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1 year ago

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A roundabout in a fairground requires an input power of 2.5 kW when operating at a constant angular velocity of 0.47 rad s–1 . (

a) Show that the frictional torque in the system is about 5 kN m. (3) 1. (b) When the power is switched off, the roundabout decelerates uniformly because the

1 answer:

natita [175]1 year ago

5 0

Answer:

Explanation:

Since the roundabout is rotating with uniform velocity ,

input power = frictional power

frictional power = 2.5 kW

frictional torque x angular velocity = 2.5 kW

frictional torque x .47 = 2.5 kW

frictional torque = 2.5 / .47 kN .m

= 5.32 kN . m

= 5 kN.m

b )

When power is switched off , it will decelerate because of frictional torque .

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On a horizontal, linear track lies a cart that has a fan attached to it. The mass of the cart plus fan is 364 g. The cart is pos

mamaluj [8]

Answer:

6.62s

Explanation:

Metric unit conversion:

364 g = 0.364 kg

789 g = 0.789 kg

Starting from rest, the cart takes 4.49 s to travel a distance of 1.43 m. We can use the following equation of motion to calculate the constant acceleration

$s = a_1t_1^2/2$

$a_1 = \frac{2s}{t_1^2} = \frac{2*1.43}{4.49^2} = 0.142 m/s^2$

Using Newton's 2nd law, we can calculate the force generated by the fan to push the 0.364 kg cart forward

$F = a_1m_1 = 0.142*0.364 = 0.052 N$

Now that more mass is added, the new acceleration of the 0.789 kg cart is

$a_2 = F/m_2 = 0.052 / 0.789 = 0.065 m/s^2$

We can reuse the same equation of motion to calculate the time it takes to travel 1.43 m from rest

$s = a_2t_2^2/2$

$t_2^2 = 2s/a_2 = 2*1.43/0.065 = 43.7$

$t_2 = \sqrt{43.7} = 6.62s$

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2 years ago

Read 2 more answers

8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s

allsm [11]

Answer:

a) $\omega = 50\,\frac{rad}{s}$ , b) $\omega = 0\,\frac{rad}{s}$

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

$\tau = F \cdot r$

$\tau = m\cdot a \cdot r$

$\tau = m \cdot \alpha \cdot r^{2}$

Where $\alpha$ is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

$\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}$

$\alpha = -3\,\frac{rad}{s^{2}}$

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

$\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)$

$\omega = 50\,\frac{rad}{s}$

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

$t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}$

$t = 66.667\,s$

Since $t > 66.667\,s$ , then the angular velocity is equal to zero. Therefore:

$\omega = 0\,\frac{rad}{s}$

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2 years ago

Some gliders are launched from the ground by means of a winch, which rapidly reels in a towing cable attached to the glider. Wha

aliina [53]

Answer:

P=627.47W

Explanation:

To solve this problem we have to take into account, that the work done by the winch is

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the force, at least must equal the gravitational force

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with force the tension in the cable makes the winch go up.

The work done is

$W=(1258.8N)(58.0m)=73010.4J$

To calculate the power we need to know what is the time t. But first we have to compute the acceleration

The acceleration will be

$v_f^2=v_0+2ah\\a=\frac{v_f^2}{2h}=\frac{(24.9\frac{m}{s})}{2(58.0m)}=0.214\frac{m}{s^2}$

and the time t

$v_f=v_0+at\\t=\frac{v_f}{a}=116.35s$

The power will be

$P=\frac{W}{t}=\frac{73010.4J}{116.35s}=627.47W$

HOPE THIS HELPS!!

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2 years ago

A penny falls from a windowsill which is 25.0 m above the sidewalk. How much time does a passerby on the sidewalk below have to

Yakvenalex [24]

T = √(h)/(0.5)(9.81)
t = √(25)/(4.905)
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hope this helps and have a great day :)

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2 years ago

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Makovka662 [10]

Answer:

the net force is 101587.5 N

Explanation:

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v = 30 m/s

The area of roof,

A = 175 m 2

The expression for the Bernoulli's theorem.

P = 12 ρv 2 ...... (1)

Here,

P is the pressure difference,

ρ is the density of air and

v is the speed of wind.

The expression for the pressure.

P = F A ..... (2)

Here,

F is the force and

A is the area of roof.

Part (a)

Substitute the values for the pressure difference in equation (1)

P = 12 × 1.29 × (30) 2 P = 580.5 Pa

Thus, the pressure difference at the roof between the inside and outside air is

580.5 Pa

Part (b)

Substitute the values for the net force in equation (2)

580.5 = F 175 F = 101587.5 N

Thus, the net force is 101587.5 N.

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2 years ago