Explanation:
Given that,
Mass of the block, m = 5 kg
Spring constant, k = 2000 N/m
The block is pulled down 5.0 cm from the equilibrium position and given an initial velocity of 1.0 m/s back towards equilibrium.
(a) The angular frequency is given by :
Since, , f is frequency
(b) The velocity of particle executing SHM is given by :
x is displacement from equilibrium position
(c) The total mechanical energy of the motion in SHM is given by :
Hence, this is the required solution.
Answer:
- The gravitational force by the Earth on the object, and by the object on the Earth is
- = 6.674×10−11 m3⋅kg−1⋅s−2 × 6 × 10^24 kg × 44.5 kg/(6.4 × 10^6 m)²
<u>Please note that the ration between the gravitation force 435 and the mass 44.5</u>
- should be gravitational acceleration
- I attribute the discrepancy between 9.78 and the usual 9.81 to rounding off in the
- Earth's weight and radius.
The mass of the Moon is M / 81.3.
The radius of the Moon is R × 0.27.
The gravitational force on the moon would be
G(M/81.3)m/(R×0.27)² = 0.17×GMm/R²
The gravitational force on the moon is smaller by the factor of about 0.17.
Answer:
.
Explanation:
Refer to the velocity-time diagram attached. The displacement of an object over a period of time is equal to the area of the region between the velocity-time plot and the horizontal time axis. That is the case even if the velocity of the object is changing over time.
For the car in this question, the distance travelled in that two seconds should be equal to the trapezoid highlighted in green. That's the region bounded with:
- on the top: the velocity-time function of this car,
- on the two sides: the start and end of the acceleration ( and ,) as well as
- on the bottom: the horizontal time axis.
The formula for the area of a trapezoid is:
.
For the imaginary trapezoid on this velocity-time graph:
- Height: .
- Upper base and lower base: and .
Therefore:
.