When a slice of buttered toast is accidentally pushed over the edge of a counter, it rotates as it falls. If the distance to the
floor is 52 cm and for rotation less than 1 rev, what are the (a) smallest and (b) largest angular speeds that cause the toast to hit and then topple to be butter-side down
1 answer:
Answer: 4.83rad/s and 14.50 rad/s
Explanation:
The distance from the center to the floor d = 52cm = 0.52m
Rotation is less than 1 rev
The toast rotates at a constant angular speed
Using kinematic d = Vit + 1/2gt^2
d = 0+1/2gt^2
t = (2d/g) square root
t = square root of 2× 0.52/ 9.8
t = 0.325s
The toast is accidentally pushed over the edge of the centre with butter side up, then the toast rotates as it falls . If the toast hits the ground and topples, the smallest angle will be less than 1/4 rev with correspondence to the smallest angular speed
Wmin = change in tetha/ change in time
= 0.25 rev/ change in t
= 0.25×2pii / change in t
= 0.5pii/0.325
= 0.5×3.142/0.325
= 4.83 rad/s
The toast is accidentally pushed over the edge of the centre with butter side up, then the toast rotates as it falls . If the toast hits the ground and topples, the maximum angle will be less than 3/4 rev with correspondence to the maximum angular speed
Wmax = change in tetha/ change in time
= 0.75 rev/ Change in time
= 0.75 ×2pii/change in tetha
= 1.5pii/0.325
= 1.5 ×3.142/0.325
= 14.50 rad/s
Note the following
Vi is the initial speed of the toast which is zero because it was initially at rest
t is the time taken for the toast to hit the floor
g is the gravitational acceleration
d is the distance between the counter and the floor. There is an attachment for you as well.
Thanks
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Complete Question
The complete question is shown on the first uploaded image
Answer:
a
b
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Explanation:
From the question we are told that
The refractive index of the core is
The refractive index of the cladding is
Generally according to Snell's law
Where is the largest angle a largest angle a ray will make with respect to the interface of the fiber and experience total internal reflection
Given from the question the the largest angle is 5°
Generally the refraction index of the cladding is mathematically represented as
Answer:
a) We kindly invite you to see the explanation and the image attached below.
b) The acceleration of the masses is 4.203 meters per square second.
c) The tension force in the cord is 28.02 newtons.
d) The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.
e) The final speed of the system is 3.551 meters per second.
Explanation:
a) At first we assume that pulley and cord are both ideal, that is, masses are negligible and include the free body diagrams of each mass and the pulley in the image attached below.
b) Both masses are connected to each other by the same cord, the direction of acceleration will be dominated by the mass of greater mass (mass A) and both masses have the same magnitude of acceleration. By the 2nd Newton's Law, we create the following equation of equilibrium:
Mass A
(1)
Mass B
(2)
Where:
- Tension force in the cord, measured in newtons.
,
- Masses of blocks A and B, measured in kilograms.
- Gravitational acceleration, measured in meters per square second.
- Net acceleration of the each block, measured in meters per square second.
By subtracting (2) by (1), we get an expression for the acceleration of each mass:
If we know that ,
and
, then the acceleration of the masses is:
The acceleration of the masses is 4.203 meters per square second.
c) From (2) we get the following expression for the tension force in the cord:
If we know that ,
and
, then the tension force in the cord:
The tension force in the cord is 28.02 newtons.
d) Given that system starts from rest and net acceleration is constant, we determine the time taken by the block to cover a distance of 1.5 meters through the following kinematic formula:
(3)
Where:
- Net acceleration, measured in meters per square second.
- Time, measured in seconds.
- Covered distance, measured in meters.
If we know that and
, then the time taken by the system is:
The system will take approximately 0.845 seconds to cover a distance of 1.5 meters.
e) The final speed of the system is calculated by the following formula:
(4)
Where is the final speed of the system, measured in meters per second.
If we know that and
, then the final speed of the system is:
The final speed of the system is 3.551 meters per second.
