Using conservation of energy and momentum we get m1*v1=(m1+m2)*v2 so rearranging for v2 and plugging the given values in we get:
(200000kg*1.00m/s)/(21000kg)=.952m/s
Answer:
The energy level is 5.
Explanation:
Given that,
n = 3
l = 2
We know that,
l shows the number of sub-shells and define the number of angular nodes.
n shows the number of electron shell.
is a quantum number. It is define the number of energy level in a sub-shells .
is define the spin of the electron.
So, The quantum number is
is 
and 
for every energy level.
The energy level is 5.
Hence, The energy level is 5.
<span>1.0x10^3 Joules
The kinetic energy a body has is expressed as the equation
E = 0.5 M V^2
where
E = Energy
M = Mass
V = Velocity
Since the shot was at rest, the initial energy is 0. Let's calculate the energy that the shot has while in motion
E = 0.5 * 7.2 kg * (17 m/s)^2
E = 3.6 kg * 289 m^2/s^2
E = 1040.4 kg*m^2/s^2
E = 1040.4 J
So the work performed on the shot was 1040.4 Joules. Rounding the result to 2 significant figures gives 1.0x10^3 Joules</span>
Answer:
The component form will be;
In the x-axis = 121.73 due west
In the y-axis = 690.35 due south
Explanation:
An image of the calculation has been attached
Answer:
The tangential speed at Livermore is approximately 284.001 meters per second.
Explanation:
Let suppose that the Earth rotates at constant speed, the tangential speed (
), measured in meters per second, at Livermore (37.6819º N, 121º W) is determined by the following expression:
(1)
Where:
- Rotation time, measured in seconds.
- Radius of the Earth, measured in meters.
- Latitude of the city above the Equator, measured in sexagesimal degrees.
If we know that 
, 
and 
, then the tangential speed at Livermore is:
The tangential speed at Livermore is approximately 284.001 meters per second.
