Compared to microwaves, ultraviolet waves have __________________________

Two satellites are in circular orbits around a planet that has radius 9.00x10^6 m. One satellite has mass 53.0 kg , orbital radi

Andreas93 [3]

Answer:

The orbital speed of this second satellite is 5195.16 m/s.

Explanation:

Given that,

Orbital radius of first satellite $r_{1}= 8.20\times10^{7}$

Orbital radius of second satellite $r_{2}=7.00\times10^{7}\ m$

Mass of first satellite $m_{1}=53.0\ kg$

Mass of second satellite $m_{2}=54.0\ kg$

Orbital speed of first satellite = 4800 m/s

We need to calculate the orbital speed of this second satellite

Using formula of orbital speed

$v=\sqrt{\dfrac{GM}{r}}$

From this relation,

$v_{1}\propto\dfrac{1}{\sqrt{r}}$

Now, $\dfrac{v_{1}}{v_{2}}=\sqrt{\dfrac{r_{2}}{r_{1}}}$

$v_{2}=v_{1}\times\sqrt{\dfrac{r_{1}}{r_{2}}}$

Put the value into the formula

$v_{2}=4800\times\sqrt{\dfrac{ 8.20\times10^{7}}{7.00\times10^{7}}}$

$v_{2}=5195.16\ m/s$

Hence, The orbital speed of this second satellite is 5195.16 m/s.

4 0

3 years ago

These are nontoxic, nonflammable chemicals containing atoms of carbon, chlorine, and fluorine that have created a hole in the oz

yarga [219]

Hello! The nontoxic, nonflammable chemicals containing atoms of carbon, chlorine, and fluorine that have created a hole in the ozone layer are the Chlorofluorocarbons (CFCs)

These are compounds developed and improved by Thomas Midgley in the late 1920s. They were used as refrigerants and aerosol propellants.

These compounds created a hole in the ozone layer by the following reactions:

CCl₃F → CCl₂F· + Cl· (In the presence of light. Radical Reaction)
Cl· + O₃ → ClO + O₂
ClO + O₃ → Cl· + O₂

The last 2 reactions can repeat in a radical mechanism and explain why these compounds are so harmful to the ozone layer.

4 0

3 years ago

An object travels with a constant speed in a circular path. The net force on the object is

Pepsi [2]

Answer:

toward the center

Explanation:

Before answering, let's remind the first two Newton Laws:

1) An object at rest tends to stay at rest and an object moving at constant velocity tends to continue its motion at constant velocity, unless acted upon a net force

2) An object acted upon a net force F experiences an acceleration a according to the equation

$F=ma$

where m is the mass of the object.

In this problem, we have an object travelling at constant speed in a circular path. The fact that the trajectory of the object is circular means that the direction of motion of the object is constantly changing: this means that its velocity is changing, so it has an acceleration. And therefore, a net force is acting on it. The force that keeps the object travelling in the circular path is called centripetal force, and it is directed towards the center of the circle (because it prevents the object from continuing its motion straight away).

So, the correct answer is

toward the center

8 0

3 years ago

2. A 2000 kg car with speed 12.0 m/s hits a tree. The tree does not move or

krek1111 [17]

a) The work done by the tree is $-1.44\cdot 10^5 J$

b) The amount of force applied is 2880 N

Explanation:

a)

According to the work-energy theorem, the work done on the car is equal to the change in kinetic energy of the car. Therefore, we can write:

$W=K_f - K_i = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

where

W is the work done on the car

m is the mass of the car

u is its initial speed

v is its final speed

For the car in this problem, we have:

m = 2000 kg

u = 12.0 m/s

v = 0 (since the car comes to a stop, after the crash)

Therefore, the work done by the tree on the car is:

$W=0-\frac{1}{2}(2000)(12.0)^2=-1.44\cdot 10^5 J$

The work is negative because it is done in the direction opposite to the direction of motion of the car.

b)

The work done by the tree on the car can also be rewritten as

$W=Fd$

where

F is the force applied on the car

d is the displacement of the car during the collision

In this situation, we have:

$W=-1.44\cdot 10^5 J$ is the work done

$d=50.0 cm = 0.50 m$ is the displacement of the car during the collision

Solving the equation for F, we find the force exerted by the tree on the car:

$F=\frac{W}{d}=\frac{-1.44\cdot 10^5 J}{0.50}=-2880 N$

Where the negative sign means the force is applied opposite to the direction of motion of the car. Therefore, the magnitude of the force applied is 2880 N.

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

3 0

3 years ago

Describe what it means if a chemical or physical property is periodic

vesna_86 [32]

It means that you consider the elements as a list organized by atomic number, the property is seen to repeat over and over as you move through that list.

6 0

3 years ago

Compared to microwaves, ultraviolet waves have ___________________________ frequency.