PLEASE HELP John is rollerblading down a long, straight path. At time zero, there is a mailbox about 1 m in front of him. In the
5 s time period that follows, John's velocity is given by the velocity versus time graph in the figure. Taking the mailbox to mark the zero location, with positions beyond the mailbox as positive, plot his position versus time in the given position versus time graph. John is rollerblading down a long, straight path. At time zero, there is a mailbox about 1 m in front of him. In the 5 s time period that follows, John's velocity is given by the velocity versus time graph in the figure. Taking the mailbox to mark the zero location, with positions beyond the mailbox as positive, plot his position versus time in the given position versus time graph. Assuming that all the numbers given are exact, what is John's position at a time of 4.79 s ? Enter your answer to at least three significant digits.
1 answer:
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Kepler's third law states that, for a planet orbiting around the Sun, the ratio between the cube of the radius of the orbit and the square of the orbital period is a constant:
(1)
where
r is the radius of the orbit
T is the period
G is the gravitational constant
M is the mass of the Sun
Let's convert the radius of the orbit (the distance between the Sun and Neptune) from AU to meters. We know that 1 AU corresponds to 150 million km, so
so the radius of the orbit is
And if we re-arrange the equation (1), we can find the orbital period of Neptune:
We can convert this value into years, to have a more meaningful number. To do that we must divide by 60 (number of seconds in 1 minute) by 60 (number of minutes in 1 hour) by 24 (number of hours in 1 day) by 365 (number of days in 1 year), and we get
where
r is the radius of the orbit
T is the period
G is the gravitational constant
M is the mass of the Sun
Let's convert the radius of the orbit (the distance between the Sun and Neptune) from AU to meters. We know that 1 AU corresponds to 150 million km, so
so the radius of the orbit is
And if we re-arrange the equation (1), we can find the orbital period of Neptune:
We can convert this value into years, to have a more meaningful number. To do that we must divide by 60 (number of seconds in 1 minute) by 60 (number of minutes in 1 hour) by 24 (number of hours in 1 day) by 365 (number of days in 1 year), and we get
Answer:
428.59 N
Explanation:
Buoyant force, where V is volume, g is gravitational constant and \rho is density
where
is upward force
where
is the density of hippo
Using g as 9.81
Therefore, the upward force=428.59 N
consider the right direction as positive and left direction as negative.
M = mass of the ball = 5 kg
m = mass of stone = 1.50 kg
= initial velocity of the ball before collision = 0 m/s
= initial velocity of the stone before collision = 12 m/s
= final velocity of the ball after collision = ?
= final velocity of the stone after collision = - 8.50 m/s
using conservation of momentum
M + m
= M
+ m
(5) (0) + (1.5) (12) = 5 + (1.50) (- 8.50)
= 6.15 m/s
h = height gained by the ball
using conservation of energy
Potential energy gained by ball at Top = kinetic energy at the bottom
Mgh = (0.5) M
(9.8) h = (0.5) (6.15)²
h = 1.93 m