Answer:(a) With our choice for the zero level for potential energy of the car-Earth system when the car is at point B ,
U
B
=0
When the car is at point A, the potential energy of the car-Earth system is given by
U
A
=mgy
where y is the vertical height above zero level. With 135ft=41.1m, this height is found as:
y=(41.1m)sin40.0
0
=26.4m
Thus,
U
A
=(1000kg)(9.80m/s
2
)(26.4m)=2.59∗10
5
J
The change in potential energy of the car-Earth system as the car moves from A to B is
U
B
−U
A
=0−2.59∗10
5
J=−2.59∗10
5
J
(b) With our choice of the zero configuration for the potential energy of the car-Earth system when the car is at point A, we have U
A
=0. The potential energy of the system when the car is at point B is given by U
B
=mgy, where y is the vertical distance of point B below point A. In part (a), we found the magnitude of this distance to be 26.5m. Because this distance is now below the zero reference level, it is a negative number.
Thus,
The three parts of the Earth are Atmosphere, Hydrosphere and Lithosphere.
Atmosphere is the blanket of air that surrounds the earth. It is densest close to the surface and thins out as one moves higher. Atmosphere of Earth contains mainly Nitrogen, followed by Oxygen and small amounts of water vapor, Carbondioxide and other gases.
Lithosphere is the outer most part of the earth's surface. The Earth's crust and the mantle form Lithosphere.
Hydrosphere is the part of the Earth that has water. The Oceans, seas, rivers, lakes and other water bodies constitute the Hydrosphere.
Stratosphere, Mesosphere and Ionosphere are different layers of atmosphere.
Hence, for the study of the Earth, one needs to consider earth to be made of three parts- atmosphere, Lithosphere and Hydrosphere.
Answer:
Two of Einstein’s influential ideas introduced in 1905 were the theory of special relativity and the concept of a light quantum, which we now call a photon. Beyond 1905, Einstein went further to suggest that freely propagating electromagnetic waves consisted of photons that are particles of light in the same sense that electrons or other massive particles are particles of matter. A beam of monochromatic light of wavelength \lambda (or equivalently, of frequency f) can be seen either as a classical wave or as a collection of photons that travel in a vacuum with one speed, c (the speed of light), and all carrying the same energy, {E}_{f}=hf. This idea proved useful for explaining the interactions of light with particles of matter.
Answer:
a.) magnitude __49.7__ unit(s)
b.) direction __123.6°_ counterclockwise from the +x axis
Explanation:
Let Vector is v
x-component of Vector v = x = -27.5 units (minus sign indicate that x-component is along the minus x-axis )
y-component of Vector v = y = 41.4 units
Magnitude of v = ?
Direction of v = ?
To find the magnitude of the vector
v =
v = 
v = 49.7 units
To find direction
θ = tan⁻¹(y/x)
θ = tan⁻¹(41.4/-27.5)
θ = -56.4°
This Angle is in the clockwise direction with respect to -x axis.
We need to find Angle counterclockwise from the +x axis.
So,
θ = 180° - 56.4°
θ = 123.6°
The given vector is in 2nd quadrant
Answer:
infrared, visible and ultraviolet regions.
Explanation:
