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inna [77]
4 years ago
14

Without using the word or word forms of "motion",describe how you know and object is in motion

Physics
1 answer:
adoni [48]4 years ago
4 0

You must look at the object either two or three times,
as follows:

1). Look at it the first time.  Make a mental note, or take a photo,
of its location.

2). Wait a short time.

3). Look at the object the second time.  Make another mental note,
or take another photo, of its position.

4). Compare the two photos or mental notes.

5). If the two images show the object in different places, then the object
is in motion RELATIVE TO YOU, and your investigation is finished.  STOP.

6). If the two images show the object in the same identical place, then
you don't know the answer yet.  It might be moving in a circle.  You must
check it out.

7).  Repeat step-1.

8).  Repeat step-2, but wait a different length of time this time.

9).  Repeat step-3.  This gives you a third photo.

10).  Compare the first and third photos.
. . . . . If these are still identical, you can be pretty sure that
the object is not in motion RELATIVE TO YOU.
. . . . . If these are different, then the object is in motion RELATIVE TO YOU.

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A 1300 kg steel beam is supported by two ropes. (Figure
Dmitriy789 [7]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

  • the net horizontal force acting on the beam is

R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0

where R_1,R_2 are the magnitudes of the tensions in ropes 1 and 2, respectively;

  • the net vertical force acting on the beam is

R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0

where m=1300\,\rm kg and g=9.8\frac{\rm m}{\mathrm s^2}.

Eliminating R_2, we have

\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)

R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2

R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2

-R_1 \sin(50^\circ) = -\dfrac{mg}2

R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}

Solve for R_2.

\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0

\dfrac{R_2}2 = -mg\cot(110^\circ)

R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}

8 0
2 years ago
Identify the palindromic base sequence in the DNA donor molecule shown.
Airida [17]

Explanation:

Since a double helix is formed by two paired antiparallel strands of nucleotides that run in opposite directions, and the nucleotides always pair in the same way (adenine (A) with thymine (T) in DNA or uracil (U) in RNA; cytosine (C) with guanine (G)), a (single-stranded) nucleotide sequence is said to be a palindrome

6 0
3 years ago
An object of mass 30 kg is in free fall in a vacuum where there is no air resistance. Determine the acceleration of the object.
velikii [3]

Answer:

In free fall, mass is not relevant and there's no air resistance, so the acceleration the object is experimenting will be equal to the gravity exerted. If the object is falling on our planet, the value of gravity is approximately 9.81ms2 .

7 0
3 years ago
41. Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 h
romanna [79]

Answer:

138.3 days

Explanation:

Given that a Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 has been discovered and has a radius of 7.8 X 10 meters.

The period of time for Clayton J-21 to orbit Dayli can be calculated by using Kepler law.

T^2 is proportional to r^3

That is,

T^2/r^3 = constant

98^2 / 62^3 = T^2 / 78^3

Make T^2 the subject of formula.

T^2 = 98^2 / 62^3 × 78^3

T^2 = 19123.2

T = sqrt ( 19123.2 )

T = 138.2867 days

Therefore, the period of time for Clayton J-21 to orbit Dayli is 138.3 days approximately.

4 0
3 years ago
Compared with a force, a torque involves
ycow [4]
A rotational movement.
6 0
3 years ago
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