(1 parsec) is the distance at which an object has a parallax of 1 arcsecond. The distance is about 3.26 light years.
Another way to understand it is: The distance from which the Earth's orbit appears 1 arcsecond across.
For a parallax angle of 1/2 arcsecond, the distance is <em>2 parsecs </em>(about 6.52 light years).
1 arcsecond is 1/3600 of a degree, 0.00028 degree.
Answer:
P = 100 lb
Explanation:
For an ideal machine:
where,
P = Effort = Input Force = ?
L = Effort Arm = Length of inclined plane = 10 ft
W = Load to be lifted = 200 lb
H = Load Arm = Height = 5 ft
Therefore,
<u>P = 100 lb</u>
X can probably be vacuum through space through air and or solid objects
Answer:
a. 25 N/m
b. 0.8886 s
c. 0.707 m/s
d. At the equilibrium point
Explanation:
m = 500 g = 0.5 kg
L = 20 cm = 0.2 m
A = 10 cm = 0.1 m
a. Let g = 10 m/s2, then the gravity of the 0.5 kg book acting on the spring is
F = mg = 0.5*10 = 5 N
If the spring is stretched L = 0.2 m under 5N load, then the spring constant k is:
k = F/l = 5 / 0.2 = 25 N/m
b. We can treat this as simple harmonic motion with magnitude A = 0.1 cm. The period of this motion is
c. The book maximum speed:
d. Due to the law of energy conservation, the maximum speed would occur at the equilibrium point. This is where the potential energy, elastic energy is 0 and the kinetic energy is greatest.
Answer:
<em>The crane lifts the crate up to 76 m high</em>
Explanation:
<u>Work Done by a Force</u>
The work done by a force of magnitude F that displaces an object by a distance y is given by
We know a crane does a work of 9,500 Joule to lift a crate and is using a force of 125 Nw.
The above equation can be solved to know the value of y in terms of the work and the force:
Plugging in the given values
The crane lifts the crate up to 76 m high