Answer:
141.78 ft
Explanation:
When speed, u = 44mi/h, minimum stopping distance, s = 44 ft = 0.00833 mi.
Calculating the acceleration using one of Newton's equations of motion:
Note: The negative sign denotes deceleration.
When speed, v = 79mi/h, the acceleration is equal to when it is 44mi/h i.e. -116206.48 mi/h^2
Hence, we can find the minimum stopping distance using:
The minimum stopping distance is 141.78 ft.
Explanation:
The moment of inertia of each disk is:
Idisk = 1/2 MR²
Using parallel axis theorem, the moment of inertia of each rod is:
Irod = 1/2 mr² + m (R − r)²
The total moment of inertia is:
I = 2Idisk + 5Irod
I = 2 (1/2 MR²) + 5 [1/2 mr² + m (R − r)²]
I = MR² + 5/2 mr² + 5m (R − r)²
Plugging in values:
I = (125 g) (5 cm)² + 5/2 (250 g) (1 cm)² + 5 (250 g) (5 cm − 1 cm)²
I = 23,750 g cm²
Answer:
0.09 x10^-10m
Explanation:
Using wavelength=( 12.27 A)/√V
= 12.27 x 10^-10/ √1.6x10^2
= 0.09x10^-10m
Since my givens are x = .550m [Vsub0] = unknown
[Asubx] = =9.80
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]
[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0])
Vsubx is the final velocity, which at the max height is 0, and Xsub0 is just 0 as that's where it starts so I just plug the rest in
0^2 = [Vsub0x]^2 + 2[-9.80]*(.550)
0 = [Vsub0x]^2 -10.78
10.78 = [Vsub0x]^2
Sqrt(10.78) = 3.28 m/s
Explanation:
Since, it is given that the magnet drops and falls lengthwise towards the canter of the ring. As a result, change in magnetic flux will occur which tends to induce an electric current in the ring.
Therefore, a magnetic field is also produced by the ring itself which will actually oppose or repel the magnet.
Thus, we can conclude that the falling magnet be repelled by the ring due to the magnetic interaction of the magnet and the ring.
