I think B is the most correct, because logically it's harder to bend a stiffer spring than it is to bend a softer one. Also, I don't think length comes into play. So B.
Answer:
77J
Explanation:
Not really an explanation to this, I just had this lesson last year and remembered it.
Hope I helped! ☺
#1
As we are increasing the frequency in the simulation the wavelength is decreasing
So if speed remains constant then wavelength and frequency depends inversely on each other
If we are in boat and and moving over very small wavelengths then these small wavelength will be encountered continuously by the boat in short interval of times
#2
As we are changing the amplitude in the simulation there is no change in the speed frequency and wavelength.
So amplitude is independent of all these parameter
Amplitude of wave will decide the energy of wave
So light of greater intensity is the light of larger amplitude
#3
In our daily life we deal with two waves
1 sound waves
2 light waves
Complete question :
NASA is concerned about the ability of a future lunar outpost to store the supplies necessary to support the astronauts the supply storage area of the lunar outpost where gravity is 1.63m/s/s can only support 1 x 10 over 5 N. What is the maximum WEIGHT of supplies, as measured on EARTH, NASA should plan on sending to the lunar outpost?
Answer:
601000 N
Explanation:
Given that :
Acceleration due to gravity at lunar outpost = 1.6m/s²
Supported Weight of supplies = 1 * 10^5 N
Acceleration due to gravity on the earth surface = 9.8m/s²
Maximum weight of supplies as measured on EARTH :
Ratio of earth gravity to lunar post gravity:
(Earth gravity / Lunar post gravity) ;
(9.8 / 1.63) = 6.01
Hence, maximum weight of supplies as measured on EARTH should be :
6.01 * (1 × 10^5)
6.01 × 10^5
= 601000 N
The distance travelled by the ball that is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s is 5.34 m
s = ut + 1 / 2 at²
s = Distance
u = Initial velocity
t = Time
a = Acceleration
Vertically,
s = 15.4 m
u = 0
a = 9.8 m / s²
15.4 = 0 + ( 1 / 2 * 9.8 * t² )
t² = 3.14
t = 1.77 s
Horizontally,
u = 3.01 m / s
a = 0 ( Since there is no external force )
s = ( 3.01 * 1.77 ) + 0
s = 5.34 m
Therefore, the distance travelled by the ball before hitting the ground is 5.34 m
To know more about distance travelled
brainly.com/question/12696792
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