All categories

3 years ago

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

2 answers:

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.

You might be interested in

What is one property of a suspension that is different from that of a solution or a colloid?

Step2247 [10]

Answer:

Explanation:

If left to rest it will separate.

3 0

3 years ago

The greatest speed with which an athlete can jump vertically is around 5 m/sec. Determine the speed at which Earth would move do

katrin2010 [14]

Answer:

Approximately $2.0 \times 10^{-23}\; \rm m \cdot s^{-1}$ if that athlete jumped up at $1.8\; \rm m \cdot s^{-1}$ . (Assuming that $g = 9.81\; \rm m\cdot s^{-1}$ .)

Explanation:

The momentum $p$ of an object is the product of its mass $m$ and its velocity $v$ . That is: $p = m \cdot v$ .

Before the jump, the speed of the athlete and the earth would be zero (relative to each other.) That is: $v(\text{athlete, before}) = 0$ and $v(\text{earth, before}) = 0$ . Therefore:

$\begin{aligned}& p(\text{athlete, before}) = 0\end{aligned}$ and $p(\text{earth, before}) = 0$ .

Assume that there is no force from outside of the earth (and the athlete) acting on the two. Momentum should be conserved at the instant that the athlete jumped up from the earth.

Before the jump, the sum of the momentum of the athlete and the earth was zero. Because momentum is conserved, the sum of the momentum of the two objects after the jump should also be zero. That is:

$\begin{aligned}& p(\text{athlete, after}) + p(\text{earth, after}) \\ & =p(\text{athlete, before}) + p(\text{earth, before}) \\ &= 0\end{aligned}$ .

Therefore:

$p(\text{athelete, after}) = - p(\text{earth, after})$ .

$\begin{aligned}& m(\text{athlete}) \cdot v(\text{athelete, after}) \\ &= - m(\text{earth}) \cdot v(\text{earth, after})\end{aligned}$ .

Rewrite this equation to find an expression for $v(\text{earth, after})$ , the speed of the earth after the jump:

$\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \end{aligned}$ .

The mass of the athlete needs to be calculated from the weight of this athlete. Assume that the gravitational field strength is $g = 9.81\; \rm N \cdot kg^{-1}$ .

$\begin{aligned}& m(\text{athlete}) = \frac{664\; \rm N}{9.81\; \rm N \cdot kg^{-1}} \approx 67.686\; \rm N\end{aligned}$ .

Calculate $v(\text{earth, after})$ using $m(\text{earth})$ and $v(\text{athlete, after})$ values from the question:

$\begin{aligned} &v(\text{earth, after}) \\ &= -\frac{m(\text{athlete}) \cdot v(\text{athlete, after})}{m(\text{earth})} \\ &\approx -2.0 \times 10^{-23}\; \rm m \cdot s^{-1}\end{aligned}$ .

The negative sign suggests that the earth would move downwards after the jump. The speed of the motion would be approximately $2.0 \times 10^{-23}\; \rm m \cdot s^{-1}$ .

3 0

3 years ago

What determines if a star becomes a dwarf, neutron star or a black hole?

faust18 [17]

I'm pretty sure it's the neutron star.

3 0

3 years ago

Why are neutral objects attracted to charged objects?

kow [346]

Explanation:

When neutral objects are placed in the vicinity of charged objects,they get attracted.

The isolated neutral object has positive charge and negative charge spread throughout it completely.

When a charged particle is brought,the opposite charges in the neutral object occupy the positions near to the charged particle and the like charges occupy the positions far from the charged particle.

This creates a dipole with some dipole moment.

This dipole attracts to the field of the charged particle.

5 0

4 years ago

Marlon is studying a crab population. He has a large batch of crabs that were captured in the ocean. He places a plastic tag on

tatiyna

First time I did it I got 2143658709

7 0

3 years ago