⇒ $GM=gR^{2}$ where g is the acceleration due to gravity of that central object.

Hence $V=\sqrt{\frac{gR^{2} }{R} }$

⇒ $V=\sqrt{gR}$

By knowing the value of g due to that central object we can also calculate its tangential speed.

7 0

3 years ago

Read 2 more answers

You are standing on a sidewalk. There is a car in the distance. The horn on the car sounds. You hear it at a pitch that correspo

AnnyKZ [126]

Answer: The answer: The car is moving away from you.

Both A and C are true as Car can be moving in line away from you or has component of velocity in opposite direction.

Explanation:The decrease in the frequency of the sound is the result of Doppler's effect. A/c to Doppler's effect the frequency of received sound of source is changed if it is moving relative to the receiver, i.e. the distance between them is changing due to motion.

The general formula of Doppler's Effect is attached as the picture.

In this formula v_D is the velocity of Detector i.e the receiver relative to wind. While v_s is the velocity of source relative to wind and v is the velocity of sound.

The Doppler's effect is not effected by the velocity of wind as the wind itself could not change the distance between the two objects i.e. you and the car. Wind velocity can change the speed of sound and its wavelength but the change does not effect the frequency.

Hence if we assume the car to be moving with velocity v_c and you are stationary

$f'=f_s*\frac{v}{v-v_c}$

hence the frequency is reduced.

4 0

3 years ago