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BaLLatris [955]
3 years ago
14

Newtons law of garvitation is known as universal law. Why​

Physics
1 answer:
xeze [42]3 years ago
5 0

Answer:

It is called so because it is applicable on all bodies having mass, and the bodies will be governed by the same law, that is newton's law of gravitation. Thus, as it is applicable universally, it is called as universal law.

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. Calculate the efficiency of a bicycle if the input work to turn the pedal is 45J and the output work is 20J. * 1 point 2.25 2.
cestrela7 [59]

20/45=0.4*100= 44.4 so the answer is..................................................

Answer: 44.4%

8 0
3 years ago
When we kept the Earth's mass the same, but shrank its size, we saw that had an effect on its escape speed. Albert Einstein used
nadezda [96]

Answer:

Check the explanation

Explanation:

The escape velocity is the velocity needed by any object to overcome the gravitational force of the planet on which it’s present. Now we know that the gravitational force is directly proportional to the mass of the planet and inversely proportional to the distance of the object from the center of planet.

If we keep the mass of earth constant and decrease the size of the earth than this will decrease the distance between the object and the center of the earth and thus the gravitational force that will act on the body will increase substantially which will in turn increase the value of the escape velocity.

The value of escape velocity will keep on increasing as the size of the earth will shrink till it reaches to a point when the value of escape velocity becomes more than the speed of light and since it’s impossible to travel with a speed greater than the speed of light and therefore at this point it will become impossible for a spacecraft to escape the earth.

7 0
3 years ago
During the middle of a family picnic, Barry Allen received a message that his friends Bruce and Hal
weeeeeb [17]

The kinematics of the uniform motion and the addition of vectors allow finding the results are:

  • The  Barry's initial trajectory is 94.30 10³ m with n angles of θ = 138.8º
  • The return trajectory and speed are v = 785.9 m / s, with an angle of 41.2º to the South of the East

Vectors are quantities that have modulus and direction, so they must be added using vector algebra.

A simple method to perform this addition in the algebraic method which has several parts:

  • Vectors are decomposed into a coordinate system
  • The components are added
  • The resulting vector is constructed

 Indicate that Barry's velocity is constant, let's find using the uniform motion thatthe distance traveled in ad case

              v = \frac{\Delta d}{t}

              Δd = v t

Where  v is the average velocity, Δd the displacement and t the time

We look for the first distance traveled at speed v₁ = 600 m / s for a time

          t₁ = 2 min = 120 s

          Δd₁ = v₁ t₁

          Δd₁ = 600 120

          Δd₁ = 72 10³ m

Now we look for the second distance traveled for the velocity v₂ = 400 m/s    

  time t₂ = 1 min = 60 s

          Δd₂ = v₂ t₂

          Δd₂ = 400 60

          Δd₂ = 24 103 m

   

In the attached we can see a diagram of the different Barry trajectories and the coordinate system for the decomposition,

We must be careful all the angles must be measured counterclockwise from the positive side of the axis ax (East)

Let's use trigonometry for each distance

Route 1

          cos (180 -35) = \frac{x_1}{\Delta d_1}

          sin 145 = \frac{y_1}{\Delta d1}

          x₁ = Δd₁ cos 125

          y₁ = Δd₁ sin 125

          x₁ = 72 103 are 145 = -58.98 103 m

          y₁ = 72 103 sin 155 = 41.30 10³ m

Route 2

          cos (90+ 30) = \frac{x_2}{\Delta d_2}

          sin (120) = \frac{y_2}{\Delta d_2}

          x₂ = Δd₂ cos 120

          y₂ = Δd₂ sin 120

          x₂ = 24 103 cos 120 = -12 10³ m

           y₂ = 24 103 sin 120 = 20,78 10³ m

             

The component of the resultant vector are

              Rₓ = x₁ + x₂

              R_y = y₁ + y₂

              Rx = - (58.98 + 12) 10³ = -70.98 10³ m

              Ry = (41.30 + 20.78) 10³ m = 62.08 10³ m

We construct the resulting vector

Let's use the Pythagoras' Theorem for the module

             R = \sqrt{R_x^2 +R_y^2}

             R = \sqrt{70.98^2 + 62.08^2}   10³

             R = 94.30 10³ m

We use trigonometry for the angle

             tan θ ’= \frac{R_y}{R_x}

             θ '= tan⁻¹ \frac{R_y}{R_x}

             θ '= tan⁻¹ \frac{62.08}{70.98}

             θ ’= 41.2º

Since the offset in the x axis is negative and the displacement in the y axis is positive, this vector is in the second quadrant, to be written with respect to the positive side of the x axis in a counterclockwise direction

            θ = 180 - θ'

            θ = 180 -41.2

            θ = 138.8º

Finally, let's calculate the speed for the way back, since the total of the trajectory must be 5 min and on the outward trip I spend 3 min, for the return there is a time of t₃ = 2 min = 120 s.

The average speed of the trip should be

             v = \frac{\Delta R}{t_3}  

             v = \frac{94.30}{120}  \ 10^3

              v = 785.9 m / s

in the opposite direction, that is, the angle must be

               41.2º to the South of the East

In conclusion, using the kinematics of the uniform motion and the addition of vectors, results are:

  • To find the initial Barry trajectory is 94.30 10³ m with n angles of  138.8º
  • The return trajectory and speed is v = 785.9 m / s, with an angle of 41.2º to the South of the East

Learn more here:  brainly.com/question/15074838

4 0
2 years ago
Use the SI prefixes in Table 3 of this chapter to convert these hypothetical units of measure into appropriate quantities: a. 10
8090 [49]
A. 10 rations = 1 deca-ration.

b. 2000 mockingbirds = 2 x 10³ = 2 kilo-mockingbirds.

c. 10⁻⁵ phones = 1 micro-phones.

d. 10⁻⁹ goats = 1 nano-goats.

e. 1018 miners = 1.018 x 10³ = 1.018 kilo-miners.
4 0
2 years ago
Hey can anyone help me out in dis pls!
patriot [66]

Answer:

D

Explanation:

3 0
3 years ago
Read 2 more answers
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