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mr Goodwill [35]
2 years ago
6

A particle moves in three-dimensional space with a constant acceleration. Can the z component of the acceleration affect the x c

omponent of its location
Physics
1 answer:
denpristay [2]2 years ago
5 0

The components of the velocity vector that represent the change in the particle's position vector will be independent of one another as long as the components of the particle's position vector are independent of one another.

Can the z component of the acceleration affect the x and y component?

  • Similar to position and velocity vector components, acceleration vector components are also independent of one another.
  • For example, any change in the z component of the acceleration vector would not affect the x or y components of the object's position and velocity vector.
  • However, if one makes the assumption that, let's say, the particle's x position is dependent on its z position The x component of the position vector can then be connected to any change in the z component of the velocity vector.
  • The x component of the position vector can be adjusted to account for any changes in the z component of the acceleration vector.

Learn more about the acceleration vector with the help of the given link:

brainly.com/question/13590043

#SPJ4

I understand that the question you are looking for is "A particle moves in three-dimensional space with a constant acceleration. Can the z component of the acceleration affect the x component of its location? Can the z component of the acceleration affect the y component of the velocity?"

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The initial velocity of the ball is 1.01 m/s

Explanation:

The motion of the ball rolling off the desk is a projectile motion, which consists of two independent motions:

- A uniform horizontal motion with constant horizontal velocity

- A vertical accelerated motion with constant acceleration (g=9.8 m/s^2, acceleration due to gravity)

We start by analyzing the vertical motion: we can find the time of flight of the ball by using the following suvat equation

s=ut+\frac{1}{2}gt^2

where

s = 1.20 m is the vertical displacement (the height of the desk)

u = 0 is the initial vertical velocity

g=9.8 m/s^2

t is the time of flight

Solving for t,

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(1.20)}{9.8}}=0.495 s

Now we analyze the horizontal motion. We know that the ball covers a horizontal distance of

d = 0.50 m

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t = 0.495 s

Therefore, since the horizontal velocity is constant, we can calculate it as

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So, the ball rolls off the table at 1.01 m/s.

Learn more about projectile motion:

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4 0
3 years ago
A team of eight dogs pulls a sled with waxed wood runners on wet snow (mush!). The dogs have average masses of 18.5 kg, and the
meriva

Answer:

a.2.86 m/s^2

b.1058 N

Explanation:

We are given that

Mass of each dog,M=18.5 kg

Mass of sled with rider,m=250 kg

a.Average force,F=185 N

\mu_s=0.14

g=9.8 m/s^2

By Newton's second law

8F-f=(8M+m)a

a=\frac{8F-f}{8M+m}=\frac{8(185)-(0.14)(9.8)(250)}{8(18.5)+250}

a=2.86 m/s^2

b.By Newton's second law

T=ma+\mu_s mg

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T=250\times 2.86+0.14(250)(9.8)=1058 N

Hence, the force in the coupling between the dogs and the sled=1058 N

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Answer:

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