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Grace [21]
3 years ago
7

An aeroplane, flying in a straight line at a constant

Physics
1 answer:
dedylja [7]3 years ago
3 0

Answer:

200

Explanation:

You might be interested in
how much centripetal force is needed to make a body of mass 0.5 kg to move in a circle of radius 50 cm with a speed 3ms-1
Alex17521 [72]

Answer:

9 N

Explanation:

The centripetal force F is F = mrω^2 = (mv^2)/r where m is mass, r is radius of the curve, ω is angular velocity and v is tangential velocity.

In this case, m = 0.5kg, r = 0.5m, v = 3m/s

So F = [0.5kg(3m/s)^2]/0.5m = 9kg-m/s^2 which is 9N

6 0
3 years ago
Read 2 more answers
What do these weapon stats mean?
kaheart [24]

Answer:

its sort of like how Pounds are measured in lbs

Explanation:

Pounds are measured in lbs when they dont sound anything similar, same thing applies to gun technology.

5 0
3 years ago
The x-coordinates of two objects moving along the x-axis are given as a function of time (t). x1= (4m/s)t x2= -(161m) + (48m/s)t
Kitty [74]
The two displacement functions are
x₁ = 4t
x₂ = -161 + 48t - 4t²
where
x₁, x₂ are in meters
t is time, s

The distance between the two objects is
x = x₁ - x₂
   =  4t + 161 - 48t + 4t²
x = 4t² - 44t + 161

Write this equation in the standard form for a parabola.
x = 4[t² - 11t] + 161
  = 4[ (t - 5.5)² - 5.5² ] + 161
 x = 4(t-5)² + 40

Ths is a parabola that faces up and has its vertex (lowest point) at (5, 40).
Therefore the closest approach of the two objects is 40 m.
The graph of x versus t confirms the result.

Answer: The distance of the closest approach is 40 m.

5 0
3 years ago
#26 question
4vir4ik [10]

So, the time that taken for the astronaut to fall to the surface of the moon is <u>2.5 s.</u>

<h3>Introduction</h3>

Hi ! In this question, I will help you. In this question, you will learn about the fall time of the free fall motion. Free fall is a downward vertical motion without being preceded by an initial velocity. When moving in free fall, the time required can be calculated by the following equation:

\sf{h = \frac{1}{2} \cdot g \cdot t^2}

\sf{\frac{2 \cdot h}{g} = t^2}

\boxed{\sf{\bold{t = \sqrt{\frac{2 \cdot h}{g}}}}}

With the following condition :

  • t = interval of the time (s)
  • h = height or any other displacement at vertical line (m)
  • g = acceleration of the gravity (m/s²)

<h3>Problem Solving</h3>

We know that :

  • h = height = 5.00 m
  • g = acceleration of the gravity = 1.6 m/s²

What was asked :

  • t = interval of the time = ... s

Step by step :

\sf{t = \sqrt{\frac{2 \cdot h}{g}}}

\sf{t = \sqrt{\frac{2 \cdot 5}{1.6}}}

\sf{t = \sqrt{6.25}}

\boxed{\sf{t = 2.5 \: s}}

<h3>Conclusion</h3>

So, the time that taken for the astronaut to fall to the surface of the moon is 2.5 s.

<h3>See More</h3>
  • Time that needed for hearing the splash of fallen rock in the well brainly.com/question/26485521
  • The speed of the object at a certain height (free fall motion) brainly.com/question/26377041
  • The relationship between acceleration and the change in velocity and time in free fall brainly.com/question/26486625
3 0
3 years ago
Two motors in a factory are running at slightly different rates. One runs at 825.0 rpm and the other at 786.0 rpm. You hear the
mixas84 [53]

Answer:

T=1.54s

Explanation:

From the question we are told that:

Speed of Motor 1 \omega_1=825rpm=>2 \pi 13.75

Speed of Motor 2 \omega_1=786rpm=>2 \pi 13.1

Therefore

Frequency of Motor 1 f_1=13.75

Frequency of Motor 2  f_2= 13.1

Generally the equation for Time Elapsed is mathematically given by

T=\frac{1}{df}

Where

df=f_1-f_2

df=13.75-13.1

df=0.65Hz

Therefore

T=\frac{1}{65}

T=1.54s

5 0
3 years ago
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