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Fynjy0 [20]
4 years ago
6

Suppose the initial position of an object is zero, the starting velocity is 3 m/s and the final velocity was 10 m/s. The object

moves with constant acceleration. Which part of a velocity vs. time graph can be used to calculate the displacement of the object?
the area of the rectangle under the line
the area of the rectangle above the line
the area of the rectangle plus the area of the triangle under the line
the area of the rectangle plus the area of the triangle above the line
Physics
1 answer:
svlad2 [7]4 years ago
5 0

Answer:

The area of the rectangle plus the area of the triangle under the line

Explanation:

In this case, the object moves with constant acceleration. Initial speed of the object is 0.

Equation of kinematics :

v = u +at

v = at

v is directly proportional to time. If means that the graph is straight line passing through origin. We know that area under v-t graph gives displacement. So, the correct option is (c).

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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}

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3 years ago
What’s the most distant object in the solar system
Scrat [10]

Answer:

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Explanation:

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What does the index of refraction directly measure? the angle between the incident ray and the normal line the angle between the
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<span>The bending of light in a medium</span>
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The top of a tower much like the leaning bell tower at Pisa, Italy, moves toward the south at an average rate of 1.4 mm/y. The t
Gemiola [76]

Answer:

\omega=7.16*10^{-13}\frac{rad}{s}

Explanation:

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\omega=\frac{v}{r}

Here v is the linear speed and r is the radius of the circular motion. The height of the tower is equal to the radius of the circular motion of the top of the tower, since is rotating about its base. We need to convert the given linear speed to \frac{m}{s}:

1.4\frac{mm}{y}*\frac{10^{-3}m}{1mm}*\frac{1y}{3.154*10^7s}=4.44*10^{-11}\frac{m}{s}

Now, we calculate the angular speed:

\omega=\frac{4.44*10^{-11}\frac{m}{s}}{62m}\\\omega=7.16*10^{-13}\frac{rad}{s}

8 0
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It was once recorded that a Jaguar left skid marks that were 290 m in length. Assuming that the Jaguar had a constant accelerati
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