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never [62]
3 years ago
9

when astronauts work in space, they are sometimes attached to the spacecraft by a line called an umbilical. Why do you think the

line has been given this name?
Physics
2 answers:
wariber [46]3 years ago
3 0
Probably for the umbilical cord that connects babies (from their early stages in the womb to their removal) to their mothers. The cord is cut, forming the belly button. This is analogous to astronauts in space.
Zigmanuir [339]3 years ago
3 0
Because it is like the umbilical cord on a baby to its mother this is why they call it umbilical
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A 16.0 kg child on roller skates, initially at rest, rolls 2.0 m down an incline at an angle of 20.0° with the horizontal. If th
enyata [817]

The kinetic energy of the child at the bottom of the incline is 106.62 J.

The given parameters:

  • <em>Mass of the child, m = 16 kg</em>
  • <em>Length of the incline, L = 2 m</em>
  • <em>Angle of inclination, θ = 20⁰</em>

The vertical height of fall of the child from the top of the incline is calculated as;

sin(20) = \frac{h}{2} \\\\h = 2 \times sin(20)\\\\h = 0.68 \ m

The gravitational potential energy of the child at the top of the incline is calculated as;

P.E = mgh\\\\P.E = 16 \times 9.8 \times 0.68\\\\P.E = 106.62 \ J

Thus, based on the principle of conservation of mechanical energy, the kinetic energy of the child at the bottom of the incline is 106.62 J since no energy is lost to friction.

Learn more about conservation of mechanical energy here: brainly.com/question/332163

7 0
2 years ago
An astronaut on an alien planet drops a rock into a crater which is 100 meters deep. The rock hits the bottom of the crater 4 se
Semmy [17]

Answer:

The gravity on this planet is stronger than that of earth.

Explanation:

First we need to find the acceleration due to gravity value of this planet to compare its gravity force with that of the earth. Hence, we will use second equation of motion:

h = Vi t + (0.5)gt²

where,

h = height or depth of crater = 100 m

Vi = Initial Velocity of rock = 0 m/s

t = time = 4 s

g = acceleration due to gravity on this planet = ?

Therefore,

100 m = (0 m/s)(4 s) + (0.5)(g)(4 s)²

g = (200 m)/(16 s²)

g = 12.5 m/s²

on earth:

ge = 9.8 m/s²

Since,

ge < g

Therefore,

<u>The gravity on this planet is stronger than that of earth.</u>

6 0
3 years ago
Modern vehicles are designed ________ in a crash to absorb kinetic energy?
nekit [7.7K]
<span>Modern vehicles are designed to crush or crumple to absorb kinetic energy.</span>
5 0
2 years ago
How long will it take for a boat traveling at 36 k/h to go 126 km?
brilliants [131]
3 hours 30 min i believe because 126/36 is 3.5 and 3.5 hours is 3 hours and 30 min. Brainliest pls
5 0
2 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
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