A dolphin emits a sound wave that hits a target 120 m away. The wave reflects back from the target to the dolphin. If the Bulk m
1 answer:
Answer:
The time taken for the sound wave to make the round trip is 0.16 s.
Explanation:
Given;
distance traveled by the sound wave, d = 120 m
bulk modulus of sea water, B = 2.3 x 10⁹ N/m²
density of sea water, ρ = 1022 kg
The speed of the wave is given by;
Speed is given by;
total distance of the round trip = 2 x 120m = 240 m
Time taken for the sound wave to make the round trip is given by;
Therefore, the time taken for the sound wave to make the round trip is 0.16 s.
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Answer:
The initial velocity of the softball is 14.711 meters per second.
Explanation:
This is a case of an object which experiments a free fall, that is, an uniform accelerated motion due to gravity and in which effects from air friction and Earth's rotation can be neglected.
From statement we must understand that the student threw the softball upwards and it is caught at original position 3.56 seconds later. Initial and final heights, time and gravitational acceleration are known and initial speed is unknown. The following equation of motion is used:
(Eq. 1)
Where:
- Initial height of the softball, measured in meters.
- Final height of the softball, measured in meters.
- Initial velocity of the softball, measured in meters per second.
- Time, measured in seconds.
- Gravitational acceleration, measured in meters per square second.
If we know that ,
and
, the initial velocity of the softball is:
The initial velocity of the softball is 14.711 meters per second.
Answer:
atomic concentration = 2 atoms/unit cell
lattice parameter: a= 3.22 x 10⁻¹⁰ m
atomic radius: r= 1.39 x 10⁻¹⁰m
Explanation:
The atomic concentration is the number of atoms that can fit into a unit cell. It is a known number for each unit cell crystal structure. For a BCC (body-centered cube) crystal structure, atomic concentration is 2 atoms/unit cell because there are a 1/8 part of an atom in each corner of the cube (1/8 x 8= 1 atom) and 1 central atom in the central position of the cube ⇒ n= 1 atom + 1 atom= 2 atoms/unit cell
In order to calculate the lattice parameter a, we introduce the atomic mass 95.94 g/mol and the density 10.22 g/cm³ in the expression for the volume of the cube:
Vc= a³=
a³= 3.12 x 10⁻²³ m³
⇒ a = ∛(3.12 x 10⁻²³ m³) = 3.22 x 10⁻¹⁰m
Once we know the lattice parameter a, we can calculate the atomic radius r by using the expression of a for a BCC structure:
a=
⇒ r= a x √3/4= (3.22 x 10⁻¹⁰ m) x √3/4 = 1.39 x 10⁻¹⁰ m