Answer:
Increasing speed.
Explanation:
In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.
This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.
Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.
Mathematically, acceleration is given by the equation;
In this scenario, an object moves with a positive acceleration. Thus, the object is moving with an increasing speed and as such it has acceleration in the same direction as its velocity with respect to time.
Answer: the volleyball
Explanation:
The volleyball, in contrast, is full of air, which contains fewer, more widely spaced particles of matter. In other words, the matter inside the bowling ball is denser than the matter inside the volleyball. A bowling ball is denser than a volleyball.
Q: A rock is thrown off of a 100 foot cliff with an upward velocity of 45 m/s. As a result its height after t seconds is given by the formula:
h(t)=100+45t−4.9t2
(a)
What is its height after 3 seconds?
(b)What is its velocity after 3 seconds?
Answer:
(a) 190.9 m.
(b) 15.6 m/s upward
Explanation:
Given:
h(t) = 100 + 45t - 4.9t²
The height after 3 seconds,
t = 3 s
Substitute the value of t in to the equation above.
h(3) = 100+45(3)-4.9(3)²
h(3) = 100+135-44.1
h(3) = 190.9 m
Therefore the height after 3 seconds = 190.9 m.
(b) Velocity after 3 seconds
The velocity is obtained by differentiating h(t) with respect to time
v = dh(t)/dt
dh(t)/dt = 45-9.8t
v = 45 - 9.8t ......................................... Equation 1
t = 3 s.
Substitute the value of t into the equation above,
v = 45 - 9.8(3)
v = 45- 29.4
v = 15.6 m/s
Thus the velocity after 3 seconds = 15.6 m/s upward
Answer:
V=22.4m/s;T=2.29s
Explanation:
We will use two formulas in order to solve this problem. To determine the velocity at the bottom we can use potential and kinetic energy to solve for the velocity and use the uniformly accelerated displacement formula:
Solving for velocity using equation 1:
Solving for time in equation 2:
Heya user☺☺
All options are wrong here.
The correct answer is..
Work/Time.
Hope this will help☺☺