Answer:
14,700 N
Explanation:
The hyppo is standing completely submerged on the bottom of the lake. Since it is still, it means that the net force acting on it is zero: so, the weight of the hyppo (W), pushing downward, is balanced by the upward normal force, N:
(1)
the weight of the hyppo is
where m is the hyppo's mass and g is the gravitational acceleration; therefore, solving eq.(1) for N, we find
Answer: 11,100 ft/s^2
1) Constant acceleration=> uniformly accelerated motion.
2) Formula for uniformly accelerated motion:
Vf = Vo + at
3) Data:
Vo = 1,100 ft/s
a = 1,000 ft/s^2
t = 10.0 s
4) Solution:
Vf = 1,100 ft/s + 1,000 ft/s^2 * 10.0 s = 1,100 ft/s + 10,000 ft/s
Vf = 11,100 ft/s
The time taken to hit the ground is 3.9 s, the range is 18m and the final velocity is 42.82 m/s
<h3>
Motion Under Gravity</h3>
The motion of an object under gravity is the vertical motion of the object under the influence of acceleration due to gravity.
Given that a ball is thrown horizontally from the roof of a building 75 m tall with a speed of 4.6 m/s.
a. how much later does the ball hit the ground?
The time can be calculated by considering the vertical component of the motion with the use of formula below.
h = ut + 1/2gt²
Where
- Initial velocity u = 0 ( vertical velocity )
- Acceleration due to gravity g = 9.8 m/s²
Substitute all the parameters into the formula
75 = 0 + 1/2 × 9.8 × t²
75 = 4.9t²
t² = 75/4.9
t² = 15.30
t = √15.3
t = 3.9 s
b. how far from the building will it land?
The range can be found by using the formula
R = ut
Where u = 4.6 m/s ( horizontal velocity )
R = 4.6 × 3.9
R = 18 m
c. what is the velocity of the ball just before it hits the ground?
The final velocity will be
v = u + gt
v = 4.6 + 9.8 × 3.9
v = 4.6 + 38.22
v = 42.82 m/s
Therefore, the answers are 3.9 s, 18 m and 42.82 m/s
Learn more about Vertical motion here: brainly.com/question/24230984
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