Answer:
The time taken by the rock to reach the ground is 0.569 seconds.
Explanation:
Given that,
A student throws a rock horizontally off a 5.0 m tall building, s = 5 m
The initial speed of the rock, u = 6 m/s
We need to find the time taken by the rock to reach the ground. Using second equation of motion to find it. We get :

So, the time taken by the rock to reach the ground is 0.569 seconds. Hence, this is the required solution.
To solve the problem we will simply perform equivalence between both expressions. We will proceed to place your units and develop your internal operations in case there is any. From there we will compare and look at its consistency


At the same time we have that



Therefore there is not have same units and both are not consistent and the correct answer is B.
Answer:
11.09 m/s
Explanation:
Given that an object is thrown vertically up and attains an upward velocity of 9.6 m/s when it reaches one fourth of its maximum height above its launch point.
The parameters given are:
Initial velocity U = ?
Final velocity V = 9.6 m/s
Acceleration due to gravity g = 9.8m/s^2
Let first assume that the object is thrown from rest with the velocity U, at maximum height V = 0
Using third equation of motion
V^2 = U^2 - 2gH
0 = U^2 - 2 × 9.8H
U^2 = 19.6H ........ (1)
Using the formula again for one fourth of its maximum height
9.6^2 = U^2 - 2 × 9.8 × H/4
92.16 = U^2 - 19.6/4H
92.16 = U^2 - 4.9H
U^2 = 92.16 + 4.9H ...... (2)
Substitute U^2 in equation (1) into equation (2)
19.6H = 92.16 + 4.9H
Collect the like terms
19.6H - 4.9H = 92.16
14.7H = 92.16
H = 92.16/14.7
H = 6.269
Substitute H into equation 2
U^2 = 92.16 + 4.9( 6.269)
U^2 = 92.16 + 30.72
U^2 = 122.88
U = 11.09 m/s
Therefore, the initial velocity of the object is 11.09 m/s
Answer:
Explanation:
An example of an intense aerobic activity would be running/ sprinting sprinting targets six specific muscle groups: hamstrings, quadriceps, glutes, hips, abdominals and calves. Sprinting is a total body workout featuring short, high-intensity repetitions and long, easy recoveries.
Answer:Surface wave
Explanation:
Longitudinal wave is perpendicular to the wave motion and transverse wave is the same as the wave motion