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AlexFokin [52]
3 years ago
15

What are you experiencing if you close your eyes and feel as if you are not moving

Physics
1 answer:
vlabodo [156]3 years ago
8 0
Well i would feel i am moving i also would be scared if i didnt know i was moving even with my eyes closed 

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What is the net force on this object f air = 400n (up) fgrav=600n (down)
n200080 [17]
Fnet=F1+F2 or Fnet=F1-F2
So 400n up - 600n down
Fnet= 400-600= -200N
5 0
3 years ago
Read 2 more answers
12. A rocket, initially at rest on the ground, accelerates vertically. It accelerates uniformly until it
vodomira [7]

Answer:

We kindly invite you to read carefully the explanation and check the image attached below.

Explanation:

According to this problem, the rocket is accelerated uniformly due to thrust during 30 seconds and after that is decelerated due to gravity. The velocity as function of initial velocity, acceleration and time is:

v_{f} = v_{o}+a\cdot (t-t_{o}) (1)

Where:

v_{o} - Initial velocity, measured in meters per second.

v_{f} - Final velocity, measured in meters per second.

a - Acceleration, measured in meters per square second.

t_{o} - Initial time, measured in seconds.

t - Final time, measured in seconds.

Now we obtain the kinematic equations for thrust and free fall stages:

Thrust (v_{o} = 0\,\frac{m}{s}, a = 30\,\frac{m}{s^{2}}, t_{o} = 0\,s, 0\,s\le t< 30\,s)

v = 30\cdot t (2)

Free fall (v_{o} = 900\,\frac{m}{s}, a = -9.807\,\frac{m}{s}, t_{o} = 30\,s, 30\,s \le t \le 120\,s)

v = 900-9.81\cdot (t-30) (3)

Now we created the graph speed-time, which can be seen below.

5 0
2 years ago
When have you experienced an increase in kinetic<br> energy within a system?
Mars2501 [29]

Answer:

If a man starts running on a boat with an acceleration a with respect to the boat, there is no external force that acts on the Boat+Man system

8 0
2 years ago
Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference
strojnjashka [21]

Answer:

1.7323

Explanation:

To develop this problem, it is necessary to apply the concepts related to refractive indices and Snell's law.

From the data given we have to:

n_{air}=1

\theta_{liquid} = 19.38\°

\theta_{air}35.09\°

Where n means the index of refraction.

We need to calculate the index of refraction of the liquid, then applying Snell's law we have:

n_1sin\theta_1 = n_2sin\theta_2

n_{air}sin\theta_{air} = n_{liquid}sin\theta_{liquid}

n_{liquid} = \frac{n_{air}sin\theta_{air}}{sin\theta_{liquid}}

Replacing the values we have:

n_{liquid}=\frac{(1)sin(35.09)}{sin(19.38)}

n_liquid = 1.7323

Therefore the refractive index for the liquid is 1.7323

6 0
3 years ago
Which of these is equivalent to the law of conservation of energy?
Korolek [52]
The Law of the Conservation of Energy is stating that the total mechanical energy is always conserved or in simpler terms, not used or saved.
7 0
3 years ago
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