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2 years ago

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an object moves 15.0 m north and then 11.0 m south. find both the distance it has traveled and the magnitude of its displacement

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1 answer:

Naddika [18.5K]2 years ago

7 0

Distance is 15 m + 11 m =26 m
Displacement is North 15 - 11 south= 4 m north.

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A solid sphere is released from the top of a ramp that is at a height

Alexus [3.1K]

Hi there!

We can use the work-energy theorem and apply it to this situation.

At the top of the ramp, the ball only has gravitational potential energy, and at the bottom of the ramp, the ball has BOTH translational and rotational kinetic energy.

We must use the following equations:
$GPE = mgh \\KE_T = \frac{1}{2}mv^2\\\\KE_R = \frac{1}{2}I \omega^2$

m = mass of sphere (kg)
g = acceleration due to gravity (m/s²)

h = height of ramp (m)

v = final velocity (m/s)
I = Moment of Inertia (kgm²)

ω = angular velocity (rad/sec)

Since:
$E_i = E_f\\\\mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega ^2$

In order to make things easier, since the ball is not slipping, we can relate angular velocity to translational velocity:
$\omega = \frac{v}{r}$

Also, recall the equation for the moment of inertia for a solid sphere:
$I = \frac{2}{5}mr^2$

We can use these to simply our equation:
$KE_R = \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = \frac{1}{5}mv^2$

Now, we can rewrite the equation and solve for 'v'.

$mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2\\\\mgh = \frac{7}{10}mv^2\\\\v = \sqrt{\frac{10gh}{7}} = \sqrt{\frac{10(9.8)(2.2 - 1.87)}{7}} = 2.149\frac{m}{s}$

a)

We can begin by solving for the time taken for the ball to land on the ground. The ball only has a horizontal velocity, so this is essentially a free-fall situation. Use the rearranged kinematic equation:
$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(1.87)}{9.8}} = .6178 s$

Now, use the following equation to solve for horizontal distance given horizontal velocity and time:
$d_x = v_x t\\\\d_x = 2.149 * .6178 = \boxed{1.328 m}$

b)
We can use the previously-stated relationship between translational and angular velocity to solve for the angular velocity.

$\omega = \frac{v}{r}$
It is given that the diameter is 0.14 m, so the radius is 1/2th the diameter, or 0.07 m.

Solve for the angular velocity:
$\omega = \frac{2.149}{0.07} = 30.706 \frac{rad}{sec}}$

Using the above fall time and dimensional analysis to convert from rad/sec to revolutions, we can solve for the # of revolutions made by the ball:
$\frac{30.706rad}{sec} * .6178 sec * \frac{1 rev}{2\pi rad} = \boxed{3.019 rev}$

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OverLord2011 [107]

Answer:

True

Explanation:

When no net force is applied to a moving object, it still comes to rest because of its inertia.

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3 years ago

A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,

viktelen [127]

Part A)
First of all, let's convert the radii of the inner and the outer sphere:
$r_A = 11.0 cm = 0.110 m$
$r_B = 16.5 cm=0.165 m$
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
$C=4 \pi \epsilon _0 \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=$
$=3.67\cdot 10^{-11}F$

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
$Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C$

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
$E\cdot (4 \pi r^2) = \frac{Q}{\epsilon _0}$
from which
$E(r) = \frac{Q}{4 \pi \epsilon_0 r^2}$
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
$E(0.111m)=2680 N/C$

And so, the energy density at r=0.111 m is
$U= \frac{1}{2} \epsilon _0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3$

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: $Q=3.67 \cdot 10^{-9}C$ . We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
$E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C$

And therefore, the energy density at this distance from the center is
$U= \frac{1}{2}\epsilon_0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3$

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3 years ago