Answer:
B. The image is right-left reversed and about as far away from the mirror as the reflected object.
Explanation:
A plane mirror is used in several ways in our life, from the mirror that we have on the bathroom to the mirrors we use on cars, the main difference is that the mirror reflects the image reversed on the Vertical axis, which means that the left is the right and the right is the left, a mirror also protrays objects at a certain distance, about as far away from the mirror as the reflected object but they are actually closer in real life.
Answer:
22.15 N/m
Explanation:
As we know potential energy = m*g*h
Potential energy of spring = (1/2)kx^2
m*g*h = (1/2)kx^2
Substituting the given values, we get -
(400)*(9.8)*(10) = (0.5)*(k)*(2.0^2)
k = 39200/2.645
k = 19600 N/m
For safety reasons, this spring constant is increased by 13 % So the new spring constant is
k = 19600 * 1.13 = 22148 N/m = 22.15 N/m
Answer:
If we consider a system where the y-axis as the South-North line, and the x-axis as the West-East line (where North and East are the positive sides)
We know that the sun goes from East to West, so in our system, the sun goes from the positive side of the x-axis to the negative side of the x-axis.
Where we would see this if we were standing right in the equator line.
If we where in other point of the planet, the Sun will stil move from East to West, but it will have a little tilt along the path, so we will have a little displacement in the y-axis. This displacement will depend on where we are, if we are at the North of the equator, we will se that the sun seems to go a little towards South as it goes to the West side.
Answer:
Theta1 = 12° and theta2 = 168°
The solution procedure can be found in the attachment below.
Explanation:
The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).
In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.