Answer:
The phenomenon known as "tunneling" is one of the best-known predictions of quantum physics, because it so dramatically confounds our classical intuition for how objects ought to behave. If you create a narrow region of space that a particle would have to have a relatively high energy to enter, classical reasoning tells us that low-energy particles heading toward that region should reflect off the boundary with 100% probability. Instead, there is a tiny chance of finding those particles on the far side of the region, with no loss of energy. It's as if they simply evaded the "barrier" region by making a "tunnel" through it.
Explanation:
Before you start working on any motion problem, YOU decide which direction you're going to call 'positive'. Everybody almost always calls UP positive, and the acceleration of gravity points down, so it winds up negative. But you could just as well call DOWN the positive direction. Then, the cannonball is fired with a negative vertical speed, and the acceleration of gravity eventually robs all of its negative speed, and makes it start falling in the positive direction. The whole thing is your choice.
Answer: a) 139.4 μV; b) 129.6 μV
Explanation: In order to solve this problem we have to use the Ohm law given by:
V=R*I whre R= ρ *L/A where ρ;L and A are the resistivity, length and cross section of teh wire.
Then we have:
for cooper R=1.71 *10^-8* 1.8/(0.001628)^2= 11.61 * 10^-3Ω
and for silver R= 1.58 *10^-8* 1.8/(0.001628)^2=10.80 * 10^-3Ω
Finalle we calculate the potential difference (V) for both wires:
Vcooper=11.62* 10^-3* 12 * 10^-3=139.410^-6 V
V silver= 10.80 10^-3* 12 * 10^-3=129.6 10^-6 V
Answer:
0.3 m
Explanation:
Initially, the package has both gravitational potential energy and kinetic energy. The spring has elastic energy. After the package is brought to rest, all the energy is stored in the spring.
Initial energy = final energy
mgh + ½ mv² + ½ kx₁² = ½ kx₂²
Given:
m = 50 kg
g = 9.8 m/s²
h = 8 sin 20º m
v = 2 m/s
k = 30000 N/m
x₁ = 0.05 m
(50)(9.8)(8 sin 20) + ½ (50)(2)² + ½ (30000)(0.05)² = ½ (30000)x₂²
x₂ ≈ 0.314 m
So the spring is compressed 0.314 m from it's natural length. However, we're asked to find the additional deformation from the original 50mm.
x₂ − x₁
0.314 m − 0.05 m
0.264 m
Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.
Answer:
52 mm/s (approximately)
Explanation:
Given:
Initial speed of the projectile is, 
Angle of projection is, 
Time taken to land on the hill is, 
In a projectile motion, there is acceleration only in the vertical direction which is equal to acceleration due to gravity acting vertically downward. There is no acceleration in the horizontal direction.
So, the velocity in the horizontal direction always remains the same.
The horizontal component of initial velocity is given as:
Now, the velocity in the vertical direction goes on decreasing and becomes 0 at the highest point of the trajectory. So, at the highest point, only horizontal component acts.
Therefore, the projectile's velocity at the highest point of its trajectory is equal to the horizontal component of initial velocity and thus is equal to 52 mm/s.
