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vfiekz [6]
3 years ago
9

A particle of mass M moves along a straight line with initial speed vi. A force of magnitude F pushes the particle a distance D

along the direction of its motion.
Part A: Find vf, the final speed of the particle after it has traveled a distance D. Express the final speed in terms of vi, M, F, and D.
Physics
1 answer:
LenaWriter [7]3 years ago
6 0

Answer:

vf = √(vi²+2*(F/m)*D)

Explanation:

Given

Mass of the particle: M

Initial speed of the particle: vi

Force: F

Distance: D

We can apply the formula

F = M*a   ⇒    a = F/m

then we use the equation

vf = √(vi²+2*a*D)

⇒  vf = √(vi²+2*(F/m)*D)

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Can anyone please help me with the steps
KIM [24]

Answer:

A) v_average = - 10 km / h,  B)    v = 1.6 m / s, v = 17.6 m / s

Explanation:

A) the average speed is the average speed of a body, if we assume that the direction of going up the hill is positive

    v₁ = 40 km / h

    v₂ = - 60 km / h

the average speed is

         v_average = \frac{v_1 + v_2}{2}

         v_average = ( 40 - 60)/2

         v_average = - 10 km / h

B) in this case they indicate the acceleration a = 3.2 m / s² and the velocity vo = 9.6 m / s

i) the speed for 2.5 s above

         v = v₀ + a t

as the time is earlier

          t = - 2.5 s

we substitute

          v = 9.6 - 3.2 2.5

          v = 1.6 m / s

ii) the velocity for a subsequent time of 2.5 s

          t = 2.5 s

           

we substitute

          v = 9.6 + 3.2 2.5

          v = 17.6 m / s

3 0
3 years ago
What happens to the force between charges when: (a) the distance between them is halved?<br>​
charle [14.2K]

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Which of the following best describes the way that scientists make observations
horrorfan [7]
I don’t see any answer choice but the best way is asking questions about the natural phenomenon, making hypothesis, and predicting the consequences in the hypothesis.
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2 years ago
To start a lawn mower, you must pull on a rope wound around theperimeter of a flywheel. After you pull the rope for 0.95 s, thef
Molodets [167]

Answer:

29.76245 rad/s², -117.80972 rad/s²

28.2743 rad/s

3.95833

Explanation:

\omega_f = Final angular velocity

\omega_i = Initial angular velocity

\alpha = Angular acceleration

\theta = Angle of rotation

t = Time taken

Equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{4.5\times 2\pi-0}{0.95}\\\Rightarrow \alpha=29.76245\ rad/s^2

Angular acceleration during speed up is 29.76245 rad/s²

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-4.5\times 2\pi}{0.24}\\\Rightarrow \alpha=-117.80972\ rad/s^2

Angular acceleration during spin down is -117.80972 rad/s²

Angular speed is given by

\omega=2\pi 4.5=28.2743\ rad/s

Maximum angular speed reached by the flywheel is 28.2743 rad/s

\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=0\times t+\frac{1}{2}\times 29.76245\times 0.95^2\\\Rightarrow \theta=13.4303\ rad

\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow \theta=2\pi 4.5\times 0.24+\frac{1}{2}\times -117.80972\times 0.24^2\\\Rightarrow \theta=3.39292\ rad

The ratio would be \dfrac{13.4303}{3.39292}=3.95833

3 0
3 years ago
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