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mihalych1998 [28]
3 years ago
8

The former soviet union launched the first artificial earth satellite. What is the name of this satelitte?

Physics
1 answer:
faltersainse [42]3 years ago
7 0

Answer:

Sputnik I.

Explanation:

The Soviet Union (former) launched the first artificial earth satellite called Sputnik I on October 4, 1957.

It had a diameter of 58 cm and it weighed 83.6 kg. It had an orbital period of 5880 seconds around the earth. It orbited the 11 weeks (officially 3 weeks but then, its battery died and it orbited for 8 weeks before falling back into the earth's atmosphere).

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What distance should be used to pattern a shotgun hunter ed?
Murljashka [212]
I think the distance that should be used is the distance that one expects to be from the game you are hunting. Before taking a shotgun for a gobbler or even for ducks or other animals, you need to see how your gun performs by patterning it at various ranges with the load you want to use.
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How can you describe the motion of an object in a race?
Hitman42 [59]
You can describe the motion of an object by its position, speed, direction, and acceleration
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Car A has a mass of 1,200 kg and is traveling at a rate of 22 km/hr. It collides with car B. Car B has a mass of 1,900 kg and is
anastassius [24]

The car A has a mass of 1200 kg.

The car B has the mass of 1900 kg.

It is given that velocity of car  A is given as 22 Km/hr

The car B has the velocity of 25 Km/hr.

Let the mass of two bodies are denoted as  m_{1} \ and\ m_{2}

Let the velocity of cars A and B are denoted as v_{1} \ and\ v_{2}

The momentum before collision is-

                                                  p_{i} =m_{1} v_{1} +m_{2} v_{2}

[Here p stand for momentum.]

We are asked to calculate the final momentum of the system after collision.

The answer of the question is based law of conservation of  linear momentum.

As per law of conservation of linear momentum the sum total linear momentum for an isolated system is always constant.Hence irrespective of the type of collision[elastic and inelastic],the momentum of the system is always constant which is a universal truth.

Let after the collision the velocity of A and B are v'_{1} \ and\ v'_{2}

Hence the final momentum of the system is-

                                                        p_{f} = m_{1} v'_{1} +m_{2} v'_{2}

As per the law of conservation of linear momentum, the initial and final momentum must be equal i.e      

                              p_{i} =p_{f}

                               m_{1} v_{1} +m_{2}v_{2} =m_{1} v'_{1} +m_{2} v'_{2}

Hence the option A  is right.

7 0
3 years ago
Read 2 more answers
The scale is 1:30 and their are 10 boulders. How many boulders are their in the real pond
Alenkasestr [34]

Answer:

the correct answer is C

Explanation:

When we express that the scale is 1:30 we mean that the objects of the realization are reduced by a factor of 30 in the graph, for example a distance of 30 cm in the graph is represented by a distance of 1 cm.

Therefore something that in the graph has n value to bring it to real size must be multiplied by the scale.

Applying this to our case if there is

               10 boulder on the chart

in reality there are #_boulder = 10 30

                                #_boulder = 300 boulder

so the correct answer is C

7 0
3 years ago
At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?
sergey [27]

1) Initial upward acceleration: 6.0 m/s^2

2) Mass of burned fuel: 0.10\cdot 10^4 kg

Explanation:

1)

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude F_g = mg, in the downward direction, where

m=1.9\cdot 10^4 kg is the rocket's mass

g=9.8 m/s^2 is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

T=3.0\cdot 10^5 N

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

T-mg=ma (1)

where a is the acceleration of the rocket.

Solving for a, we find the initial acceleration:

a=\frac{T-mg}{m}=\frac{3.0\cdot 10^5-(1.9\cdot 10^4)(9.8)}{1.9\cdot 10^4}=6.0 m/s^2

2)

When the rocket reaches an altitude of 5000 m, its acceleration has increased to

a'=6.9 m/s^2

The reason for this increase is that the mass of the rocket has decreased, because the rocket has burned some fuel.

We can therefore rewrite eq.(1) as

T-m'g=m'a'

where

m' is the new mass of the rocket

Re-arranging the equation and solving for m', we find

m'=\frac{T}{g+a}=\frac{3.0\cdot 10^5}{9.8+6.9}=1.8\cdot 10^4 kg

And since the initial mass of the rocket was

m=1.9 \cdot 10^4 kg

This means that the mass of fuel burned is

\Delta m = m-m'=1.9\cdot 10^4 - 1.80\cdot 10^4 = 0.10\cdot 10^4 kg

3 0
3 years ago
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