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Bess [88]
3 years ago
6

1- How are translations represented as a function?. . . 2- What is the relationship between a translation and a rigid motion?

Physics
2 answers:
Darya [45]3 years ago
7 0
(1). Every translation is a function of a direction and a distance, it is represented by a direction and a distance
(2). Translation is a form of rigid motion. it involves the action of moving an object to a different point without altering its shape and size. Reflection, rotation, translation and glide reflection are the four types of rigid motion that exist.
Amiraneli [1.4K]3 years ago
4 0

Explanation:

1. Translations are a function of distance and direction. It is mathematically represented by,

f( distance, displacement )

It is a form rigid motion.

2. Translation is one of the four form of rigid motions. In translation, every point on the object move by the same and equal amount in the same direction. This is nothing but a rigid motion as in rigid motion the objects moves to a different place without changing its shape and size.

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Which part of the eye controls whether or not your pupil dilates or restricts?
valentinak56 [21]
The answer would be, the iris. Hope this helps:)
5 0
4 years ago
Find the resistance of a stereo: current is 7 A voltage is 49 V.
pochemuha

Answer:

R = 7 [amp]

Explanation:

To solve this problem we must use ohm's law which tells us that the voltage is equal to the product of the current by the resistance. In this way, we have the following equation.

V = I*R

where:

V = voltage = 49 [V] (units of volts)

I = current = 7 [amp] (amperes)

R = resistance [ohms]

Now clearing R.

R =V/I

R = 49/7

R = 7 [amp]

5 0
3 years ago
Two people are pushing a couch from opposite sides. one person pushing to the right with a force of 3 N. the other person pushes
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8 0
3 years ago
A plane takes off in san francisco at noon and flies toward the southeast. an hour later, it is 400 kilometers east and 300 kilo
eduard
This first step involves a right triangle. If the plane is 400 km east and 300 km south of the origin, and it flew in a straight line, then you can construct a right triangle with side lengths 300, 400, and c. You may recognize that these are multiples of the Pythagorean triple 3, 4, 5, so the side length c is 500 km. Otherwise, you would write 
c^2 = 300^2+400^2 \\ c = 500. 

This second step is, if I am correctly interpreting "degrees south of east," to find the angle formed by the horizontal line representing the east and the path of the plane. I made a diagram that does just that (see attached). You can use a trig function of one of the angles to solve. I chose
tan(C)= \frac{3}{4}  \\  arctan( \frac{3}{4})=36.870. Thus, I believe it is 37° south of east.

5 0
3 years ago
A brave child decides to grab onto an already spinning merry‑go‑round. The child is initially at rest and has a mass of 25.5 kg.
Svetllana [295]

Answer:

72.75 kg m^2

Explanation:

initial angular velocity, ω = 35 rpm

final angular velocity, ω' =  19 rpm

mass of child, m = 15.5 kg

distance from the centre, d = 1.55 m

Let the moment of inertia of the merry go round is I.

Use the concept of conservation of angular momentum

I ω = I' ω'

where I' be the moment of inertia of merry go round and child

I x 35 = ( I + md^2) ω'

I x 35 = ( I + 25.5 x 1.55 x 1.55) x 19

35 I = 19 I + 1164

16 I = 1164

I = 72.75 kg m^2

Thus, the moment of inertia of the merry go round is 72.75 kg m^2.

5 0
3 years ago
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