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2 years ago

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Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of

22 N is applied tangentially to a sprocket of radius 6 cm for 6 s, what angular speed does the wheel achieve, assuming it rolls without slipping?

1 answer:

vredina [299]2 years ago

7 0

Answer:

The angular speed after 6s is $\omega = 1466.67s^{-1}$ .

Explanation:

The equation

$I\alpha = Fd$

relates the moment of inertia $I$ of a rigid body, and its angular acceleration $\alpha$ , with the force applied $F$ at a distance $d$ from the axis of rotation.

In our case, the force applied is $F = 22N$ , at a distance $d = 6cm =0.06m$ , to a ring with the moment of inertia of $I =mr^2$ ; therefore, the angular acceleration is

$\alpha =\frac{Fd}{I}$

$\alpha =\frac{22N*0.06m}{(1.5kg)*(0.06)^2}$

$\alpha = 244.44\: s^{-2}$

Therefore, the angular speed $\omega$ which is

$\omega = \alpha t$

after 6 seconds is

$\omega = 244.44$\: s^{-2}* 6s$

$\boxed{\omega = 1466.67s^{-1}}$

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If the cold temperature reservoir of a Carnot engine is held at a constant 306 K, what temperature should the hot reservoir be k

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Efficiency η of a Carnot engine is defined to be:
<span>η = 1 - Tc / Th = (Th - Tc) / Th </span>
<span>where </span>
<span>Tc is the absolute temperature of the cold reservoir, and </span>
<span>Th is the absolute temperature of the hot reservoir. </span>

<span>In this case, given is η=22% and Th - Tc = 75K </span>
<span>Notice that although temperature difference is given in °C it has same numerical value in Kelvins because magnitude of the degree Celsius is exactly equal to that of the Kelvin (the difference between two scales is only in their starting points). </span>

<span>Th = (Th - Tc) / η </span>
<span>Th = 75 / 0.22 = 341 K (rounded to closest number) </span>
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<span>Lower temperature is Tc = 266 K </span>
<span>Higher temperature is Th = 341 K</span>

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2 years ago

In the absence of air resistance, at what other angle will a thrown ball go the same distance as one thrown at an angle of 75 de

snow_tiger [21]

As we know that range of the projectile motion is given by

$R = \frac{v^2 sin(2\theta)}{g}$

here we know that range will be same for two different angles

so here we can say the two angle must be complementary angles

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$\theta, 90 - \theta$

so it is given that one of the projection angle is 75 degree

so other angle for same range must be 90 - 75 = 15 degree

so other projection angle must be 15 degree

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A student claims that earthquakes are more destructive than tsunamis. In one or two sentences, explain at least two reasons that

Klio2033 [76]

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When a 75.0-kg man slowly adds his weight to a vertical spring attached to the ceiling, he reaches equilibrium when the spring i

Sergio039 [100]

Answer:

1) $k=11.319kN/m$

2)displacement $=13.02cm$

3) $k_{eq}=5.65kN/m$

Explanation:

At equilibrium position the weight of the man should be balanced by force in the spring

thus we have at equilibrium

$kx=mg\\\\k=\frac{mg}{x}$

Applying values we get

$k=\frac{75\times 9.81}{0.065}\\\\k=11.319kN/m$

2)

When we add another identical spring we get an equivalent spring with spring constant as

$\frac{1}{k_{eq}}=\frac{1}{k_1}+\frac{1}{k_2}$

Applying values we get

$\frac{1}{k_{eq}}=\frac{1}{11.319}+\frac{1}{11.319}\\\\k_{eq}=5.65kN/m$

Thus at equilibrium we have

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Two speedboats are traveling at the same speed relative to the water in opposite directions in a moving river. An observer on th

DENIUS [597]

Answer:

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vb₁s = vb₁w + vrs

In one case, the boat is moving in the same direction as the water:

vb₁s = vb₁w + vrs = 7.0 m/s (1)

For the other boat, it is clear that is moving in an opposite direction:

vb₂s = vb₂w - vrs = 4.9 m/s (2)

As we know that vb₁w = vb₂w, adding both sides, we can remove the river speed from the equation, as follows:

vb₁w = vb₂w = $\frac{7.0 m/s + 4.9 m/s}{2} =5.95 m/s$

b) Replacing this value in (1) and solving for vriver, we have:

vriver = 7.0 m/s - 5.95 m/s = 1.05 m/s

(we could have arrived to the same result subtracting both sides in (1), and (2))

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