Answer:
0.4778 m/s
Explanation:
To solve this question, we will make use of law of conservation of momentum.
We are given that the rock's velocity is 12 m/s at 35°. Thus, the horizontal component of this velocity is;
V_x = (12 m/s)(cos(35°)) = 9.83 m/s.
Thus, the horizontal component of the rock's momentum is;
(3.5 kg)(9.83 m/s) = 34.405 kg·m/s.
Since the person is not pushed up off the ice or down into it, his momentum will have no vertical component and so his momentum will have the same magnitude as the horizontal component of the rock's momentum.
Thus, to get the person's speed, we know that; momentum = mass x velocity
Mass of person = 72 kg and we have momentum as 34.405 kg·m/s
Thus;
34.405 = 72 x velocity
Velocity = 34.405/72
Velocity = 0.4778 m/s
Answer:
v = 7.4 m/s
Explanation:
Given that,
Mass if a volleyball, m = 5 kg
The ball reaches a height of 2.8 m
We need to find how fast the ball is going as it bumped into the air. Ket the velocity is v. Using the conservation of energy to find it as follows :
So, the required speed is 7.4 m/s. Hence, the correct option is (b).
Answer: Normal fault
Explanation:
The type of fault that is explained above is a normal fault. We should note that normal faults typically takes place in a divergent boundary in a scenario where the crusts may have been pulled apart.
Since the crust is pulled apart in this case, it leads to the downward movement of the hanging wall which leads to the football being above the hanging wall.
Explanation:
Given that,
Work done to stretch the spring, W = 130 J
Distance, x = 0.1 m
(a) We know that work done in stretching the spring is as follows :
(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m
So,
So, the new work is more than 130 J.
