Answer:
a) V_f = 25.514 m/s
b) Q =53.46 degrees CCW from + x-axis
Explanation:
Given:
- Initial speed V_i = 20.5 j m/s
- Acceleration a = 0.31 i m/s^2
- Time duration for acceleration t = 49.0 s
Find:
(a) What is the magnitude of the satellite's velocity when the thruster turns off?
(b) What is the direction of the satellite's velocity when the thruster turns off? Give your answer as an angle measured counterclockwise from the +x-axis.
Solution:
- We can apply the kinematic equation of motion for our problem assuming a constant acceleration as given:
V_f = V_i + a*t
V_f = 20.5 j + 0.31 i *49
V_f = 20.5 j + 15.19 i
- The magnitude of the velocity vector is given by:
V_f = sqrt ( 20.5^2 + 15.19^2)
V_f = sqrt(650.9861)
V_f = 25.514 m/s
- The direction of the velocity vector can be computed by using x and y components of velocity found above:
tan(Q) = (V_y / V_x)
Q = arctan (20.5 / 15.19)
Q =53.46 degrees
- The velocity vector is at angle @ 53.46 degrees CCW from the positive x-axis.
Energy to lift something =
(mass of the object) x (gravity) x (height of the lift).
BUT ...
This simple formula only works if you use the right units.
Mass . . . kilograms
Gravity . . . meters/second²
Height . . . meters
For this question . . .
Mass = 55 megagram = 5.5 x 10⁷ grams = 5.5 x 10⁴ kilograms
Gravity (on Earth) = 9.8 m/second²
Height = 500 cm = 5.0 meters
So we have ...
Energy = (5.5 x 10⁴ kilogram) x (9.8 m/s²) x (5 m)
= 2,696,925 joules .
That's quite a large amount of energy ... equivalent to
straining at the rate of 1 horsepower for almost exactly an
hour, or burning a 100 watt light bulb for about 7-1/2 hours.
The reason is the large mass that's being lifted.
On Earth, that much mass weighs about 61 tons.
<span>Melting of ice is an endothermic process, meaning that energy is absorbed. When ice spontaneously melts, ΔH (change in enthalpy) is "positive". ΔS (entropy change) is also positive, because, becoming a liquid, water molecules lose their fixed position in the ice crystal, and become more disorganized. ΔG (free energy of reaction) is negative when a reaction proceeds spontaneously, as it happens in this case. Ice spontaneously melts at temperatures higher than 0°C. However, liquid water also spontaneously freezes at temperatures below 0°C. Therefore the temperature is instrumental in determining which "melting" of ice, or "freezing" of water becomes spontaneous. The whole process is summarized in the Gibbs free energy equation:
ΔG = ΔH – TΔS</span>
Answer:
1. 0.45 s.
2. 4.41 m/s
Explanation:
From the question given above, the following data were obtained:
Height (h) = 1 m
Time (t) =?
Velocity (v) =?
1. Determination of the time taken for the pencil to hit the floor.
Height (h) = 1 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
1 = ½ × 9.8 × t²
1 = 4.9 × t²
Divide both side by 4.8
t² = 1/4.9
Take the square root of both side
t = √(1/4.9)
t = 0.45 s.
Thus, it will take 0.45 s for the pencil to hit the floor.
2. Determination of the velocity with which the pencil hit the floor.
Initial velocity (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) = 0.45 s.
Final velocity (v) =?
v = u + gt
v = 0 + (9.8 × 0.45)
v = 0 + 4.41
v = 4.41 m/s
Thus, the pencil hit the floor with a velocity of 4.41 m/s
A solar eclipse will be visible over a wide area of the north polar region
on Friday, March 20.
England is not in the path of totality, but it's close enough so that a large
part of the sun will be covered, and it will be a spectacular sight.
For Londoners, the eclipse begins Friday morning at 8:25 AM,when the
moon just begins to eat away at the sun's edge. It advances slowly, as more
and more of the sun disappears, and reaches maximum at 9:31 AM. Then
the obscured part of the sun begins to shrink, and the complete disk is
restored by the end of the eclipse at 10:41AM, after a period of 2 hours
16 minutes during which part of the sun appears to be missing.
The catch in observing the eclipse is:
<em><u>YOU MUST NOT LOOK AT THE SUN</u></em>.
Staring at the sun for a period of time can cause permanent damage to
your vision, even though <em><u>you don't feel it while it's happening</u></em>.
This is not a useful place to try and give you complete instructions or
suggestions for observing the sun over a period of hours. Please look
in your local newspaper, or search online for phrases like "safe eclipse
viewing".
