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3 years ago

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An object mass 5kg, moving at a velocity of 10metre per seconds is suddenly heat by a force of 2N for a time 3sec. Find its new

velocity motion,, this question is under newton laws of motion..

1 answer:

marin [14]3 years ago

7 0

Answer:

11.2 m/s

Explanation:

Newton's second law:

F = ma

F = m Δv / Δt

2 N = (5 kg) (v − 10 m/s) / 3 s

v = 11.2 m/s

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stellarik [79]

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Explanation:

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3 years ago

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3 years ago

Read 2 more answers

As a result of friction, the angular speed of a wheel changes with time according to dθ/dt = ω0e^−σt where ω0 and σ are constant

NNADVOKAT [17]

Hi there!

a.

We can use the initial conditions to solve for w₀.

It is given that:

$\frac{d\theta}{dt} = w_0e^{-\sigma t}$

We are given that at t = 0, ω = 3.7 rad/sec. We can plug this into the equation:

$\omega(0)= \omega_0e^{-\sigma (0)}\\\\3.7 = \omega_0 (1)\\\\\omega_0 = \boxed{3.7 rad/sec}$

Now, we can solve for sigma using the other given condition:

$2 = 3.7e^{-\sigma (8.6)}\\\\.541 = e^{-\sigma (8.6)}\\\\ln(.541) = -\sigma (8.6)\\\\\sigma = \frac{ln(.541)}{-8.6} = \boxed{0.0714s^{-1}}$

b.

The angular acceleration is the DERIVATIVE of the angular velocity function, so:

$\alpha(t) = \frac{d\omega}{dt} = -\sigma\omega_0e^{-\sigma t}\\\\\alpha(t) = -(0.0714)(3.7)e^{-(0.0714) (3)}\\\\\alpha(t) = \boxed{-0.213 rad\sec^2}$

c.

The angular displacement is the INTEGRAL of the angular velocity function.

$\theta (t) = \int\limits^{t_2}_{t_1} {\omega(t)} \, dt\\\\\theta(t) = \int\limits^{2.5}_{0} {\omega_0e^{-\sigma t}dt\\\\$

$\theta(t) = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=2.5} \atop {t_1=0}} \right.$

$\theta = -\frac{3.7}{0.0714}e^{-0.0714 t}\left \| {{t_2=2.5} \atop {t_1=0}} \right. \\\\\theta= -\frac{3.7}{0.0714}e^{-0.0714 (2.5)} + \frac{3.7}{0.0714}e^{-0.0714 (0)}$

$\theta = 8.471 rad$

Convert this to rev:

$8.471 rad * \frac{1 rev}{2\pi rad} = \boxed{1.348 rev}$

d.

We can begin by solving for the time necessary for the angular speed to reach 0 rad/sec.

$0 = 3.7e^{-0.0714t}\\\\t = \infty$

Evaluate the improper integral:

$\theta = \int\limits^{\infty}_{0} {\omega_0e^{-\sigma t}dt\\\\$

$\lim_{a \to \infty} \theta = -\frac{\omega_0}{\sigma}e^{-\sigma t}\left \| {{t_2=a} \atop {t_1=0}} \right.$

$\lim_{a \to \infty} \theta = -\frac{3.7}{0.0714}e^{-0.0714a} + \frac{3.7}{0.0714}e^{-0.0714(0)}\\\\ \lim_{a \to \infty} \theta = \frac{3.7}{0.0714}(1) = 51.82 rad$

Convert to rev:

$51.82 rad * \frac{1rev}{2\pi rad} = \boxed{8.25 rev}$

8 0

3 years ago

PLS HELP FOR THIS PHYSICS TEST <br><br> (pls only answer if u rlly know, this is an important test)

lesya692 [45]

Answer:

C. 8.01 m/s²

Explanation:

vf²= vi² + 2 • a • d

2ad = vf² - vi²

a = (vf²- vi²)/2d

d=25.00 -5.00=20.00 m

vi =0

vf=17.90 m/s

a =(17.90² -0²)/(2*20) = 8.01 m/s²

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4 0

3 years ago

Can you help me with this?

Natasha2012 [34]

Answer:

no

Explanation:

so basically I am domb so I can not help you

4 0

3 years ago