Answer:B
Explanation:
For the ball dropped From a tall ball building the direction of acceleration is downward and its magnitude is
For the ball thrown upward towards the dropped ball the direction and magnitude of acceleration is same i.e. downwards as gravity always acting downward with constant magnitude.
Thus option B is correct
If time is the x axis and distance is the y axis then yes, in the case that time is going by but distance remains the same.
Answer: 0.687 m/s
Explanation:
Given
Mass of object, m = 76.7 g = 0.767 kg
Force constant of the spring, k = 3.34 N/m
Amplitude of the spring, x = 38 cm = 0.38 m
EPE(i) = KE(f) + EPE(f)
1/2kx(i)² = 1/2mv² + 1/2kx(f)²
1/2 * 3.34 * 0.38² = (1/2 * 0.767 * v²) + [1/2 * 3.34 * (0.38/2)²]
1/2 * 3.34 * 0.1444 = (1/2 * 0.767v²) + 1/2 * 3.34 * 0.0361
1/2 * 0.482 = 1/2 * 0.767v² + 1/2 * 0.121
0.241 = 1/2 * 0.767v² + 0.06
1/2 * 0.767v² = 0.181
0.767v² = 0.181 * 2
0.767v² = 0.362
v² = 0.362 / 0.767
v² = 0.472
v = √0.472
v = 0.687 m/s
Answer:
The smallest possibility is 0.01E-22kgm/s
Explanation:
Using
Momentum= h/4πx
= 6.6x 10^-34Js/ 4(3.142* 50*10-12m)
= 0.01*10^-22kgm/s