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2 years ago

A 2.0 kg particle moving along the z-axis experiences the

Physics

1 answer:

DochEvi [55]2 years ago

6 0

At point x = 0, the particle accelerates. Since there will be change of velocity at that point. The the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.

Given that a 2.0 kg particle moving along the z-axis experiences the force shown in a given figure.

Force is the product of mass and acceleration. While acceleration is the rate of change of velocity. Both the force and acceleration are vector quantities. They have both magnitude and direction.

If the particle's velocity is 3.0 m/s at x = 0 m, that mean that the particle experience change of velocity at point x = 0. Since the the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.

Learn more here: brainly.com/question/20366032

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A donkey starts from rest and accelerates for 6.7 s at a rate of 1.6 m/s².

vitfil [10]

The maximum speed of the donkey is 10.72m/s

The question is based on the principle of motion in one dimension and hence formulas of motion in one dimension can be applied.

It is given that donkey attains an acceleration of 1.6 m/s^2

The time taken to accelerate to given speed is 6.7 seconds

We use the formula v=u + at to find the fastest speed

v is the final or maximum speed

u is the initial speed which in this case is 0 as the donkey is at rest

a is the acceleration of the donkey

t is the time taken in seconds

v = u + at

v= 0 + 1.6 x 6.7

= 10.72 m/s

Hence the donkey obtains the speed of 10.72 m/s

For further reference:

brainly.com/question/24478168?referrer=searchResults

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1 year ago

PLEASE HELP!!!<br>How are middle latitude cyclones and tropical cyclones different from each other?

defon

In contrast, extratropical cyclones have their strongest winds near the tropopause, which is about 8 miles above the surface. These differences are due to the tropical cyclone being “warm-core” in the troposphere, whereas extra-tropical cyclones are “warm-core” in the stratosphere and “cold-core” in the troposphere.

8 0

3 years ago

A 4.00-kg particle moves along the x axis. Its position varies with time according to x 5 t 1 2.0t 3, where x is in meters and t

White raven [17]

Answer:

Explanation:

Given equation is ,

x =t + 2 t³ ,

dx/dt = velocity ( v ) = 1 + 6 t²

a) kinetic energy = 1/2 m v² = .5 x 4 ( 1 + 6 t² )² = 2 ( 1 + 6 t²)²

b ) Acceleration = dv /dt = 12 t .

force( F ) = mass x acceleration = 4 x 12 t = 48 t

Power = force x velocity = 48 t x ( 1 + 6 t²). = 48 t + 288 t³ )

work done = ∫ F dx =∫ 48 t x( 1 + 6t² )dt ; = [48t²/2 + 48 x 6 x t³ /3 = 24 t² + 96 t³ )]₀² = 864 J

6 0

2 years ago

Read 2 more answers

It is observed that the time for the ball to strike the ground at b is 2.5 s. determine the speed at which the ball was thrown.

Alika [10]

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2 years ago

flipper (the dolphin) is out in the open ocean hunting tuna avec. he emits his pulse at 22khz and .42 seconds later he hears it

marusya05 [52]

First we need to find the speed of the dolphin sound wave in the water. We can use the following relationship between frequency and wavelength of a wave:
$v=\lambda f$
where
v is the wave speed
$\lambda$ its wavelength
f its frequency
Using $\lambda = 2 cm = 0.02 m$ and $f=22 kHz = 22000 Hz$ , we get
$v=(0.02 m)(22000 Hz)=440 m/s$

We know that the dolphin sound wave takes t=0.42 s to travel to the tuna and back to the dolphin. If we call L the distance between the tuna and the dolphin, the sound wave covers a distance of S=2 L in a time t=0.42 s, so we can write the basic relationship between space, time and velocity for a uniform motion as:
$v= \frac{S}{t}= \frac{2L}{t}$
and since we know both v and t, we can find the distance L between the dolphin and the tuna:
$L= \frac{vt}{2}= \frac{(440 m/s)(0.42 s)}{2}=92.4 m$

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3 years ago