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1 year ago

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Two wheels can rotate freely about fixed axles through their centers. The wheels have the same mass, but one has twice the radiu

s of the other. Forces F1 and F2 are applied as shown such that the angular acceleration of the two wheels is the same.
1)Compare the magnitudes of the two forces
F2 = F1
F2 = 2 F1
F2 = 4 F1

1 answer:

Murrr4er [49]1 year ago

7 0

Answer:

The force on second wheel is twice off the force on first wheel.

Explanation:

In this case, two wheels can rotate freely about fixed axles through their centers. We know that, in rotational mechanics, the torque is given by :

$\tau=I\alpha$

Also, $\tau=Fr$

And moment of inertia is, $I=mr^2$

It implies,

$Fr=mr^2\alpha \\\\F=mr\alpha$

Here, one has twice the radius of the other. Ratio of forces will be :

$\dfrac{F_1}{F_2}=\dfrac{mr_1\alpha }{mr_2\alpha }\\\\\dfrac{F_1}{F_2}=\dfrac{r_1 }{2r_1}\\\\\dfrac{F_1}{F_2}=\dfrac{1 }{2}\\\\F_2=2F_1$

So, the force on second wheel is twice off the force on first wheel.

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Answer:

torque is 1.7 * $10^{-2}$ Nm

Explanation:

Given data

turns n = 1000 turns

radius r = 12 cm

current I = 15A

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angle θ = 25°

to find out

the torque on the loop

solution

we know that torque on the loop is

torque = N* I* A*B* sinθ

here area A = πr² = π(0.12)²

put all value

torque = N* I* A*B* sinθ

torque = 1000* 15* π(0.12)² *5.8 x 10-5 * sin25

torque = 0.0166 N m

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2 years ago

A car drives at a constant speed around a banked circular track with a diameter of 136 m . The motion of the car can be describe

galina1969 [7]

Answer:

speed = 44.9m/s

x = 35.5 m, y = 58.0m

Explanation:

A car on a circular track with constant angular velocity ω can be described by the equation of position r:

$\overrightarrow {r(t)} = Rsin(\omega t)\hat{i} + Rcos(\omega t)\hat{j}$

The velocity v is given by:

$\overrightarrow {v(t)} = \overrightarrow{\frac{dr}{dt}}= \omega Rcos(\omega t)\hat{i} - \omega Rsin(\omega t)\hat{j}$

The acceleration a:

$\overrightarrow {a(t)} = \overrightarrow{\frac{dv}{dt}}= -\omega^2 Rsin(\omega t)\hat{i} - \omega^2 Rcos(\omega t)\hat{j}$

From the given values we get two equations:

$-\omega^2 Rsin(\omega t)=-15.4\\-\omega^2 Rcos(\omega t)=-25.4$

We also know:

$\overrightarrow {a(t)} = -\omega^2 Rsin(\omega t)\hat{i} - \omega^2 Rcos(\omega t)\hat{j}=-\omega^2\overrightarrow{r(t)}$

The magnitude of the acceleration a is:

$a=\sqrt{(-15.4)^2+(-25.4)^2}=29.7$

The magnitude of position r is:

$r=R=68m$

Plugging in to the equation for a(t):

$\overrightarrow {a(t)} =-\omega^2\overrightarrow{r(t)}$

and solving for ω:

$|\omega|=0.66$

Now solve for time t:

$\frac{sin(0.66t)}{cos(0.66t)}=tan(0.66t)=\frac{15.4}{25.4}\\t=0.83$

Using the calculated values to compute v(t):

$\overrightarrow {v(t)}= \omega Rcos(\omega t)\hat{i} - \omega Rsin(\omega t)\hat{j}\\\overrightarrow {v(t)}=44.88cos(0.55)\hai{i}-44.88sin(0.55)\hat{j}\\\overrightarrow {v(t)}=38.3\hat{i}-23.5\hat{j}$

The speed of the car is:

$\sqrt{38.3^2 + (-23.5)^2} = 44.9$

The position r:

$\overrightarrow {r(t)} = Rsin(\omega t)\hat{i} + Rcos(\omega t)\hat{j}\\\overrightarrow {r(t)} = 68sin(0.55)\hat{i} + 68cos(0.55)\hat{j}\\\overrightarrow {r(t)} = 35.5{i} + 58.0\hat{j}$

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2 years ago

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. A girl runs and jumps horizontally off a platform 10m above a pool with a speed of 4.0m/s. As soon as she leaves the platform,

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Answer:

2.39 revolutions

Explanation:

As she jumps off the platform horizontally at a speed of 10m/s, the gravity is the only thing that affects her motion vertically. Let g = 10m/s2, the time it takes for her to fall 10m to water is

$h = gt^2/2$

$10 = 10t^2/2$

$t^2 = 2$

$t = \sqrt{2} = 1.414 s$

Knowing the time it takes to fall to the pool, we calculate the angular distance that she would make at a constant acceleration of 15 rad/s2:

$\theta = \alpha t^2/2$

$\theta = 15 * 2/2 = 15 rad$

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Answer:

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Explanation:

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v = c/n

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v = 239 833 966 m/s

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