A person who weighs 614 N is riding a 85-N mountain bike. Suppose the entire weight of the rider plus bike is supported equally
1 answer:
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Answer:
(A) 9.5 m/s
(B) 5.225 m
Explanation:
vertical height (h) = 4.7 m
horizontal distance (d) = 9.3 m
acceleration due to gravity (g) = 9.8 m/s^{2}
initial speed of the fish (u) = 0 m/s
(A) what is the pelicans initial speed ?
- lets first calculate the time it took the fish to fall
s = ut +
since u = 0
s =
t =
where a = acceleration due to gravity and s = vertical height
t = = 0.98 s
- pelicans initial speed = speed of the fish
speed of the fish = distance / time = 9.3 / 0.98 = 9.5 m/s
initial speed of the pelican = 9.5 m/s
(B) If the pelican was traveling at the same speed but was only 1.5 m above the water, how far would the fish travel horizontally before hitting the water below?
vertical height = 1.5 m
pelican's speed = 9.5 m/s
- lets also calculate the time it will take the fish to fall
s = ut +
since u = 0
s =
t =
where a = acceleration due to gravity and s = vertical height
t = = 0.55 s
distance traveled by the fish = speed x time = 9.5 x 0.55 = 5.225 m
Answer: Hello there!
We know this:
The distance between the cars at t= 0 is D.
car 2 has an initial velocity of v0 and no acceleration.
car 1 has no initial velocity and a acceleration of ax that starts at t = 0
then we could obtain the acceleration of the car 1 by integrating the acceleration over the time; this is v(t) = ax*t where there is not a constant of integration because the car 1 has no initial velocity.
Because the cars are moving against each other, we want to se at what time t they meet, this is equivalent to see:
position of car 1 + position of car 2 = D
and in this way we could ignore constants of integration :D
for the position of each car we integrate again:
P1(t) = (1/2)ax*t^2 and P2(t) = v0t
v0t + (1/2)ax*t^2 = D
v0t + (1/2)ax*t^2 - D = 0
now we can solve it for t using the Bhaskara equation.
that we cant solve witout knowing the values for v0, D and ax. But you could replace them in that equation and obtain the time, where you must remember that you need to choose the positive solution (because this quadratic equation has two solutions).
Now we want to know the velocity of car 1 just before the impact, this can be calculated by valuating the time in the as the time that we just found in the velocity equation for the car 1, this is:
where again, you need to replace the values of v0, D and ax.
Explanation:
I remember that notation! The expression
is the 1st law of thermodynamics and it refers to the heat supplied to the system dQ which is also a change in its internal energy dU. The first term is the <u>partial</u> derivative of the internal energy U with respect to temperature T while the volume V is kept constant, as denoted by the subscript V. The 2nd term is similar but this time, temperature is kept constant while its volume partial derivative is being taken.
Ah, memories!