The answer would be Gravity.
Gravity is pulling the weight down, which is pulling the car up the ramp.
Answer:
Δu=1300kJ/kg
Explanation:
Energy at the initial state
Is saturated vapor at initial pressure we have
Process 2-3 is a constant volume process
The overall in internal energy
Δu=u₁-u₃
We replace the values in equation
Δu=u₁-u₃
Δu=1300kJ/kg
Answer:
Two stars, each of mass M, form a binary system. ... used is the distance between the centers of the planets; here that distance is 2R. ... r appears in the denominator of Newton's law of gravitation, the force of ... The orbital speed of a satellite orbiting the earth in a circular orbit at the ... is undergoing uniform circular motion?
Explanation:
Two stars, each of mass M, form a binary system. ... used is the distance between the centers of the planets; here that distance is 2R. ... r appears in the denominator of Newton's law of gravitation, the force of ... The orbital speed of a satellite orbiting the earth in a circular orbit at the ... is undergoing uniform circular motion?
(a) 3.56 m/s
(b) 11 - 3.72a
(c) t = 5.9 s
(d) -11 m/s
For most of these problems, you're being asked the velocity of the rock as a function of t, while you've been given the position as a function of t. So first calculate the first derivative of the position function using the power rule.
y = 11t - 1.86t^2
y' = 11 - 3.72t
Now that you have the first derivative, it will give you the velocity as a function of t.
(a) Velocity after 2 seconds.
y' = 11 - 3.72t
y' = 11 - 3.72*2 = 11 - 7.44 = 3.56
So the velocity is 3.56 m/s
(b) Velocity after a seconds.
y' = 11 - 3.72t
y' = 11 - 3.72a
So the answer is 11 - 3.72a
(c) Use the quadratic formula to find the zeros for the position function y = 11t-1.86t^2. Roots are t = 0 and t = 5.913978495. The t = 0 is for the moment the rock was thrown, so the answer is t = 5.9 seconds.
(d) Plug in the value of t calculated for (c) into the velocity function, so:
y' = 11 - 3.72a
y' = 11 - 3.72*5.913978495
y' = 11 - 22
y' = -11
So the velocity is -11 m/s which makes sense since the total energy of the rock will remain constant, so it's coming down at the same speed as it was going up.
Do you remember this formula for the distance traveled while accelerated ?
<u>Distance = (initial speed) x (t) plus (1/2) x (acceleration) x (t²)</u>
I think this is exactly what we need for this problem.
initial speed = 20 m/s down
acceleration = 9.81 m/s² down
t = 3.0 seconds
Distance down = (20) x (3) plus (1/2) x (9.81) x (3)²
Distance = (60) plus (4.905) x (9)
Distance = (60) plus (44.145) = 104.145 meters
Choice <em>D)</em> is the closest one.
