The parcel will undergo projectile motion, which means that it will have motion in both the horizontal and vertical direction.
First, we determine how long the parcel will fall using:
s = ut + 1/2 at²
where s will be the height, u is the initial vertical velocity of the parcel (0), t is the time of fall and a is the acceleration due to gravity.
5.5 = (0)(t) + 1/2 (9.81)(t)²
t = 1.06 seconds
Now, we may use this time to determine the horizontal distance covered by the parcel by using:
distance = velocity * time
The horizontal velocity of the parcel will be equal to the horizontal velocity of the cruise liner.
Distance = 10 * 1.06
Distance = 10.6 meters
The boat should be 10.6 meters away horizontally from the point of release.
Answer:
H = 109.14 cm
Explanation:
given,
Assume ,
Total energy be equal to 1 unit
Balance of energy after first collision = 0.78 x 1 unit
= 0.78 unit
Balance after second collision = 0.78 ^2 unit
= 0.6084 unit
Balance after third collision = 0.78 ^3 unit
= 0.475 unit
height achieved by the third collision will be equal to energy remained
H be the height achieved after 3 collision
0.475 ( m g h) = m g H
H = 0.475 x h
H = 0.475 x 2.3 m
H = 1.0914 m
H = 109.14 cm
Answer:
The height of the wave is determined by the wind strength and fetch.
Explanation:
The height of the wave is determined by the wind strength and fetch.
The more the strength and the more the fetch size the more will be the height of the wave.
Remember as the wave approaches the coast its wavelength decreases and the wave height increases, whereas when the wave goes away from the coast its wavelength increases and height decreases.
Answer:
A. Create radioactive waste i believe
Explanation:
Answer:
A sample of 5.2 mg decays to .65 mg or to 1/8 of its original amount.
1/8 = 1/2 * 1/2 * 1/2 or 3 half-lives.
3 * 30.07 = 90 yrs for 5.2 mg to decay to .65 mg
You can get these other numbers similarly:
5.2 / .0102 = 510 requires about 9 half-lives which is 30 * 9 = 270 yrs
