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Deffense [45]
3 years ago
11

A ladybug sits 10.8 cm from the center of a beatles music album spinning at 33.33 rpm. What is the maximum velocity (in m/s) of

its shadow on the wall behind the turntable, if illuminated parallel to the record by the parallel rays of the setting sun?
Physics
2 answers:
snow_lady [41]3 years ago
7 0

Answer:

The maximum velocity is 0.377 m/s

Explanation:

Please, the solution is in the Word file attached

Download docx
trapecia [35]3 years ago
5 0

solution;

The max^m liner velocity v_{max}=r_{w}\\here r is the radius of circular track first convert rpm to radian pedr second\\1rpm=\frac{2\pi }{60}rad/s\\33.33rpm=\frac{2\pi }{60}\times33.33rad/s\\3.49rad/s\\substitute 10.8cm for r and 3.49 ras/s for w\\v_{max}=(0.108m)(3.49rad/s)\\0.377m/s\\hence, the max^m velocity of the shadow of the ladybug is 0.377m/s

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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}

7 0
3 years ago
A VW Beetle goes from 0 to 60.0 mi/h with an acceleration of +2.35 m/s^2. (a) How much time does it take for the beetle to reach
Firlakuza [10]
Im going to tell you what to do but not the result. So pay close attention: the first thing you need to do is convert miles/h to m/s. Then for the part a) <span>divide the final velocity by the initial velocity. That will give you the amount of it will take to accelerate to the final velocity.Now for the part b you </span>use the formula v=vo+at. I hope this can help you
8 0
3 years ago
car is moving at 40 m/s. At 10 meters the driver spots a deer on the road and instantly steps on the brakes. If the car is 400 k
Mice21 [21]

Answer:

32000 N

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 40 m/s

Distance (s) = 10 m

Final velocity (v) = 0 m/s

Mass (m) of car = 400 Kg

Force (F) =?

Next, we shall determine the acceleration of the the car. This can be obtained as follow:

Initial velocity (u) = 40 m/s

Distance (s) = 10 m

Final velocity (v) = 0 m/s

Acceleration (a) =?

v² = u² + 2as

0² = 40² + (2 × a × 10)

0 = 1600 + 20a

Collect like terms

0 – 1600 = 20a

–1600 = 20a

Divide both side by –1600

a = –1600 / 20

a = –80 m/s²

The negative sign indicate that the car is decelerating i.e coming to rest.

Finally, we shall determine the force needed to stop the car. This can be obtained as follow:

Mass (m) of car = 400 Kg

Acceleration (a) = –80 m/s²

Force (F) =?

F = ma

F = 400 × –80

F = – 32000 N

NOTE: The negative sign indicate that the force is in opposite direction to the motion of the car.

7 0
2 years ago
How does the amount of friction affect the sum of forces
iren [92.7K]

Answer:

<h3>Newton's 2nd law states acceleration is proportional to the net force acting on an object. The net force is the vector sum of all the forces applied to the object. ... In this case the acceleration (slowing down) of the puck is proportional to the amount of friction.</h3>

Explanation:

<h3>mark as brainliast</h3>

6 0
2 years ago
A rock thrown straight up takes 4.2 s to reach its maximum height what was its initial velocity
liubo4ka [24]

Consider the upward direction of motion as positive and downward direction of motion as negative.

a = acceleration due to gravity in downward direction = - 9.8 \frac{m }{s^{2}}

v₀ = initial velocity of rock in upward direction = ?

v = final velocity of rock at the highest point = 0 \frac{m }{s}

t = time to reach the maximum height = 4.2 sec

Using the kinematics equation

v = v₀ + a t

inserting the values

0 = v₀ + (- 9.8) (4.2)

v₀ = 41.2 \frac{m }{s}


8 0
3 years ago
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