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3 years ago

7. A 3.0 kg object travels vertically at a constant

Physics

1 answer:

vladimir2022 [97]3 years ago

7 0

Answer:

220.7

Explanation:

distance traveled is 1.5*5=7.5m

PE gain is MGH=3*9.81*7.5=220.7

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The position of an object that is oscillating on a spring is given by the equation x = (17.4 cm) cos[(5.46 s-1)t]. what is the a

bulgar [2K]

The angular frequency of this motion is 5.46 rad/s.

The oscillation of spring is an example of Simple Harmonic Motion(SHM).

The general equation of an SHM is given by the formula.

X = Acos(wt)

Here A is the amplitude

ω is the angular frequency

T is the time

Comparing the above equation with the given condition,

X = 17.4 cm cos(5.46t)

A = 17.4 cm

ω = 5.46 rad/s

T = 1 s

Hence, the angular frequency of this motion is 5.46 rad/s.

To know more about the "general equation of SHM", refer to the link below:

brainly.com/question/14869852?referrer=searchResults

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6 0

2 years ago

_____ electrons are the most important electrons to an atom .

nalin [4]

Answer:

stable is the answer

Explanation:

stable electrons

8 0

3 years ago

Which one is the full moon?

Svetlanka [38]

Answer:

The one to the left

Explanation:

6 0

3 years ago

The power in our homes and buildings cycles at a frequency of 60Hz. If someone accidentally dropped a power line into a holding

sp2606 [1]

Answer:

The answer is "Rigor mortis".

Explanation:

When the very large volume will be that the progress of its postponed by calcium and magnesium. The causes brain intensification of a liver due to the change throughout the myofibrils within a week of dying of its organism. When anyone accidentally spilled a power cable into a rank for toads, one's tissue will be disrupted instantly by stringent deaths.

8 0

3 years ago

Two soccer players start from rest, 36 m apart. They run directly toward each other, both players accelerating. The first player

Cerrena [4.2K]

A) Both players are moving by uniformly accelerated motion, and we can write the position at time t of each of the two players as follows:
$x_1(t)= \frac{1}{2}a_1 t^2$
$x_2(t)=d- \frac{1}{2}a_2 t^2$
where
$a_1 = 0.58 m/s^2$ is the acceleration of the first player
$a_2=0.42 m/s^2$ is the acceleration of the second player
$d=36 m$ is the initial distance between the two players
and where I put a negative sign in front of the acceleration of the second player, since he's moving in the opposite direction of the first player.

The time t at which the two players collide is the time t at which $x_1 = x_2$ , therefore:
$\frac{1}{2}a_1 t^2 = d- \frac{1}{2}a_2 t^2$
from whic we find
$t= \sqrt{ \frac{2d}{a_1+a_2} }= \sqrt{ \frac{2 \cdot 36 m}{0.58 m/s^2+0.42 m/s^2} }=8.5 s$

b) We can use the equation of $x_1(t)$ to find how far the first player run in t=8.5 s:
$x_1(t)= \frac{1}{2}a_1 t^2= \frac{1}{2}(0.58 m/s^2)(8.5 s)^2=21.0 m$

8 0

3 years ago