<span>Germanium
To determine which melts first, convert their melting temperatures so they're both expressed on same scale. It doesn't matter what scale you use, Kelvin, Celsius, of Fahrenheit. Just as long as it's the same scale for everything. Since we already have one substance expressed in Kelvin and since it's easy to convert from Celsius to Kelvin, I'll use Kelvin. So convert the melting point from Celsius to Kelvin for Gold by adding 273.15
1064 + 273.15 = 1337.15 K
So Germanium melts at 1210K and Gold melts at 1337.15K. Germanium has the lower melting point, so it melts first.</span>
The car bounces off and moves in the opposite direction
Answer:
442.5 rad
Explanation:
w₀ = initial angular velocity of the disk = 7.0 rad/s
α = Constant angular acceleration = 3.0 rad/s²
t = time period of rotation of the disk = 15 s
θ = angular displacement of the point on the rim
Angular displacement of the point on the rim is given as
θ = w₀ t + (0.5) α t²
inserting the values
θ = (7.0) (15) + (0.5) (3.0) (15)²
θ = 442.5 rad
Answer:
For elliptical orbits: seldom
For circular orbits: always
Explanation:
We start by analzying a circular orbit.
For an object moving in circular orbit, the direction of the acceleration (centripetal acceleration) is always perpendicular to the direction of motion of the object.
Since acceleration has the same direction of the force (according to Newton's second law of motion), this means that the direction of the force (the centripetal force) is always perpendicular to the velocity of the object.
So for a circular orbit,
the direction of the velocity of the satellite is always perpendicular to the net force acting upon the satellite.
Now we analyze an elliptical orbit.
An elliptical orbit correponds to a circular orbit "stretched". This means that there are only 4 points along the orbit in which the acceleration (and therefore, the net force) is perpendicular to the direction of motion (and so, to the velocity) of the satellite. These points are the 4 points corresponding to the intersections between the axes of the ellipse and the orbit itself.
Therefore, for an elliptical orbit,
the direction of the velocity of the satellite is seldom perpendicular to the net force acting upon the satellite.
